SCHEME OF WORK
WEEKS TOPICS
1. Revision of s. s. one work (approximation ,significant figures and decimal places)
2. Revision of logarithm of numbers greater than one and logarithm of numbers less than one, reciprocal and accuracy of results using straight calculation.
3. Approximations, calculations using standard form, significant figure
4. Approximations and Percentage error
5. Sequence and series: concept of sequence and series, terms of Arithmetic progressions and sum : Solving problem on A.P
6. Geometric progressions; the nth term and sum of the first n- terms. Problem solving on G.P and geometric mean
7. Simultaneous Equations: concept, review of simultaneous linear equations and word problems on simultaneous linear equation.
8. Simultaneous Equations-One linear; one quadratic(solution by substitution method)
9. Graphical Solution to simultaneous equations, one linear, one quadratic.
10. Surds: addition, and subtraction, multiplication and division, Rationalization of surds
11. Revision
1ST TERM
WEEK 1
TOPIC: APPROXIMATION: DECIMAL PLACES AND SIGNIFICANT FIGURE
CONTENT:
1. Place value and rounding off number approximation
2. Approximation :significant figures
3. Application of approximations to mathematical problems.
Place Value and rounding off number (approximations)
Approximation is very useful when calculating. Numbers can be approximated in order to obtain rough estimates of calculation e.g the tithe of a man’s income is N172.38 per month, the total tithe to pay a year is N172.38 x 12.
A rough estimate is N200 x 10 N2,000
Rough estimates are not accurate. They only give an idea of the size or order of magnitude of the correct result of calculation.
https://youtu.be/T5Qf0qSSJFI
ROUNDING OF NUMBERS
Numbers can be rounded off to a given number of decimal place to the nearest (particular)
Place value e.g nearest hundred nearest tens etc.
Place values for the digits in 3875.29
Example 1:
Round off the number 341.77
(a) 2 d.p = 341.77
(b) 1 d.p = 341.8
(c) Nearest hundred = 300
(d) Nearest whole no = 342
(e) Nearest tenths = 341.8
Example 2
Round off 86.12 to
(i) 1 d.p = 86.1
(ii) Nearest whole no. = 86
(iii) Nearest tens = 90
https://youtu.be/fd-E18EqSVk
DECIMAL PLACES
In order to determine the number of decimal places contained in a figure, we count the number of digits after the decimal point. Zeros after the decimal are counted if found in between non-zero digits. The last zero in a decimal number is not counted. Decimal places are shortened to d.p
Example 1: Write the following in 4 d.p (i) 0.00630427 (ii) 15.300649
Solution
(i) 0.00630427 = 0.0063 4 d.p
(ii) 15.300649 = 15.3006 4 d.p
Example 2: Add up 3.42, 4.761, 3.04, 6.3, 11.304
Solution
3.42
4.761
3.04
6.3
11.304
28.825
Example 3: Simplify 17.36 x 4.65
Solution
17.36
X 4.65
8680
10416
6944
80.7240
EVALUATION
Round off
1 (a) 3.24m to the nearest whole number
(b) 129.7 litres to the nearest tens
(c) 18.64 to 1 d.p
(d) 2.997 to 2 d.p
2 In 1982 the currency in circulation in Nigeria was N4728.8 million give four different ways in which a newspaper night have reported this amount.
3. (i) Add up 5.67, 0.453,14.056,4.780
(ii) Simplify 15.326 x 2.15
https://youtu.be/KG6ILNOiMgM
ASSIGNMENT:
Evaluation 0.071685 ÷ 5.36 without using tables
APPROXIMATION: SIGNIFICANT FIGURE
The significance of a digit depends on its position in the number. i.e. in 137.86, 3 is more significant than 7 and 7 is more significant than 8, and the first figure (non zero) is the most significant.
Significant figures are the positioning of non-zero digits in a numeral. It is usually applicable to both decimals and whole numbers. The first significant figure in any numerical is thus the first non-zero digit when counted from the left-hand side. For example in 0.007368, the first significant figure (s.f) is 7, the second is 3, the third is 6, while the fourth is 8.
Any time a figure is taken to any number of significant figure, the convention is to round off digit 5 and above into a whole number and add to the next digit.
In 305.8 , 0 is a significant figure and it is more significant than 5. The first significant figure in a decimal fraction is the first non-zero digit in the fraction e.g
In .000832, 8 is the first significant figure.
Example 1
Round off 183.81 and 0.0093815 to 1, 2, 3, 4 significant figures
183.81
(a) 183.81 = 200 to 1 s.f
183.81 = 180 to 2 s.f
183.81 = 184 to 3 s.f
183.81 = 183.8 to 4 s.f
(b) 0.009381 = 0.009 (1 s.f)
0.0094 (2 s.f)
0.0093 (3 s.f)
0.009382 (4 s.f)
Example 2: Express 0.006457 to (a) 3 s.f (b) 2 s.f (c) 1 s.f
SOLUTION
(a) 0.00646 = 3 s.f
(b) 0.0065 = 2 s.f
(c) 0.006 = 1 s.f
Example 3: Express 45682 to (a) 2 s.f (b) 3 s.f (c) 4 s.f
SOLUTION
(a) 46000 = 2 s.f
(b) 45700 = 3 s.f
(c) 45680 = 4 s.f
https://youtu.be/fVAMKbBGzTY
EVALUATION
1. Round off to 2 s.f
(i) 20.08
(ii) 109.4
(iii) 0.0819
(iv) Round off 0.8094 to 3 s.f
2. Write each of the following correct to the number of significant figure shown in front of each of them.
(a) 0.0973 (2 s.f) (b) 14.0675 (3 s.f) (c) 2.00584 (4 s.f) (d) 0.008406 (2 s.f)
ASSIGNMENT: Correct the following into the number of significant figure shown in front of them
(a) 4384467 (3 s.f) (b) 28.0059 (4 s.f) (c) 0.35481 (2 s.f) (d) 0.002635 (1 s.f)
SPECIFIC TOPIC: Place values
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on place value
PLACE VALUES AND ESTIMATION
Sometime we just need to have an idea of the size or value of an operation involving multiplication, addition, division or subtraction. We can just use the method of estimation to determine a rough value for the operation.
Example 1: Estimate the following (i) 2.17 x 5.21 (ii) 43.26 + 19.06
Solution
(i) The estimate will be 2 x5 = 10
The real value is 11.3 to 3 d.p
(ii) 43.26 + 19.06
The estimate is 43 + 19 = 62
The real value is 62.32
Example 2: Write out the place value of 8 and 5 in the numbers given below
264153.078
Solution
The place value of 5 = tens
The place value of 8 = thousandth
https://youtu.be/LuyNdVw5f4o
EVALUATION:
Solve the following questions below
(i) Subtract 0.003649 from0.0316
(ii) Correct the following to 2 significant figures (a) 0.86439 (b) 35864
(iii) Correct the following figures to the number of decimal places indicated (i) 3.0561(2 d.p) (ii) 0.008153 (1 d.p) (iii) 0.09634 (3 d.p) (iv) 0.0006005 (4 d.p)
(iv) Estimate 21.42 x 2.43
ASSIGNMENT;
Solve the following questions
(i) Express 0.0006457 to (i) 3 s.f (ii) 2 s.f (iii)1 s.f (iv) 4 s.f
(ii) Round off the following to the nearest hundred (a) 243.36 (ii) 643.79 (iii) 3428km (iv) 236.95m
CONTENT: APPROXIMATIONS
Approximations are often linked to such terms as decimal places, significant figures, standard forms and so on. But since all these aforementioned above are definite terms with their own unique names.
Hence, approximation is a method of assuming precise values to figures. It is a method or system of counting to the nearest whole. Such wholes here could be tens, hundreds, thousands, distances, weights, and other values.
https://youtu.be/EyiC6cGjIhg
Example 1: 2 hours 12mins
Solution
2hours 12mins = 2 12/60 hours
= 2.2hours
= 2hours (to the nearest hour)
Example 2: Round off 241.863 to the nearest whole number)
Solution
241.863 = 242 (to the nearest whole number)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(a) Take 173.245 to the nearest hundred.
(b) Take 146.8432 to the nearest hundred
ASSIGNMENT: Take the following to the nearest whole number:
(a) 6.32km
(b) 486.4m
(c) 827.305
(d) 154.81cm
APPLICATION OF APPROXIMATIONS TO MATHEMATICAL PROBLEMS
Degree of accuracy
The accuracy of answers depends on the accuracy of the values used in the calculation. Thus for the final answer to any calculation to be accurate, the answer must agree with the number of significant figures in which the original data were given.
Do not give the final result of a calculation to more significant figures than those given in the initial data (otherwise stated).
In arbitrary, a better degree of accuracy of any calculation, it is generally not advisable to round off intermediate values before further calculations.
Example
In an electrical calorimetric experiment, the specific heat capacity S units of a metal is given by the formula
S = 5VIt / 9mθ
Where V is the potential difference, I is the current, t is the time, m is the mass and θ is the change in temperature.
In one such experiment, the values obtained were V = 2.8, I = 1.6, S= 0.226, m = 130.8 and θ = 18.8. Calculate the value of t correct to two significant figures
https://youtu.be/pwF7UcaHEOk
ASSIGNMENT
THEORY
1 The volume of water which fills a cylindrical container is 1.25m3. If the height of the cylinder is 1.50m, calculate the radius of the circular base of the cylinder take to be 3.14 (express your answer in 2 p.d
2 Express the square root of 0.0676 in standard form correct to 3 s.f.
Objectives
1 The index of the standard form of 2.5735 is
(a) 2 (b) 1 (c) 0 (d) -1
2 The integer of the logarithm of 1.8932 x 10-3 is
(a) -3 (b) 3 (c) 3 (d) 0
3 Round off 0.0837 to 2 s.f
(a) 0.084 (b) 0.08 (c) 0.83 (d) 0.083
4 Round off 0.0832 to 2 d.p
(a) 0.08 (b) 0.08 (c) 0.83 (d) 0.082
5 Approximate 41728.8 million to the nearest billion
(a) 41739 billion (b) 4 billion (c) 4173 billion (d) 42 billion.
READING ASSIGNMENT
1 New General mathematics for SS 2 by J.B Channon and Co page 44-45
2 Distinction in Mathematics by A.A Adelodun 3rd Edition page 20-22
CONTENT:
1. Place value and rounding off number approximation
2. Approximation :significant figures
3. Application of approximations to mathematical problems.
Place Value and rounding off number (approximations)
Approximation is very useful when calculating. Numbers can be approximated in order to obtain rough estimates of calculation e.g the tithe of a man’s income is N172.38 per month, the total tithe to pay a year is N172.38 x 12.
A rough estimate is N200 x 10 N2,000
Rough estimates are not accurate. They only give an idea of the size or order of magnitude of the correct result of calculation.
https://youtu.be/T5Qf0qSSJFI
ROUNDING OF NUMBERS
Numbers can be rounded off to a given number of decimal place to the nearest (particular)
Place value e.g nearest hundred nearest tens etc.
Place values for the digits in 3875.29
Example 1:
Round off the number 341.77
(a) 2 d.p = 341.77
(b) 1 d.p = 341.8
(c) Nearest hundred = 300
(d) Nearest whole no = 342
(e) Nearest tenths = 341.8
Example 2
Round off 86.12 to
(i) 1 d.p = 86.1
(ii) Nearest whole no. = 86
(iii) Nearest tens = 90
https://youtu.be/fd-E18EqSVk
DECIMAL PLACES
In order to determine the number of decimal places contained in a figure, we count the number of digits after the decimal point. Zeros after the decimal are counted if found in between non-zero digits. The last zero in a decimal number is not counted. Decimal places are shortened to d.p
Example 1: Write the following in 4 d.p (i) 0.00630427 (ii) 15.300649
Solution
(i) 0.00630427 = 0.0063 4 d.p
(ii) 15.300649 = 15.3006 4 d.p
Example 2: Add up 3.42, 4.761, 3.04, 6.3, 11.304
Solution
3.42
4.761
3.04
6.3
11.304
28.825
Example 3: Simplify 17.36 x 4.65
Solution
17.36
X 4.65
8680
10416
6944
80.7240
EVALUATION
Round off
1 (a) 3.24m to the nearest whole number
(b) 129.7 litres to the nearest tens
(c) 18.64 to 1 d.p
(d) 2.997 to 2 d.p
2 In 1982 the currency in circulation in Nigeria was N4728.8 million give four different ways in which a newspaper night have reported this amount.
3. (i) Add up 5.67, 0.453,14.056,4.780
(ii) Simplify 15.326 x 2.15
https://youtu.be/KG6ILNOiMgM
ASSIGNMENT:
Evaluation 0.071685 ÷ 5.36 without using tables
APPROXIMATION: SIGNIFICANT FIGURE
The significance of a digit depends on its position in the number. i.e. in 137.86, 3 is more significant than 7 and 7 is more significant than 8, and the first figure (non zero) is the most significant.
Significant figures are the positioning of non-zero digits in a numeral. It is usually applicable to both decimals and whole numbers. The first significant figure in any numerical is thus the first non-zero digit when counted from the left-hand side. For example in 0.007368, the first significant figure (s.f) is 7, the second is 3, the third is 6, while the fourth is 8.
Any time a figure is taken to any number of significant figure, the convention is to round off digit 5 and above into a whole number and add to the next digit.
In 305.8 , 0 is a significant figure and it is more significant than 5. The first significant figure in a decimal fraction is the first non-zero digit in the fraction e.g
In .000832, 8 is the first significant figure.
Example 1
Round off 183.81 and 0.0093815 to 1, 2, 3, 4 significant figures
183.81
(a) 183.81 = 200 to 1 s.f
183.81 = 180 to 2 s.f
183.81 = 184 to 3 s.f
183.81 = 183.8 to 4 s.f
(b) 0.009381 = 0.009 (1 s.f)
0.0094 (2 s.f)
0.0093 (3 s.f)
0.009382 (4 s.f)
Example 2: Express 0.006457 to (a) 3 s.f (b) 2 s.f (c) 1 s.f
SOLUTION
(a) 0.00646 = 3 s.f
(b) 0.0065 = 2 s.f
(c) 0.006 = 1 s.f
Example 3: Express 45682 to (a) 2 s.f (b) 3 s.f (c) 4 s.f
SOLUTION
(a) 46000 = 2 s.f
(b) 45700 = 3 s.f
(c) 45680 = 4 s.f
https://youtu.be/fVAMKbBGzTY
EVALUATION
1. Round off to 2 s.f
(i) 20.08
(ii) 109.4
(iii) 0.0819
(iv) Round off 0.8094 to 3 s.f
2. Write each of the following correct to the number of significant figure shown in front of each of them.
(a) 0.0973 (2 s.f) (b) 14.0675 (3 s.f) (c) 2.00584 (4 s.f) (d) 0.008406 (2 s.f)
ASSIGNMENT: Correct the following into the number of significant figure shown in front of them
(a) 4384467 (3 s.f) (b) 28.0059 (4 s.f) (c) 0.35481 (2 s.f) (d) 0.002635 (1 s.f)
SPECIFIC TOPIC: Place values
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on place value
PLACE VALUES AND ESTIMATION
Sometime we just need to have an idea of the size or value of an operation involving multiplication, addition, division or subtraction. We can just use the method of estimation to determine a rough value for the operation.
Example 1: Estimate the following (i) 2.17 x 5.21 (ii) 43.26 + 19.06
Solution
(i) The estimate will be 2 x5 = 10
The real value is 11.3 to 3 d.p
(ii) 43.26 + 19.06
The estimate is 43 + 19 = 62
The real value is 62.32
Example 2: Write out the place value of 8 and 5 in the numbers given below
264153.078
Solution
The place value of 5 = tens
The place value of 8 = thousandth
https://youtu.be/LuyNdVw5f4o
EVALUATION:
Solve the following questions below
(i) Subtract 0.003649 from0.0316
(ii) Correct the following to 2 significant figures (a) 0.86439 (b) 35864
(iii) Correct the following figures to the number of decimal places indicated (i) 3.0561(2 d.p) (ii) 0.008153 (1 d.p) (iii) 0.09634 (3 d.p) (iv) 0.0006005 (4 d.p)
(iv) Estimate 21.42 x 2.43
ASSIGNMENT;
Solve the following questions
(i) Express 0.0006457 to (i) 3 s.f (ii) 2 s.f (iii)1 s.f (iv) 4 s.f
(ii) Round off the following to the nearest hundred (a) 243.36 (ii) 643.79 (iii) 3428km (iv) 236.95m
CONTENT: APPROXIMATIONS
Approximations are often linked to such terms as decimal places, significant figures, standard forms and so on. But since all these aforementioned above are definite terms with their own unique names.
Hence, approximation is a method of assuming precise values to figures. It is a method or system of counting to the nearest whole. Such wholes here could be tens, hundreds, thousands, distances, weights, and other values.
https://youtu.be/EyiC6cGjIhg
Example 1: 2 hours 12mins
Solution
2hours 12mins = 2 12/60 hours
= 2.2hours
= 2hours (to the nearest hour)
Example 2: Round off 241.863 to the nearest whole number)
Solution
241.863 = 242 (to the nearest whole number)
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(a) Take 173.245 to the nearest hundred.
(b) Take 146.8432 to the nearest hundred
ASSIGNMENT: Take the following to the nearest whole number:
(a) 6.32km
(b) 486.4m
(c) 827.305
(d) 154.81cm
APPLICATION OF APPROXIMATIONS TO MATHEMATICAL PROBLEMS
Degree of accuracy
The accuracy of answers depends on the accuracy of the values used in the calculation. Thus for the final answer to any calculation to be accurate, the answer must agree with the number of significant figures in which the original data were given.
Do not give the final result of a calculation to more significant figures than those given in the initial data (otherwise stated).
In arbitrary, a better degree of accuracy of any calculation, it is generally not advisable to round off intermediate values before further calculations.
Example
In an electrical calorimetric experiment, the specific heat capacity S units of a metal is given by the formula
S = 5VIt / 9mθ
Where V is the potential difference, I is the current, t is the time, m is the mass and θ is the change in temperature.
In one such experiment, the values obtained were V = 2.8, I = 1.6, S= 0.226, m = 130.8 and θ = 18.8. Calculate the value of t correct to two significant figures
https://youtu.be/pwF7UcaHEOk
ASSIGNMENT
THEORY
1 The volume of water which fills a cylindrical container is 1.25m3. If the height of the cylinder is 1.50m, calculate the radius of the circular base of the cylinder take to be 3.14 (express your answer in 2 p.d
2 Express the square root of 0.0676 in standard form correct to 3 s.f.
Objectives
1 The index of the standard form of 2.5735 is
(a) 2 (b) 1 (c) 0 (d) -1
2 The integer of the logarithm of 1.8932 x 10-3 is
(a) -3 (b) 3 (c) 3 (d) 0
3 Round off 0.0837 to 2 s.f
(a) 0.084 (b) 0.08 (c) 0.83 (d) 0.083
4 Round off 0.0832 to 2 d.p
(a) 0.08 (b) 0.08 (c) 0.83 (d) 0.082
5 Approximate 41728.8 million to the nearest billion
(a) 41739 billion (b) 4 billion (c) 4173 billion (d) 42 billion.
READING ASSIGNMENT
1 New General mathematics for SS 2 by J.B Channon and Co page 44-45
2 Distinction in Mathematics by A.A Adelodun 3rd Edition page 20-22
WEEK 2
TOPIC: REVISION OF LOGARITHM OF NUMBERS GREATER THAN ONE AND LOGARITHM OF NUMBERS LESS THAN ONE.
CONTENTS
• Standard forms
• Logarithm of numbers greater than one
• Multiplication and divisions of numbers greater than one using logarithm
• Using logarithm to solve problems with roots and powers (no > 1)
• Logarithm of numbers less than one.
• Multiplication and division of numbers less than one using logarithm
• Roots and powers of numbers less than one using logarithm
STANDARD FORMS
A way of expressing numbers in the form A x 10x where 1≤ A 10 and x is an integer is said to be a standard form. Numbers are grouped into two. Large and small numbers. Numbers greater than or equal to 1 are called large numbers. In this case the x, which is the power of 10 is positive. On the other hand, numbers less than 1 are called small numbers. Here, the integer is negative.
Numbers such as 1000 can be converted to its power of ten in the form 10n where n can be term as the number of times the decimal point is shifted to the front of the first significant figure
i.e. 10000 = 104
Number = Power of 10
100 =102
10= 101
1 100
0.01 10-3
0.10 10-1
Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.
https://youtu.be/VsbpBrOmr18
Example
(i) Express 0. 08356 in standard form
(ii) 832.8 in standard form
Solution
i 0. 08356 = 8.356 x 10-2
ii 832.8 = 8.328 x 102
Express the following in standard form
(a) 39.32 = 3.932 x 101
(b) 4.83 = 4.83 x 100
(c) 0.005321 = 5.321 x 10-3
If a number is in its standard form, its power is its integer i.e. the integer of its logarithm e.g. log 7853 has integer 3 because 7853 = 7.853 x 103
Examples:
Use tables (log) to find the complete logarithm of the following numbers.
(a) 80030 (b) 8 (c) 135.80
(a) 80030 = 4.9033
(b) 8 = 0.9031
(c) 13580 = 2.1329
https://youtu.be/4006pXk940I
Evaluation.
Use table to find the complete logarithm of the following:
(a) 183 (b) 89500 (c) 10.1300 (d) 7
Reading Assignment
New General Mathematics SSII Page 20 – 21, Exercise 1a & b.
Multiplication and Division of numbers greater than one using logarithm
To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.
Examples
Evaluate using logarithm.
1. 4627 x 29.3
2. 8198 ÷ 3.905
3. 48.63 x 8.53
15.39
https://youtu.be/0-kZdwxFjY4
Solutions
1. 4627 x 29.3
To find the Antilog of the log 5.1322 use the antilogarithm table:
Check 13 under 2 diff 2 (add the value of the difference) the number is 0.1356. To place the decimal point at the appropriate place, add one to the integer of the log i.e. 5 + 1 = 6 then shift the decimal point of the antilog figure to the right (positive) in 6 places.
https://youtu.be/yzuYXIoVYMw
Evaluation:
Use logarithm to calculate.
3612 x 750.9
113.2 x 9.98
Reading Assignment
Now General Mathematics SS2 Page 21 exercise 1C.
Using logarithm to solve problems with powers and root
(no. greater than one).
Examples:
Find the single logarithm representing the numerator and the single logarithm representing the denominator, subtract the logarithm then find the anti log.
(Numerator – Denominator).
Example: Find the complete log of the following.
(a) 0.004863 (b) 0.853 (c) 0.293
Solution
Log 0.004863 = 3.6369
Log 0.0853 = 2.9309
Log 0.293 = 1.4669
https://youtu.be/N0-u57nXLQ8
Evaluation
1. Find the logarithm of the following:
(a) 0.064 (b) 0.002 (c) 0.802
2. Evaluate using logarithm.
95.3 x 318.4
1.295 x 2.03
https://youtu.be/kqVpPSzkTYA
Reading Assignment
New General mathematics for SS 2 page 22 – 24 exercise 1g.
Using logarithm to evaluate problems of Multiplication, Division, Powers and roots with numbers less than one.
OPERATION WITH BAR NOTATION
Note the following when carrying out operations on logarithm of numbers which are negative.
I. The mantissa (fractional part) is positive, so it has to be added in the usual manner.
II. The characteristic (integral part) is either positive or negative and should therefore be added or operated as directed numbers.
III. For operations like multiplication and division, separate the integer from the characteristic before performing the operation.
Examples:
Simplify the following, leaving the answers in bar notation, where necessary
I. .7675 + 2.4536
II. 6.8053 – 4.1124
III. 2.4423 x 3
IV. 2.2337 ÷ 7
Note: 3 cannot divide 2 therefore subtract 1 from the negative integer and add 1 to the positive decimal fraction so as to have 3 which is divisible
by 3 without remainder.
https://youtu.be/fnhFneOz6n8
ASSIGNMENT
Objective2
Use table to find the log of the following:
1. 900 (a) 3.9542 (b) 1.9542 (c) 2.9542 (d) 0.9542
2. 12.34 (a) 3.0899 (b) 1.089 (c) 2.0913 (d) 1.0913
3. 0.000197 (a) 4.2945 (b) 4.2945 (c) 3.2945 (d) 3.2945
4. 0.8 (a) 1.9031 (b) 1.9031 (c) 0.9031 (d) 2.9031
5. Use antilog table to write down the number whose logarithms is 3.8226.
(a) 0.6646 (b) 0.06646 (c) 0.006646 (d) 66.46
MAIN TOPIC: LOGARITHM
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
We recall that
Number | Standard form | Characteristic | Mantissa | logarithm
0.148 | 1.48 x 10-1 | - or 1 | 1703 | 1 .1703
0.065 | 6.5 x 10-2 | -2 or 2 | 8129 | 2.8129
Example 1: Simplify 2.3502 + 1.875 + 1.4625
Solution
2.3502 + 1.875 + 1. 4625 = 3.6877
Example 2: Evaluate by using logarithm tables 16.82 x 0.00635/ 0.04822 correct to three significant figures.
Solution
Number Logarithm
16.82 1.2258
0.00635 3.8028 +
1.0286 = 1.0286
0.04822 2.6830
X 2
3.3660 = 3.3660 -
Anti log = 45.98
1.6626
= 46.0 (3 sig.fig)
https://youtu.be/gOZ5wNOHaI8
EVALUATION:
Using logarithms table, evaluate 5.876 x 0.0432 / 0.00034 correct to three significant figures
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Using logarithm table, evaluate 3| 1.376 / 5| 0.007 correct to three significant figure.
Solution
Number | Logarithm
3 |1.476 | 0.1386 ÷ 3 = 0.0462
5 |0.007 | 3.8451 ÷ 5 = 1.56902 -
0.47718 = 0.4772
Anti log = 30.00
= 30.0 (3 sig. Fig)
https://youtu.be/gOZ5wNOHaI8
EVALUATION:
Using logarithms table, evaluate 3 3.543
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Evaluate 4 | 0.0763
309 x 0.008465
Solution
number log
0.0763 2.8825 = 2.8825
309 2.4900
0.008465 3.9277 +
1.4377 = 1.4377 -
1.4442
Anti log = 2781
= 2780 to 3 sig. Fig.
1
https://youtu.be/AswQ381sczk
https://youtu.be/ExbzXBS5W-k
EVALUATION:
Evaluate 4 0.3456 x 0.054
4.210 x 0.00065
SPECIFIC TOPIC: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using logarithm and straight calculation
CONTENT: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
The results obtained when calculating with 4-figure tables are accurate only to the first three figures. The 4th digit is not likely to be accurate and used for rounding off a result to 3 significant figures. Three-figure accuracy is sufficient for most practical purposes.
Example 1: Calculate to 3 significant figures, the area of a flat circular washer 6.84cm in diameter with a hole 2.96cm in diameter. Take log ∏ to be 0.4971
Solution
Outer radius = 6.84 / 2 cm = 3.42cm
Inner radius = 2.96 / 2 cm = 1.48cm
Area = ∏ x 3.422 - ∏x 1.482 cm2
= ∏(3.422 – 1.482)cm2
= ∏(3.42 + 1.48) (3.42 – 1.48)cm2
= ∏ x4.9 x 1.94cm2
= 29.86cm2
= 29.9cm2 to 3 sig.fig.
Using logarithm
Number | logarithm
∏ | 0.4971
4.9 | 0.6902
1.94 | 0.2878
Anti log = 29.86 1.4751
https://youtu.be/rBnQiLa2TYo
EVALUATION:
Calculate the area in hectares of a rectangular field 126m long and 97m wide.(1 ha = 10 000m2)
ASSIGNMENT:
(i) 3 0.023 x 0.0041
8.001 x 0.00031
(ii) Calculate the area of a circular washer of diameter 3.42cm if the hole in the centre is 1cm in diameter.
CONTENTS
• Standard forms
• Logarithm of numbers greater than one
• Multiplication and divisions of numbers greater than one using logarithm
• Using logarithm to solve problems with roots and powers (no > 1)
• Logarithm of numbers less than one.
• Multiplication and division of numbers less than one using logarithm
• Roots and powers of numbers less than one using logarithm
STANDARD FORMS
A way of expressing numbers in the form A x 10x where 1≤ A 10 and x is an integer is said to be a standard form. Numbers are grouped into two. Large and small numbers. Numbers greater than or equal to 1 are called large numbers. In this case the x, which is the power of 10 is positive. On the other hand, numbers less than 1 are called small numbers. Here, the integer is negative.
Numbers such as 1000 can be converted to its power of ten in the form 10n where n can be term as the number of times the decimal point is shifted to the front of the first significant figure
i.e. 10000 = 104
Number = Power of 10
100 =102
10= 101
1 100
0.01 10-3
0.10 10-1
Note: One tenth; one hundredth, etc are expressed as negative powers of 10 because the decimal point is shifted to the right while that of whole numbers are shifted to the left to be after the first significant figure.
https://youtu.be/VsbpBrOmr18
Example
(i) Express 0. 08356 in standard form
(ii) 832.8 in standard form
Solution
i 0. 08356 = 8.356 x 10-2
ii 832.8 = 8.328 x 102
Express the following in standard form
(a) 39.32 = 3.932 x 101
(b) 4.83 = 4.83 x 100
(c) 0.005321 = 5.321 x 10-3
If a number is in its standard form, its power is its integer i.e. the integer of its logarithm e.g. log 7853 has integer 3 because 7853 = 7.853 x 103
Examples:
Use tables (log) to find the complete logarithm of the following numbers.
(a) 80030 (b) 8 (c) 135.80
(a) 80030 = 4.9033
(b) 8 = 0.9031
(c) 13580 = 2.1329
https://youtu.be/4006pXk940I
Evaluation.
Use table to find the complete logarithm of the following:
(a) 183 (b) 89500 (c) 10.1300 (d) 7
Reading Assignment
New General Mathematics SSII Page 20 – 21, Exercise 1a & b.
Multiplication and Division of numbers greater than one using logarithm
To multiply and divide numbers using logarithms, first express the number as logarithm and then apply the addition and subtraction laws of indices to the logarithms. Add the logarithm when multiplying and subtract when dividing.
Examples
Evaluate using logarithm.
1. 4627 x 29.3
2. 8198 ÷ 3.905
3. 48.63 x 8.53
15.39
https://youtu.be/0-kZdwxFjY4
Solutions
1. 4627 x 29.3
To find the Antilog of the log 5.1322 use the antilogarithm table:
Check 13 under 2 diff 2 (add the value of the difference) the number is 0.1356. To place the decimal point at the appropriate place, add one to the integer of the log i.e. 5 + 1 = 6 then shift the decimal point of the antilog figure to the right (positive) in 6 places.
https://youtu.be/yzuYXIoVYMw
Evaluation:
Use logarithm to calculate.
3612 x 750.9
113.2 x 9.98
Reading Assignment
Now General Mathematics SS2 Page 21 exercise 1C.
Using logarithm to solve problems with powers and root
(no. greater than one).
Examples:
Find the single logarithm representing the numerator and the single logarithm representing the denominator, subtract the logarithm then find the anti log.
(Numerator – Denominator).
Example: Find the complete log of the following.
(a) 0.004863 (b) 0.853 (c) 0.293
Solution
Log 0.004863 = 3.6369
Log 0.0853 = 2.9309
Log 0.293 = 1.4669
https://youtu.be/N0-u57nXLQ8
Evaluation
1. Find the logarithm of the following:
(a) 0.064 (b) 0.002 (c) 0.802
2. Evaluate using logarithm.
95.3 x 318.4
1.295 x 2.03
https://youtu.be/kqVpPSzkTYA
Reading Assignment
New General mathematics for SS 2 page 22 – 24 exercise 1g.
Using logarithm to evaluate problems of Multiplication, Division, Powers and roots with numbers less than one.
OPERATION WITH BAR NOTATION
Note the following when carrying out operations on logarithm of numbers which are negative.
I. The mantissa (fractional part) is positive, so it has to be added in the usual manner.
II. The characteristic (integral part) is either positive or negative and should therefore be added or operated as directed numbers.
III. For operations like multiplication and division, separate the integer from the characteristic before performing the operation.
Examples:
Simplify the following, leaving the answers in bar notation, where necessary
I. .7675 + 2.4536
II. 6.8053 – 4.1124
III. 2.4423 x 3
IV. 2.2337 ÷ 7
Note: 3 cannot divide 2 therefore subtract 1 from the negative integer and add 1 to the positive decimal fraction so as to have 3 which is divisible
by 3 without remainder.
https://youtu.be/fnhFneOz6n8
ASSIGNMENT
Objective2
Use table to find the log of the following:
1. 900 (a) 3.9542 (b) 1.9542 (c) 2.9542 (d) 0.9542
2. 12.34 (a) 3.0899 (b) 1.089 (c) 2.0913 (d) 1.0913
3. 0.000197 (a) 4.2945 (b) 4.2945 (c) 3.2945 (d) 3.2945
4. 0.8 (a) 1.9031 (b) 1.9031 (c) 0.9031 (d) 2.9031
5. Use antilog table to write down the number whose logarithms is 3.8226.
(a) 0.6646 (b) 0.06646 (c) 0.006646 (d) 66.46
MAIN TOPIC: LOGARITHM
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
We recall that
Number | Standard form | Characteristic | Mantissa | logarithm
0.148 | 1.48 x 10-1 | - or 1 | 1703 | 1 .1703
0.065 | 6.5 x 10-2 | -2 or 2 | 8129 | 2.8129
Example 1: Simplify 2.3502 + 1.875 + 1.4625
Solution
2.3502 + 1.875 + 1. 4625 = 3.6877
Example 2: Evaluate by using logarithm tables 16.82 x 0.00635/ 0.04822 correct to three significant figures.
Solution
Number Logarithm
16.82 1.2258
0.00635 3.8028 +
1.0286 = 1.0286
0.04822 2.6830
X 2
3.3660 = 3.3660 -
Anti log = 45.98
1.6626
= 46.0 (3 sig.fig)
https://youtu.be/gOZ5wNOHaI8
EVALUATION:
Using logarithms table, evaluate 5.876 x 0.0432 / 0.00034 correct to three significant figures
SPECIFIC TOPIC: LOGARITHM OF NUMBER LESS THAN 1
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on logarithm of number less than 1
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Using logarithm table, evaluate 3| 1.376 / 5| 0.007 correct to three significant figure.
Solution
Number | Logarithm
3 |1.476 | 0.1386 ÷ 3 = 0.0462
5 |0.007 | 3.8451 ÷ 5 = 1.56902 -
0.47718 = 0.4772
Anti log = 30.00
= 30.0 (3 sig. Fig)
https://youtu.be/gOZ5wNOHaI8
EVALUATION:
Using logarithms table, evaluate 3 3.543
CONTENT: LOGARITHM OF NUMBER LESS THAN 1
Example 1: Evaluate 4 | 0.0763
309 x 0.008465
Solution
number log
0.0763 2.8825 = 2.8825
309 2.4900
0.008465 3.9277 +
1.4377 = 1.4377 -
1.4442
Anti log = 2781
= 2780 to 3 sig. Fig.
1
https://youtu.be/AswQ381sczk
https://youtu.be/ExbzXBS5W-k
EVALUATION:
Evaluate 4 0.3456 x 0.054
4.210 x 0.00065
SPECIFIC TOPIC: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems using logarithm and straight calculation
CONTENT: ACCURACY OF RESULT USING LOGARITHM TABLE AND STRAIGHT CALCULATION
The results obtained when calculating with 4-figure tables are accurate only to the first three figures. The 4th digit is not likely to be accurate and used for rounding off a result to 3 significant figures. Three-figure accuracy is sufficient for most practical purposes.
Example 1: Calculate to 3 significant figures, the area of a flat circular washer 6.84cm in diameter with a hole 2.96cm in diameter. Take log ∏ to be 0.4971
Solution
Outer radius = 6.84 / 2 cm = 3.42cm
Inner radius = 2.96 / 2 cm = 1.48cm
Area = ∏ x 3.422 - ∏x 1.482 cm2
= ∏(3.422 – 1.482)cm2
= ∏(3.42 + 1.48) (3.42 – 1.48)cm2
= ∏ x4.9 x 1.94cm2
= 29.86cm2
= 29.9cm2 to 3 sig.fig.
Using logarithm
Number | logarithm
∏ | 0.4971
4.9 | 0.6902
1.94 | 0.2878
Anti log = 29.86 1.4751
https://youtu.be/rBnQiLa2TYo
EVALUATION:
Calculate the area in hectares of a rectangular field 126m long and 97m wide.(1 ha = 10 000m2)
ASSIGNMENT:
(i) 3 0.023 x 0.0041
8.001 x 0.00031
(ii) Calculate the area of a circular washer of diameter 3.42cm if the hole in the centre is 1cm in diameter.
WEEK 3
SPECIFIC TOPIC: APPROXIMATION AND PERCENTAGE ERROR
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximation and percentage error
CONTENT:
APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy’s error = 15.6 – 13.4 = 2.2m
Second boy’s error = 16.9 – 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = Absolute error/Actual value x 100
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 – 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
https://youtu.be/oZsw4mNaHog
EVALUATION:
A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 – 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
https://youtu.be/By1QgpFRuJk
EVALUATION:
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 – 26.00
#374.00
Example 2: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66/100 x 6500/1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
https://youtu.be/4X_vLNeVC3k
EVALUATION:
A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: An error of 4% was made in finding the length of a rope that was actually 25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
https://youtu.be/JbEo46uV6d4
EVALUATION:
(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
TOPIC: PERCENTAGE ERROR
CONTENT
Definition of percentage error
Calculation of percentage error
Percentage error (range of values via approximations)
Calculations on percentage error in relation to approximation
Definition of Percentage Error
No measurement, however, carefully made is exact (accurate) i.e if the length of a classroom is measured as 2.8m to 2 s.f the actual length may be between 2.75 and 2.85, the error of this measurement is 2.75 – 2.8 or 2.85 – 2.8 = + 0.05.
Example 2
Suppose the length of the same room is measured to the nearest cm 280cm i.e. (280cm) calculate the percentage error.
Measurement = 280cm.
The range of measurement will be between 279.5cm or 280cm .
Error = 280 – 279.5 = 0.5cm
% error = error / Measurement x 100 / 1
% error = 0.5/ 280 x 100 / 1 = 0.178%
= 0.18% (2sf)
Example 3
The length of a field is measured as 500m, find the percentage error of the length if the room is measured to
i. nearest meter
ii. nearest 10m
iii. one significant figure.
Solutions
i. To nearest meter
Measurement =500m
Actual measurement = between 499.5 - 500.5
Error = + 0.5m
% error = error / measurement x 100
0.5 / 500 x 100 / 1 = 0.10%
= 0.10%
ii. Nearest 10m
measurement = 500m,
range= 595m – 505m
error = + 5m
error = 5/500 x 100/1 = 1%
iii. To 1 s.f.
measurement = 500m
range = 450 – 550
error = + 50
% error = 50 / 500 x 100 / 1 = 10%
https://youtu.be/_LClC1OdxXo
Evaluation
1. A bottle of Coca-Cola indicates that it contains 350ml, a food inspector carefully measured the contents of a bottle and finds that it contains 36cl, calculate the % percentage error of the content.
2. The length of a line segment is 90cm to the nearest cm, calculate the % error.
Percentage Error (range of values via approximations)
1. Range of values measured to the nearest whole number i.e. nearest tens, hundreds etc. e.g.
Find the range of values of N6000 to:
i. nearest naira = N5999.50 = 6000.50
ii. nearest N10 = N5995 = 6005
iii. nearest N100 = N5950 = 6050
iv. nearest N1000 = N5500 = 6,500
2. Range of values measured to a given significant figure. E.g. Find the range of value of 6000 to
1 sf = 5500 - 6500
2 sf = 5950 - 6050
3 sf = 5995 - 6005
5 sf = 5999.95- 6000.05
3. Range of values measured to a given decimal places e.g. 39.8 to a 1d.p = 39.75 – 39.85.
Note: if it is 1 d.p, the range of values will be in 2 d.p, if 2 d.p, the range will be in 3 d.p etc. (i.e the range = d.p + 1). The same rule is also applicable to range of values to given significant figure.
https://youtu.be/5y4BszPCBcs
Evaluation
Orally: From New General Mathematics SS 2 by J. B. Channon and Co 3rd edition exercise 46 no. 1a – f.
Reading Assignment
New General Mathematics SS 2 exercise 46 no 1g – j, then calculate their percentage error.
Calculations on percentage error:
Example:
Calculate the percentage error if
1. The capacity of a bucket is 7.5 litres to 1 d.p.
2. The mass of a student is 62kg to 2 s.f.
Solutions
1. Measurement = 7.5litres ( 1d.p)
Range of values = 7.45 - 7.55
Error = 7.5 – 7.45 = 0.05
% error = error / measurement x 100 / 1
0.05/7.5 x 100/1
= 0.67%
2. Measurement = 62kg (2 s.f)
Range of values = 61.5kg to 62.5kg
error = 6.2 - 61.5 = 0.5kg
% error = error / measurement x 100 / 1
0.5 / 62 x 100 / 1 = 0.81%
https://youtu.be/Cv6poRrgX_4
Evaluations
1. Calculate correct to 2 s.f. the percentage error in approximately 0.375 to 0.4.
2. Calculate the percentage error of a pole 125m high is measured to the nearest cm.
Reading Assignment
New General Mathematics SS 2 page 47 exercise 4b no. 4 – 10.
ASSIGNMENT
Objectives
What is the error in the following measurement.
1. The distance between two towns is 60km to the nearest km.
(a) 5km (b) 0.5km (c) 8.3km (d) 0.83km
2. The area of a classroom is 400m2 to 2 s.f.
(a) 50m2 (b) 1.25m2 (c) 2.5m2 (d) 5m2
3. A sales girl gave girl a balance of N1.15 to a customer instead of N1.25, calculate the % error.
4. A student measured the length of a room and obtained the measurement of 3.99m, if the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room (a) 3.78m (b) 3.80m (c) 4.18m (d) 4.20m
5. A man is 1.5m tall to the nearest cm, calculate his percentage error.
(a) 0.05cm (b) 0.33% (c) 0.033% (d) 0.05cm
Theory:
1. A classroom is 10m by 10m, a student measured a side as 9.5m and the other side as 10m and uses his measurement to calculate the area of the classroom. Find the percentage error in
a. the length of one of the sides
b. the area of the room
2. Instead of recording the number 1.23cm for the radius of a tube, a student recorded 1.32cm, find the percentage error correct to 1 d.p.
Reading Assignment
Read on sequence and arithmetic progression page 177 – 187.
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on approximation and percentage error
CONTENT:
APPROXIMATION AND PERCENTAGE ERROR
Hardly do we take any measurement and get exact results. The results often got are inaccurate. The importance being stressed here is not the slight error made in the measurement but to decide the degree of such error. This degree or level of error always measured in percentage will then guide any decision maker on the acceptable level of error.
The error that may be made in any measurement could either be positive or negative, e.g. the actual length of a rope is 15.6m. If the rope is measured by two boys as 13.4m and 16.9m respectively, the errors are as follows:
First boy’s error = 15.6 – 13.4 = 2.2m
Second boy’s error = 16.9 – 15.6 = 1.3m
Whether the error is against (as in first boy) or in favour (as in second boy), error is error, therefore, the following formula is used:
Percentage error = Absolute error/Actual value x 100
Example 1: A rope of length 15cm was measured by a girl to be 14.4cm. Find the percentage error.
Solution
The actual error is 15 – 14.4 = 0.6cm
Percentage error = 0.6/15 x 100%
= 4%
https://youtu.be/oZsw4mNaHog
EVALUATION:
A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1decimal place?
Solution
The actual error is 6.65 – 6.35 = 2.5
Percentage error = 2.5/6.35 x 100%
= 4.7%
https://youtu.be/By1QgpFRuJk
EVALUATION:
If the age of a man 64 years is written as 71 years, calculate error to 3 significant figures.
ASSIGNMENT:
Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: A man underestimated his expenses by 6.5%,but actually spent #400.00. What was his estimate?
Solution
6.5% of #400.00 = #26.00
Therefore, the actual estimate is 400.00 – 26.00
#374.00
Example 2: A man underestimated his expenses by 1 2/3%, but actually spent #6500.00. What is his estimate?
Solution
1 2/3% = 5/3 = 1.66%
1.66/100 x 6500/1
1.66 x 65
#107.9
Therefore the actual estimate = #6500.00 - #107.9
= #6392.1
https://youtu.be/4X_vLNeVC3k
EVALUATION:
A man underestimated his expenses by 4.8%but actually spent #2355.00. What was his estimate?
CONTENT: APPROXIMATION AND PERCENTAGE ERROR
Example 1: An error of 4% was made in finding the length of a rope that was actually 25m. By how many metres was the measurement wrong?
Solution
4% of 25 = 1
The actual length of the rope is 25m
The error is 1
Therefore, the measurement was wrong with 1m
https://youtu.be/JbEo46uV6d4
EVALUATION:
(i) What is the percentage error in an area of a lawn that actually measures 750m2 but found to be 690m2
(ii) The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope to 1 decimal place.
TOPIC: PERCENTAGE ERROR
CONTENT
Definition of percentage error
Calculation of percentage error
Percentage error (range of values via approximations)
Calculations on percentage error in relation to approximation
Definition of Percentage Error
No measurement, however, carefully made is exact (accurate) i.e if the length of a classroom is measured as 2.8m to 2 s.f the actual length may be between 2.75 and 2.85, the error of this measurement is 2.75 – 2.8 or 2.85 – 2.8 = + 0.05.
Example 2
Suppose the length of the same room is measured to the nearest cm 280cm i.e. (280cm) calculate the percentage error.
Measurement = 280cm.
The range of measurement will be between 279.5cm or 280cm .
Error = 280 – 279.5 = 0.5cm
% error = error / Measurement x 100 / 1
% error = 0.5/ 280 x 100 / 1 = 0.178%
= 0.18% (2sf)
Example 3
The length of a field is measured as 500m, find the percentage error of the length if the room is measured to
i. nearest meter
ii. nearest 10m
iii. one significant figure.
Solutions
i. To nearest meter
Measurement =500m
Actual measurement = between 499.5 - 500.5
Error = + 0.5m
% error = error / measurement x 100
0.5 / 500 x 100 / 1 = 0.10%
= 0.10%
ii. Nearest 10m
measurement = 500m,
range= 595m – 505m
error = + 5m
error = 5/500 x 100/1 = 1%
iii. To 1 s.f.
measurement = 500m
range = 450 – 550
error = + 50
% error = 50 / 500 x 100 / 1 = 10%
https://youtu.be/_LClC1OdxXo
Evaluation
1. A bottle of Coca-Cola indicates that it contains 350ml, a food inspector carefully measured the contents of a bottle and finds that it contains 36cl, calculate the % percentage error of the content.
2. The length of a line segment is 90cm to the nearest cm, calculate the % error.
Percentage Error (range of values via approximations)
1. Range of values measured to the nearest whole number i.e. nearest tens, hundreds etc. e.g.
Find the range of values of N6000 to:
i. nearest naira = N5999.50 = 6000.50
ii. nearest N10 = N5995 = 6005
iii. nearest N100 = N5950 = 6050
iv. nearest N1000 = N5500 = 6,500
2. Range of values measured to a given significant figure. E.g. Find the range of value of 6000 to
1 sf = 5500 - 6500
2 sf = 5950 - 6050
3 sf = 5995 - 6005
5 sf = 5999.95- 6000.05
3. Range of values measured to a given decimal places e.g. 39.8 to a 1d.p = 39.75 – 39.85.
Note: if it is 1 d.p, the range of values will be in 2 d.p, if 2 d.p, the range will be in 3 d.p etc. (i.e the range = d.p + 1). The same rule is also applicable to range of values to given significant figure.
https://youtu.be/5y4BszPCBcs
Evaluation
Orally: From New General Mathematics SS 2 by J. B. Channon and Co 3rd edition exercise 46 no. 1a – f.
Reading Assignment
New General Mathematics SS 2 exercise 46 no 1g – j, then calculate their percentage error.
Calculations on percentage error:
Example:
Calculate the percentage error if
1. The capacity of a bucket is 7.5 litres to 1 d.p.
2. The mass of a student is 62kg to 2 s.f.
Solutions
1. Measurement = 7.5litres ( 1d.p)
Range of values = 7.45 - 7.55
Error = 7.5 – 7.45 = 0.05
% error = error / measurement x 100 / 1
0.05/7.5 x 100/1
= 0.67%
2. Measurement = 62kg (2 s.f)
Range of values = 61.5kg to 62.5kg
error = 6.2 - 61.5 = 0.5kg
% error = error / measurement x 100 / 1
0.5 / 62 x 100 / 1 = 0.81%
https://youtu.be/Cv6poRrgX_4
Evaluations
1. Calculate correct to 2 s.f. the percentage error in approximately 0.375 to 0.4.
2. Calculate the percentage error of a pole 125m high is measured to the nearest cm.
Reading Assignment
New General Mathematics SS 2 page 47 exercise 4b no. 4 – 10.
ASSIGNMENT
Objectives
What is the error in the following measurement.
1. The distance between two towns is 60km to the nearest km.
(a) 5km (b) 0.5km (c) 8.3km (d) 0.83km
2. The area of a classroom is 400m2 to 2 s.f.
(a) 50m2 (b) 1.25m2 (c) 2.5m2 (d) 5m2
3. A sales girl gave girl a balance of N1.15 to a customer instead of N1.25, calculate the % error.
4. A student measured the length of a room and obtained the measurement of 3.99m, if the percentage error of his measurement was 5% and his own measurement was smaller than the length, what is the length of the room (a) 3.78m (b) 3.80m (c) 4.18m (d) 4.20m
5. A man is 1.5m tall to the nearest cm, calculate his percentage error.
(a) 0.05cm (b) 0.33% (c) 0.033% (d) 0.05cm
Theory:
1. A classroom is 10m by 10m, a student measured a side as 9.5m and the other side as 10m and uses his measurement to calculate the area of the classroom. Find the percentage error in
a. the length of one of the sides
b. the area of the room
2. Instead of recording the number 1.23cm for the radius of a tube, a student recorded 1.32cm, find the percentage error correct to 1 d.p.
Reading Assignment
Read on sequence and arithmetic progression page 177 – 187.
WEEK 4
MAIN TOPIC: SEQUENCE AND SERIES
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
https://youtu.be/VgBSzYBMbh4
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
https://youtu.be/m5Yn4BdpOV0
EVALUATION:
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
SPECIFIC TOPIC: SERIES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are
3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
https://youtu.be/r614AD021n0
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
https://youtu.be/Tj89FA-d0f8
EVALUATION:
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
https://youtu.be/ZZaUO_35hdk
EVALUATION:
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
https://youtu.be/SyCsmjvOODo
EVALUATION:
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
SPECIFIC TOPIC: CALCULATION INVOLVING SUM OF AN A.P
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Solve problems on sum of an A.P
(ii) Find the nth term of an A.P
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
https://youtu.be/0m8GKU-eiyQ
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
TOPIC: ARITHMETIC PROGRESSION (A. P).
CONTENTS
• Sequence
• Definition of Arithmetic Progression
• Denotations in Arithmetic progression
• Deriving formulae for the term of A. P.
• Sum of an arithmetic series
SEQUENCE
Examples:
Find the next two terms in each of the following sets of number and in each case state the rule which gives the term.
(a) 1, 5, 9, 13, 17, 21, 25 (any term +4 = next term)
(b) 2, 6, 18, 54, 162, 486, 1458 (any term x 3 = next term)
(c) 1, 9, 25, 49, 81, 121, 169, (sequence of consecutive odd no)
(d) 10, 9, 7, 4, 0, -5, -11, -18, -26, (starting from 10, subtract 1, 2, 3 from immediate no).
In each of the examples below, there is a rule which will give more terms in the list. A list like this is called a SEQUENCE in many cases; it can simply matter if a general term can be found for a sequence e.g.
1, 5, 9, 13, 17 can be expressed as
1, 5, 9, 13, 17 ……………. 4n – 3 where n = no of terms
Check: 5th term = 4(5) -3
20 – 3 = 17
10th term = 4(10) – 3
40 – 3 = 37
Example 2
Find the 6th and 9th terms of the sequence whose nth term is
(a) (2n + 1)
(b) 3 – 5n.
Solution
(a) 2n + 1
6th term = 2(6) + 1
=12 + 1 = 13
9th term = 2 (9) + 1
=18 + 1 = 19
(b) 3 – 5n
6th term = 3 – 5 (6)
=3 – 30 = -27
9th term = 3 – 5 (9)
=3 – 45 = -42
https://youtu.be/vlPHnGbKi9w
Evaluation
For each of the following sequence, find the next two terms and the rules which gives the term.
1. 1, , , , , ____, ____
2 100, 96, 92, 88, _____, ____
3. 2, 4, 6, 8, 10, ____, _____
4. 1, 4, 9, 16, 25, ____, _____
(i) Arrange the numbers in order of increasing give to a sequence.
(ii) Find the next two terms in the sequence
5. 19, 13, 16, 22, 10
6. -21/2, 51/2, 31/2, 11/2, -1/2
7. Find the 15th term of the sequence whose nth term is 3n - 5
4
Reading Assignment
New General mathematics SS II by G. B. Channon & Co. exercise 18a.
DEFINITION OF ARITHMETIC PROGRESSION
This is a sequence in which the terms either increase or decrease in equal steps is called an Arithmetic Progression.
The sequence 9, 12, 15, 18, 21, ____, _____, _____ has a first term of 9 and a common difference of +3 between the terms.
Denotations in A. P.
a = 1st term
d = common difference
n = no of terms
Un = nth term
Sn = Sum of the first n terms
Formula for nth term of Arithmetic Progression.
e.g. in the sequence 9, 12, 15, 18, 21.
a = 9
d = 12 – 9 or 18 – 15 = 3.
1st term = U1 = 9 = a
2nd term = U2 = 9 + 3 = a + d
3rd term = U3 = 9 + 3 + 3 = a + 2d
10th term = U10 = 9 + 9(3) = a + 9d
nth term = Un = 9+(n-1)3 = a + (n-1)d
nth term = Un = a + (n-1)d
Example:
Given the A.P, 9, 12, 15, 18 …… find the 50th term.
a = 9
d = 3
n = 50
Un = U50
Un = a + (n-1) d
U50 = 9 + (50-1) 3
9 + (49) 3
9 + 147
=156.
2. The 43rd term of an AP is 26, find the 1st term of the progression given that its common difference is ½ and also find the 50th term.
U43 = 26
d = ½
a = ?
n = 43
Un = a + (n-1) d
26 = a + (43-1) ½
26 = a + 42(1/2)
26 = a + 21
26 – 21 = a
5 = a
a = 5
(b) a = 5
d = ½
n = 50
U50 =?
Un = a + (n-1) d
U50 = 5 + (50-1)1/2
= 5 + 49(1/2)
U50 = 5 + 241/2
U50 = 291/2
https://youtu.be/YqP5TuwtEdg
Evaluation
1. Find the 37th term of the sequence 20, 10, 0, -10…
2. 1, 5… 69 are the 1st, 2nd, and last term of the sequence; find the common difference between them and the number of terms in the sequence.
Reading Assessment
New General Mathematics for SS 2, Exercise 18b, no 8, 10, 11, 12.
SUM OF AN ARITHMETIC SERIES
When the terms of a sequence are added, the resulting expression is called series e.g. in the sequence 1, 3, 5, 7, 9, 11.
Series = 1 + 3 + 5 + 7 + 9 + 11
When the terms of a sequence are unending the series is called infinite series, it is often impossible to find the sum of the terms in an infinite series.
e.g. 1 + 3 + 5 + 7 + 9 + 11 + …………………. Infinite
Sequence with last term or nth term is termed finite series.
e.g.
Find the sum of
1, 3, 5, 7, 9, 11, 13, 15
If sum = 2, n = 8
Then
S = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
Or S = 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
Add eqn1 and eqn 2
2s = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16
= 48 = 8(16)
2 2 = S = 64
Deriving the formula for sum of A. P. The following represent a general arithmetic series when the terms are added.
S = a + (a+d) + a + 2d + …………………………… + (L-2d) + (L-d) + L – eqn
S = L + (L-d) + L – 2d + ……………………………… a + 2d + (a+d) + a – eqn
2s = (a + L) + (a + L) + (a + L) + …………………… (a + L) + (a + L) + (a + L)
2s = n(a + L)
2
S = n(a+L)
2
L => Un = a + (n-1)d
Substitute L into eq**
S = n(a + a+(n-1)d
2
S = n(2a + (n-1)d = n ( 2a+ (n-1)d
2 2
S = n(a+L) when Un (last term L) is given
2
or
S = n(2a + (n-1)d when d is given or obtained
2
Example 2
Find the sum of the 20th term of the series 16 + 9 + 2 + …………………
a = 16
d = 9 – 16 = -7
n = 20
S = n(2a + (n-1)d)
2
S = 20 (2x16) + (20-1)(-7)
2
20 (32 + 19(-7)
2
S =10 (32 - 133) = 10(-101)
S = -1010
https://youtu.be/TtJ4dphwh0U
Evaluation
1. Find the sum of the arithmetic series with 16 and -117 as the first and 20th term respectively.
2. The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
Reading Assignment
New General Mathematics for SS 2, Page 181 – 184. Exercise 18d.
Evaluations (Exercise)
1. An A. P. has 15 terms and a common difference of -3, find its first and last term if its sum is 120.
2. On the 1st of January, a student puts N10 in a box, on the 2nd she puts N20 in the box, on the 3rd she puts N30 and so on putting on the same no. of N10 notes as the day of the month. How much will be in the box if she keeps doing this till 16th January.
Evaluation
The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
ASSIGNMENT
Theory
1. Eight wooden poles are to be used for pillars and the length of the poles form an arc Arithmetic Progression (A. P.) if the second pole is 2m and the 6th pole is 5m, give the lengths of the poles in order and sum up the lengths of the poles.
2a. Write down the 15th term of the sequence.
2_, 3 , 4 , 5
1x3 2x4 3x5 4 x6
b. An arithmetic progression (A. P.) has 3 as its term and 4 as the common difference.
c. Write an expression in its simplest form for the nth term.
d. Find the least term of the A. P. that is greater than 100.
Objectives
1. Find the 4th term of an A. P. whose first term is 2 and the common difference is 0.5 (a) 4 (b) 4.5 (c) 3.5 (d) 2.5
2. In an A. P. the difference between the 8th and 4th term is 20 and the 8th term is 11/2 times the 4th term, find the common difference
(a) 5 (b) 7 (c) 3 (d) 10
3. Find the first term of the sequence in no. 2
(a) 70 (b) 45 (c) 25 (d) 5
4. The next term of the sequence 18, 12, 60 is
(a) 12 (b) 6 (c) -6 (d) -12
5. Find the no. of terms of the sequence 1/2 , ¾, 1, ……………….. 51/2
(a) 21 (b) 43/4 (c) 1 (d) 22
SPECIFIC TOPIC: CONCEPT OF SEQUENCE AND SERIES
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Generate the sequences of their own
CONTENT: CONCEPT OF SEQUENCE AND SERIES
A Sequence is a succession of terms in such a way that the terms are related to one another according to a well defined rule. The rules may differ depending on the arrangement of such terms.
Examples are given below.
(a) 5,11,17,23,(multiply set of positive integer by 6 and subtract 1)
(b) 1,3,9,27,81 (3n-1).
In general, the rule that works for a particular sequence may not work for another. The nth term of a sequence may be represented by Tn so that the first, second, third, can be written as T1,T2, T3 etc
https://youtu.be/VgBSzYBMbh4
Example 1: The nth term of a sequence is given by 3 x 2n-2. Write down the first three terms of the sequence.
Solution
Tn = 3 x 2n-2
T1 = 3 X 21-2
= 3 X 2-1
= 3/2
T2 = 3 x 22-2
= 3 x 20
= 3 x 1
= 3
T3 = 3 x 23-2
= 3 x 2 1
= 2x3
= 6
The first three terms are 3/2, 3, 6.
https://youtu.be/m5Yn4BdpOV0
EVALUATION:
If the nth term of a sequence is denoted by the formula n(2n+1) – 3n, find the sum of the first four terms.
SPECIFIC TOPIC: SERIES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on series
CONTENT: CONCEPT OF SERIES
A series is the addition of the terms of a sequence for example for example, in example 1 under sequence the first three terms of the sequence are
3/2, 3 and 6.
Other examples of a series are as follows
(i) 3 + 5 + 7 + 9 + 11 +............
(ii) 8 + 9 + 10 + 11 + 12 + ..............
(iii) 2 + 5 + 10 + 15 + 26 + 37 + ............
(iv) -1 + 0 + 7 + 14 + 23 + 34 + ............
https://youtu.be/r614AD021n0
Example 1: Find the series of the first six terms of 2n + 4n2
Solution
Tn = 2n + 4n2
T1 = 21 + 4(1)2
T1 = 2 + 4 = 6
T2 = 22 + 4(2)2 = 20
T3 = 23 + 4(3)2 = 44
T4 = 24 + 4(4)2 = 80
T5 = 25 + 4(5)2 = 132
T6 = 26 + 4(6)2 = 208
https://youtu.be/Tj89FA-d0f8
EVALUATION:
(i) Find the sum of the series n2 + 5n up to the 4th terms.
(ii) The nth term of a sequence is denoted by 3n(2n – 1). Find the sum of the first 5terms.
ASSIGNMENT:
Given that the nth term of a sequence is given by the formula n/2(3n2 + 1). Write down the first six terms.
SPECIFIC TOPIC: SUM OF TERMS IN A.P
REFERENCE BOOK: New General Mathematics for senior school 2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on the sum of an A.P
CONTENT: SUM OF TERMS IN A.P
The sum of terms in any A.P is the addition of all the terms involved in the particular A.P.
Given a particular A.P to an nth term as, a , (a + d) , (a + 2d), (a + 3d) , (a + 4d), ....... a + (n – 1)d.
Let Sn represent the sum of the A.P above.
Therefore
(a) Sn = a + (a +d) + (a + 2d) + (a + 3d) + ... + a + (n – 1)d.
It was noted that the nth terms of an A.P is given by,
Tn = a + (n – 1)d
If we now wish to find the sum Sn of an A.P with n-terms, then the sum will be:
(b) Sn = (a + (n-1)d) + (a + (n-2)d) + (a + (n-3)d) + ......+ (a + d) + a.
Adding (a) and (b)
2Sn = (2a + (n-1)d) + (2a + 9n-1)d) + (2a + (n-1)d) + .....+ (2a + (n-1)d).
Therefore,
Sn = n/2(2a + (n-1)d)
Which is the sum of n-terms of an A.P
OR
Sn = n/2(a + L)
Where L is the last term.
This is used when the first and last terms and the number of terms are given
Example 1; Find the sum of the first 25 terms of the sequence 11,15,19,23,27,......
Solution
First term (a) = 11
Common difference (d) = 4
Sn = n/2(2a +(n -1)d)
S25 = 25/2(2 x11(25 – 1)4)
= 25/2(22 +(24)4)
= 25/2(22 +96)
= 25/2(118)
= 25/2 x 118/1
= 25 x 59
= 1475
https://youtu.be/ZZaUO_35hdk
EVALUATION:
The 9th and 22nd terms of an A.P are 29 and 55 respectively. Find the sum of its first 60 terms.
ASSIGNMENT:
Find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7
SPECIFIC TOPIC: CALCULATION INVOLVING SEQUENCE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems involving sum of A.P
CONTENT: CALCULATION INVOLVING SUM OF A.P
Example 1: The sum of 11 terms of an A.P is 891. Find the 28th and 45th terms if the common difference is 15
Solution
Sn = n/2(2a + (n-1)d)
Sn = 11/2(2a + (11-1)d)
= 11/2(2a + 10d)
But S11 = 891
Then 11/2(2a +10d) = 891
11(2a + 10d) = 1782
22a + 110d = 1782
When d = 15: 22a + 1650 1782
22a = 1782 – 1650 = 132
a= 132/22
a= 6(i.e the first term)
28th term will be a +27d
= 6 + 27(15)
= 6 + 405
= 411
45th term will be a + 44d
= 6 + 44(15)
= 6 + 660
= 666
https://youtu.be/SyCsmjvOODo
EVALUATION:
(i) The sum of 15th term of an A.P is 862. Find the 20th terms if the common difference is 12.
(ii) The first and the last term of an A.P are 67 and 171 in that order. If there are 14 terms, find the 16th term of the A.P
ASSIGNMENT:
The 9th term of an A.P is 52 while the 12th term is 70. Find the sum of its 20 terms.
SPECIFIC TOPIC: CALCULATION INVOLVING SUM OF AN A.P
REFERENCE BOOK: New General Mathematics for senior school 1
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Solve problems on sum of an A.P
(ii) Find the nth term of an A.P
CONTENT: TEST ON SEQUENCE AND SERIES
(i) Find the 6th terms of the A.P. Whose first term is 6 and common difference is 7.
(ii) The 14th term of an A.P is 96 while the 25th term is 173. Find
(a) 19th term
(b) Sum of 13th and 56th terms.
(c) Product of 6th and 13th terms
https://youtu.be/0m8GKU-eiyQ
ASSIGNMENT;
Given that the first term of an A.P is 7 an its 10th term is twice the 2nd term, Calculate the (a) 19th term (b) sum of 28th terms
TOPIC: ARITHMETIC PROGRESSION (A. P).
CONTENTS
• Sequence
• Definition of Arithmetic Progression
• Denotations in Arithmetic progression
• Deriving formulae for the term of A. P.
• Sum of an arithmetic series
SEQUENCE
Examples:
Find the next two terms in each of the following sets of number and in each case state the rule which gives the term.
(a) 1, 5, 9, 13, 17, 21, 25 (any term +4 = next term)
(b) 2, 6, 18, 54, 162, 486, 1458 (any term x 3 = next term)
(c) 1, 9, 25, 49, 81, 121, 169, (sequence of consecutive odd no)
(d) 10, 9, 7, 4, 0, -5, -11, -18, -26, (starting from 10, subtract 1, 2, 3 from immediate no).
In each of the examples below, there is a rule which will give more terms in the list. A list like this is called a SEQUENCE in many cases; it can simply matter if a general term can be found for a sequence e.g.
1, 5, 9, 13, 17 can be expressed as
1, 5, 9, 13, 17 ……………. 4n – 3 where n = no of terms
Check: 5th term = 4(5) -3
20 – 3 = 17
10th term = 4(10) – 3
40 – 3 = 37
Example 2
Find the 6th and 9th terms of the sequence whose nth term is
(a) (2n + 1)
(b) 3 – 5n.
Solution
(a) 2n + 1
6th term = 2(6) + 1
=12 + 1 = 13
9th term = 2 (9) + 1
=18 + 1 = 19
(b) 3 – 5n
6th term = 3 – 5 (6)
=3 – 30 = -27
9th term = 3 – 5 (9)
=3 – 45 = -42
https://youtu.be/vlPHnGbKi9w
Evaluation
For each of the following sequence, find the next two terms and the rules which gives the term.
1. 1, , , , , ____, ____
2 100, 96, 92, 88, _____, ____
3. 2, 4, 6, 8, 10, ____, _____
4. 1, 4, 9, 16, 25, ____, _____
(i) Arrange the numbers in order of increasing give to a sequence.
(ii) Find the next two terms in the sequence
5. 19, 13, 16, 22, 10
6. -21/2, 51/2, 31/2, 11/2, -1/2
7. Find the 15th term of the sequence whose nth term is 3n - 5
4
Reading Assignment
New General mathematics SS II by G. B. Channon & Co. exercise 18a.
DEFINITION OF ARITHMETIC PROGRESSION
This is a sequence in which the terms either increase or decrease in equal steps is called an Arithmetic Progression.
The sequence 9, 12, 15, 18, 21, ____, _____, _____ has a first term of 9 and a common difference of +3 between the terms.
Denotations in A. P.
a = 1st term
d = common difference
n = no of terms
Un = nth term
Sn = Sum of the first n terms
Formula for nth term of Arithmetic Progression.
e.g. in the sequence 9, 12, 15, 18, 21.
a = 9
d = 12 – 9 or 18 – 15 = 3.
1st term = U1 = 9 = a
2nd term = U2 = 9 + 3 = a + d
3rd term = U3 = 9 + 3 + 3 = a + 2d
10th term = U10 = 9 + 9(3) = a + 9d
nth term = Un = 9+(n-1)3 = a + (n-1)d
nth term = Un = a + (n-1)d
Example:
Given the A.P, 9, 12, 15, 18 …… find the 50th term.
a = 9
d = 3
n = 50
Un = U50
Un = a + (n-1) d
U50 = 9 + (50-1) 3
9 + (49) 3
9 + 147
=156.
2. The 43rd term of an AP is 26, find the 1st term of the progression given that its common difference is ½ and also find the 50th term.
U43 = 26
d = ½
a = ?
n = 43
Un = a + (n-1) d
26 = a + (43-1) ½
26 = a + 42(1/2)
26 = a + 21
26 – 21 = a
5 = a
a = 5
(b) a = 5
d = ½
n = 50
U50 =?
Un = a + (n-1) d
U50 = 5 + (50-1)1/2
= 5 + 49(1/2)
U50 = 5 + 241/2
U50 = 291/2
https://youtu.be/YqP5TuwtEdg
Evaluation
1. Find the 37th term of the sequence 20, 10, 0, -10…
2. 1, 5… 69 are the 1st, 2nd, and last term of the sequence; find the common difference between them and the number of terms in the sequence.
Reading Assessment
New General Mathematics for SS 2, Exercise 18b, no 8, 10, 11, 12.
SUM OF AN ARITHMETIC SERIES
When the terms of a sequence are added, the resulting expression is called series e.g. in the sequence 1, 3, 5, 7, 9, 11.
Series = 1 + 3 + 5 + 7 + 9 + 11
When the terms of a sequence are unending the series is called infinite series, it is often impossible to find the sum of the terms in an infinite series.
e.g. 1 + 3 + 5 + 7 + 9 + 11 + …………………. Infinite
Sequence with last term or nth term is termed finite series.
e.g.
Find the sum of
1, 3, 5, 7, 9, 11, 13, 15
If sum = 2, n = 8
Then
S = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
Or S = 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1
Add eqn1 and eqn 2
2s = 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16
= 48 = 8(16)
2 2 = S = 64
Deriving the formula for sum of A. P. The following represent a general arithmetic series when the terms are added.
S = a + (a+d) + a + 2d + …………………………… + (L-2d) + (L-d) + L – eqn
S = L + (L-d) + L – 2d + ……………………………… a + 2d + (a+d) + a – eqn
2s = (a + L) + (a + L) + (a + L) + …………………… (a + L) + (a + L) + (a + L)
2s = n(a + L)
2
S = n(a+L)
2
L => Un = a + (n-1)d
Substitute L into eq**
S = n(a + a+(n-1)d
2
S = n(2a + (n-1)d = n ( 2a+ (n-1)d
2 2
S = n(a+L) when Un (last term L) is given
2
or
S = n(2a + (n-1)d when d is given or obtained
2
Example 2
Find the sum of the 20th term of the series 16 + 9 + 2 + …………………
a = 16
d = 9 – 16 = -7
n = 20
S = n(2a + (n-1)d)
2
S = 20 (2x16) + (20-1)(-7)
2
20 (32 + 19(-7)
2
S =10 (32 - 133) = 10(-101)
S = -1010
https://youtu.be/TtJ4dphwh0U
Evaluation
1. Find the sum of the arithmetic series with 16 and -117 as the first and 20th term respectively.
2. The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
Reading Assignment
New General Mathematics for SS 2, Page 181 – 184. Exercise 18d.
Evaluations (Exercise)
1. An A. P. has 15 terms and a common difference of -3, find its first and last term if its sum is 120.
2. On the 1st of January, a student puts N10 in a box, on the 2nd she puts N20 in the box, on the 3rd she puts N30 and so on putting on the same no. of N10 notes as the day of the month. How much will be in the box if she keeps doing this till 16th January.
Evaluation
The salary scale for a clerical officer starts at N55,200 per annum. A rise of N3,600 is given at the end of each year, find the total amount of money earned in 12 years.
ASSIGNMENT
Theory
1. Eight wooden poles are to be used for pillars and the length of the poles form an arc Arithmetic Progression (A. P.) if the second pole is 2m and the 6th pole is 5m, give the lengths of the poles in order and sum up the lengths of the poles.
2a. Write down the 15th term of the sequence.
2_, 3 , 4 , 5
1x3 2x4 3x5 4 x6
b. An arithmetic progression (A. P.) has 3 as its term and 4 as the common difference.
c. Write an expression in its simplest form for the nth term.
d. Find the least term of the A. P. that is greater than 100.
Objectives
1. Find the 4th term of an A. P. whose first term is 2 and the common difference is 0.5 (a) 4 (b) 4.5 (c) 3.5 (d) 2.5
2. In an A. P. the difference between the 8th and 4th term is 20 and the 8th term is 11/2 times the 4th term, find the common difference
(a) 5 (b) 7 (c) 3 (d) 10
3. Find the first term of the sequence in no. 2
(a) 70 (b) 45 (c) 25 (d) 5
4. The next term of the sequence 18, 12, 60 is
(a) 12 (b) 6 (c) -6 (d) -12
5. Find the no. of terms of the sequence 1/2 , ¾, 1, ……………….. 51/2
(a) 21 (b) 43/4 (c) 1 (d) 22
WEEK 5
TOPIC: GEOMETRIC PROGRESSION
CONTENT
• Definition of Geometric Progression
• Denotations of Geometric progression
• The nth term of a G. P.
• The sum of Geometric series
• Sum of G. P. to infinity
• Geometric mean
Definition of G. P.
The sequence 5, 10, 20, 40 has a first term of 5 and the common ratio
Between the term is 2 e.g. (10/5 or 40/2o = 2).
A sequence in which the terms either increase or decrease in a common table ratio is called a Geometric Progression (G. P).
A. P.
a, ar, ar2, ar3 ………………
Denotations in G. P.
a = 1st term
r = common ratio
Un = nth term
Sn = sum.
The nth term of a G. P.
The nth term = Un
Un = arn-1
1st term = a
2nd term = a x r =ar
3rd term = a x r x r = ar2
4th term = a x r x r x r = ar3
8th term = a x r x r x r x r x r x r x r = ar7
nth term = a x r x r x r x ……….. ar n-1
Example
Given the GP 5, 10, 20, 40. Find its (a) 9th term (b) nth term
a = 5
r = 10/5 = 2
U9 = ar n-1
U9 = 5 (2) a-1
= 5 (2)8
= 5 x 256 = 1,280
(b) Un = arn-1
= 5(2) n-1
Example 2
The 8th term of a G.P is -7/32. Find its common ratio if it first term is 28.
https://youtu.be/yo2DtZDYzKY
Evaluation
1. The 6th term of a G.P is 2000. Find its first term if its common ratio is 10.
2. The third term of a Geometric Progression G.P. is 24 and its seventh term is 4 20/27. Find its first term.
THE SUM OF A GEOMETRIC SERIES
a + ar + ar2 + ar3 + ………………. ar n-1
represent a general geometric series where the terms are added.
S = a + ar + ar2 ………… arn-1 - eqn 1
Multiply through r
rs = ar + ar2 + ar3 ………. arn ……… eqn 2
subtract eqn 2 from 1
Example:
Find the sum of the series.
a. ½ + ¼ + 1/8 + …………………… as far as 6th term
b. 1 + 3 + 9 + 27 + …………………. 729
Solution
https://youtu.be/eQ5mwPI05Y0
Evaluation
Find the sum of the series 40, -4, 0.4 as far as the 7th term.
SUM OF G. P. TO INFINITY
Sum of G. P to infinity is only possible where r is < 1.
Where r is > 1 there is no sum to infinity.
Example:
Find the sum of G. P. 1 + ½ + ¼ + ……………………
2. The third term of a Geometric Progression G.P. is 24 and its seventh term is
1. To 10 terms
To 100 terms.
Hence deduct the sum of the series (formula) if it has a very large no. of term or infinity.
Example 2:
Find the sum of the series 45 + 30 + 20 + ………………
https://youtu.be/FVpAJeDof8k
Evaluation
1. The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is -1/2
2. The 3rd and 6th term of a G. P. are 48 and 142/9 respectively, write down the first four terms of the G. P.
3. The sum of a G. P. is 100, find its fist term of the common ratio is 0.8.
4. If three numbers such as x , y and z are consecutive terms of a G.P then their common ratio will be
y = z
x y
y2 = xz
y = √xz
The middle value , y is the geometric mean (GM). We can conclude by saying that the GM of two numbers is the positive square root of their products.
Example
Calculate the geometric mean of
I. 3 and 27
II. 49 and 25
Example
The first three terms of a GP are k + 1, 2k – 1, 3k + 1. Find the possible values of the common ratio.
Solution
The terms are k + 1, 2k – 1, 3k + 1
2k -1 = 3k + 1
k + 1 2k – 1
(2k-1)(2k-1) = (k+1)(3k+1)
4k2-2k-2k +1 = 3k2 +k+3k + 1
4k2-4k +1 = 3k2 +4k + 1
4k2-3k2-4k-4k+1-1 = 0
k2 -8k = 0
k(k-8) = 0
k=0 or k-8 =0
k = 0 or 8
The common ratio will have two values due to the two values of k
When k=0 when k= 8
K+1 = 0+1 =1 k+1 = 8+1 = 9
2k- 1= 2x0 – 1 = -1 2k- 1 = 2x8 – 1 = 15
3k+ 1= 3x0+ 1 = 1 3k+1 = 3x8 +1 = 25
terms are 1 , -1 , 1 terms are 9,15,25
Common ratio, r = -1/1 common ratio, r = 15/9
r = -1 r = 5/3
https://youtu.be/eRRjDCHlDwg
Evaluation
1. p-6, 2p and 8p+20 are three consecutive terms of a GP. Determine the value of
a. p
b. the common ratio
2. If 1/16 , x , 1/4 , y , ….are in GP , find the product of x and y
Reading assignment: Essential mathematics page 175—180
ASSIGNMENT
Objectives
1. In the 2nd and 4th term of a G.P are 8 and 32 respectively, what is the sum of the first four terms. (a) 28 (b) 40 (c) 48 (d) 60
2. The sum of the first five term of the G.P. 2, 6, 18, is
(a) 484 (b) 243 (c) 242 (d) 130
3. The 4th term of a GP is -2/3 and its first term is 18 what is its common ratio. (a) ½ (b) 1/3 (c) -1/3 (d) -1/2
4. If the 2nd and 5th term of a G. P. are -6 and 48 respectively, find the sum of the first four terms: (a) -45 (b) -15 (c) 15 (d) 33
5. Find the first term of the G.P. if its common ratio and sum to infinity – 3/3 and respectively (a) 48 (b) 18 (c) 40 (d) -42
Theory:-
1. The 3rd term of a GP is 360 and the 6th term is 1215. Find the
a. Common ratio
b. First term
c. Sum of the first four terms
2. a. The first term of a G. P. is 48. Find the common ratio between its term if its sum to infinity is 36.
b. If (3- x) + (6) + (7- 5x) is a geometric series, find two possible values for
I. x II. The common ratio III. the sum of the G.P
Reading Assignment:
New General Mathematics for SS 2 page 179 – 180, 185 – 187.
CONTENT
• Definition of Geometric Progression
• Denotations of Geometric progression
• The nth term of a G. P.
• The sum of Geometric series
• Sum of G. P. to infinity
• Geometric mean
Definition of G. P.
The sequence 5, 10, 20, 40 has a first term of 5 and the common ratio
Between the term is 2 e.g. (10/5 or 40/2o = 2).
A sequence in which the terms either increase or decrease in a common table ratio is called a Geometric Progression (G. P).
A. P.
a, ar, ar2, ar3 ………………
Denotations in G. P.
a = 1st term
r = common ratio
Un = nth term
Sn = sum.
The nth term of a G. P.
The nth term = Un
Un = arn-1
1st term = a
2nd term = a x r =ar
3rd term = a x r x r = ar2
4th term = a x r x r x r = ar3
8th term = a x r x r x r x r x r x r x r = ar7
nth term = a x r x r x r x ……….. ar n-1
Example
Given the GP 5, 10, 20, 40. Find its (a) 9th term (b) nth term
a = 5
r = 10/5 = 2
U9 = ar n-1
U9 = 5 (2) a-1
= 5 (2)8
= 5 x 256 = 1,280
(b) Un = arn-1
= 5(2) n-1
Example 2
The 8th term of a G.P is -7/32. Find its common ratio if it first term is 28.
https://youtu.be/yo2DtZDYzKY
Evaluation
1. The 6th term of a G.P is 2000. Find its first term if its common ratio is 10.
2. The third term of a Geometric Progression G.P. is 24 and its seventh term is 4 20/27. Find its first term.
THE SUM OF A GEOMETRIC SERIES
a + ar + ar2 + ar3 + ………………. ar n-1
represent a general geometric series where the terms are added.
S = a + ar + ar2 ………… arn-1 - eqn 1
Multiply through r
rs = ar + ar2 + ar3 ………. arn ……… eqn 2
subtract eqn 2 from 1
Example:
Find the sum of the series.
a. ½ + ¼ + 1/8 + …………………… as far as 6th term
b. 1 + 3 + 9 + 27 + …………………. 729
Solution
https://youtu.be/eQ5mwPI05Y0
Evaluation
Find the sum of the series 40, -4, 0.4 as far as the 7th term.
SUM OF G. P. TO INFINITY
Sum of G. P to infinity is only possible where r is < 1.
Where r is > 1 there is no sum to infinity.
Example:
Find the sum of G. P. 1 + ½ + ¼ + ……………………
2. The third term of a Geometric Progression G.P. is 24 and its seventh term is
1. To 10 terms
To 100 terms.
Hence deduct the sum of the series (formula) if it has a very large no. of term or infinity.
Example 2:
Find the sum of the series 45 + 30 + 20 + ………………
https://youtu.be/FVpAJeDof8k
Evaluation
1. The sum to infinity of a Geometric Series is 100. Find the first term if the common ratio is -1/2
2. The 3rd and 6th term of a G. P. are 48 and 142/9 respectively, write down the first four terms of the G. P.
3. The sum of a G. P. is 100, find its fist term of the common ratio is 0.8.
4. If three numbers such as x , y and z are consecutive terms of a G.P then their common ratio will be
y = z
x y
y2 = xz
y = √xz
The middle value , y is the geometric mean (GM). We can conclude by saying that the GM of two numbers is the positive square root of their products.
Example
Calculate the geometric mean of
I. 3 and 27
II. 49 and 25
Example
The first three terms of a GP are k + 1, 2k – 1, 3k + 1. Find the possible values of the common ratio.
Solution
The terms are k + 1, 2k – 1, 3k + 1
2k -1 = 3k + 1
k + 1 2k – 1
(2k-1)(2k-1) = (k+1)(3k+1)
4k2-2k-2k +1 = 3k2 +k+3k + 1
4k2-4k +1 = 3k2 +4k + 1
4k2-3k2-4k-4k+1-1 = 0
k2 -8k = 0
k(k-8) = 0
k=0 or k-8 =0
k = 0 or 8
The common ratio will have two values due to the two values of k
When k=0 when k= 8
K+1 = 0+1 =1 k+1 = 8+1 = 9
2k- 1= 2x0 – 1 = -1 2k- 1 = 2x8 – 1 = 15
3k+ 1= 3x0+ 1 = 1 3k+1 = 3x8 +1 = 25
terms are 1 , -1 , 1 terms are 9,15,25
Common ratio, r = -1/1 common ratio, r = 15/9
r = -1 r = 5/3
https://youtu.be/eRRjDCHlDwg
Evaluation
1. p-6, 2p and 8p+20 are three consecutive terms of a GP. Determine the value of
a. p
b. the common ratio
2. If 1/16 , x , 1/4 , y , ….are in GP , find the product of x and y
Reading assignment: Essential mathematics page 175—180
ASSIGNMENT
Objectives
1. In the 2nd and 4th term of a G.P are 8 and 32 respectively, what is the sum of the first four terms. (a) 28 (b) 40 (c) 48 (d) 60
2. The sum of the first five term of the G.P. 2, 6, 18, is
(a) 484 (b) 243 (c) 242 (d) 130
3. The 4th term of a GP is -2/3 and its first term is 18 what is its common ratio. (a) ½ (b) 1/3 (c) -1/3 (d) -1/2
4. If the 2nd and 5th term of a G. P. are -6 and 48 respectively, find the sum of the first four terms: (a) -45 (b) -15 (c) 15 (d) 33
5. Find the first term of the G.P. if its common ratio and sum to infinity – 3/3 and respectively (a) 48 (b) 18 (c) 40 (d) -42
Theory:-
1. The 3rd term of a GP is 360 and the 6th term is 1215. Find the
a. Common ratio
b. First term
c. Sum of the first four terms
2. a. The first term of a G. P. is 48. Find the common ratio between its term if its sum to infinity is 36.
b. If (3- x) + (6) + (7- 5x) is a geometric series, find two possible values for
I. x II. The common ratio III. the sum of the G.P
Reading Assignment:
New General Mathematics for SS 2 page 179 – 180, 185 – 187.
WEEK 6
Simultaneous Linear Equations
Simultaneous equations are equations that are solved together to get an answer to solve two equations.
Methods of solving Simultaneous equation.
i. Elimination method
ii. Substitution method
iii. Graphical method
Example:
Solve for x and y
2x + 5y = 1
3x – 2y = 30
Solution
2x + 5y = 1 ………. eqn 1 (x (3)
3x – 2y = 30 ………… eqn 2 (x (2)
To eliminate x multiply eqn 1 by 3 and eqn 2 by 2
6x + 15 y = 3 ………. eqn 3
6x – 4y = 60 ……….. eqn 4
eqn 3 – eqn 4
19y/19 = -57 /10 3
y = -3
Substitute y = -3 into eqn 1
2x + 5 (-3) = 1
2x = 1 + 15
2x/2 = 16/2 = 8
x = 8
y = -3 and x = 8
https://youtu.be/d6vyYvx8URw
Substitution method.
2x + 5y = 1……………. eq 1
3x – 2y = 30 ………….. eq 2
Make x the subject in eqn 1
2x/2 = 1 – 5y/2
x = 1 – 5y /2 ………… eqn 3
Substitute eq3 into eqn 2
3 (1-5y) /2 - 2y = 30
3 – 15y - 4y/2 = 30
Multiple through by 2 or find the LCM and cross multiply.
3 – 15y – 4y = 60
3 – 19y = 60
-19y = 60 – 3
-10y = 57 3
-10 -10
y = -3
Substitute y = -3 into eq 3
x = 1 – 5y/2
x = 1 – 5 (-3) /2 = 1 + 15 /2 = 16/2
x = 8
x = 8, y = -3
Example:
Solve the following equations simultaneously
2/x - 1/y = 3
4/x + 3/y = 16
Solution
2/x - 1/y = 3
4 /x + 3/y = 16
Instead of using x and y as the unknown, let the unknown be (1/x) an (1/y).
2(1/x) - (1/y) = 3 ……………. eq1
4 (1/x) - 3 (1/y) = 16 …………… eq 2
Using elimination method, multiply equation 1 by 2 to eliminate x.
4(1/x) – 2(1/y) = 6 …………….. eq3
4 (1/x) + 3(1/y) = 16 ……………. eq4
-5 (1/y)/-5 = -10/-5
1/y = 2
y = ½
Substitute (1/y) = 2 into eqn 1
2 (1/x) – (1/y) = 3
2 (1/x) – (2) = 3
2(1/x) = 3 + 2
2 (1/x) = 5
1/x = 5/2
x = 2/5
y = ½, x = 2/5
https://youtu.be/ZSJ32Bq9sbQ
Evaluation
New General Mathematics for SSII
Exercise 7b no. 15 – 19.
FURTHER EXAMPLES
Example:
Solve for x and y simultaneously: 2x – 3y + 2 = x + 2y – 5 = 3x + y.
Solutions.
2x – 3y + 2 = x + 2y – 5 = 3x + y
Form two equations out of the question
2x – 3y + 2 = 3x + y
x + 2y – 5 = 3x + y
OR
2x – 3y + 2 = x + 2y – 5 ------------- eq 1
x + 2y – 5 = 3x + y -------------- eq 2
Rearrange the equations to put the unknown on one side and the constant at the other side.
2x – 3y – x – 2y = - 5 – 2
2x – x – 3y – 2y = -7
x – 5y = -7 ---------------- eq 3
From eq 2.
x – 3x + 2y – y – 5
- 2x + y = 5 ------------- eq 4
Using substitution method solve eq 3 & 4
x – 5y = -7 ---------------- eq 3
-2x + y = 5 --------------- eq 4
Make y the subject in eq 4.
y = 5 + 2x --------------- eq 5
Substitute eq 5 into eq 3.
x – 5 (5 + 2x) = -7
x – 25 – 10x = -7
-9x – 25 = -7
-9x = -7 + 25
-9x = 18
x = 18
-9
x= -2
Substitute x = -2 into eq 5
y = 5 + 2x
y = 5 + 2(-2)
y = 5 – 4
y = 1
x = -2, y = 1
Example
Solve the equations
By comparison
4x – 2 = 9x – 3y
4x – 9x + 3y = 2
- 5x + 3y =2 --------- eq 4
Solve equation 3 and 4 simultaneously
2x – y = 0 --------- eq 3
-5x + 3y =2 ---------- eq 4
Using elimination method: multiply equation 3 by 3
6x – 3y = 0 -------- eq 3
-5x + 3y = 2 ---------- eq 4
eq 3 + eq 4
x = 2
Substitute x = 2 into eq 3
2x – y = 0
2 (2) – y = 0
4 – y = 0
4 = 0+y
4 = y
x = 2, y=4 Word problem
https://youtu.be/WJbeyscO6Is
Example
.Seven cups and eight plates cost N1750, eight cups and seven plates cost N1700. Calculate the cost of a cup and a plate
solution
Let cups be x and plates be y
7x + 8y = 1750 -------------- eq 1
8x + 7y = 1700 -------------- eq 2
Multiply eq 1 by 8 and eq 2 by 7 to eliminate x (cups).
56x + 64y = 14000 ---------- eq 3
56x + 49y = 11900 ---------- eq 4
Subtracting eq 4 from eq 3
15y = 2100
y = 2100/15
Substitute y = 140 into eq 2
8x + 7y = 1700
8x + 7 (140) = 1700
8x + 980 = 1700
8x = 1700 – 980
x = 720/8
x = 90
Each cup cost N90 and each plate cost N140
https://youtu.be/zS6iXOZFkG4
Example
Find a two digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.
Solution
Let the two digit number be ab, where a is the tens digit and b is the unit digit
From the first statement,
2a + 3 = 3b
2a – 3b = -3 ………….eq1
From the second statement,
4(10a + b) – 99 = 10b + a
40a + 4b – 99 = 10b + a
40a – a + 4b – 10b = 99
39a – 6b = 99
Dividing through by 3
13a – 2b = 33 ………….eq2
Solving both equations simultaneously,
a = 3 , b = 3
Hence, the two digit number is 33
https://youtu.be/9tqlzouVQfs
Evaluation
Reading assignment : Essential Mathematics, Bk2 Chapter 9
New General Mathematics, Bk2
ASSIGNMENT
OBJECTIVE
1. If (x-y) log10 6 = log10 216 and 2 x+y =32 , calculate the values of x and y
a. x=1 , y=4 b. x= 4 , y =1 c. x=-4 , y= 1 d. x=4, y= -1
2. The point of intersection of the lines 3x- 2y =-12 and x + 2y = 4 is …
a. (5, 0) b. (3, 4) c. (-2, 5) d. (-2, 3)
3. Find the value of (x - y) , if 2x + 2y =16 and 8x – 2y = 44
a. 2 b. 4 c. 5 d. 6
4. If 5 (p +2q) =5 and 4 (p+3q) =16 , the value of 3 (p+q) is …..
a.0 b. -1 c.2 d. 1
5. Given 4x – 3y = 11 , evaluate y2 – 3x
7x – 4y 23 3
a-2 b. 3 c. -3 d. 2
THEORY
1. Given that 2 1- x/y = 1/32, find x in terms of y, and hence solve the simultaneous
equation 2x + 3y – 30 = 0 ; 2 1- x/y = 1/32 .(WAEC)
2. A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the
original number. (WAEC)
Simultaneous equations are equations that are solved together to get an answer to solve two equations.
Methods of solving Simultaneous equation.
i. Elimination method
ii. Substitution method
iii. Graphical method
Example:
Solve for x and y
2x + 5y = 1
3x – 2y = 30
Solution
2x + 5y = 1 ………. eqn 1 (x (3)
3x – 2y = 30 ………… eqn 2 (x (2)
To eliminate x multiply eqn 1 by 3 and eqn 2 by 2
6x + 15 y = 3 ………. eqn 3
6x – 4y = 60 ……….. eqn 4
eqn 3 – eqn 4
19y/19 = -57 /10 3
y = -3
Substitute y = -3 into eqn 1
2x + 5 (-3) = 1
2x = 1 + 15
2x/2 = 16/2 = 8
x = 8
y = -3 and x = 8
https://youtu.be/d6vyYvx8URw
Substitution method.
2x + 5y = 1……………. eq 1
3x – 2y = 30 ………….. eq 2
Make x the subject in eqn 1
2x/2 = 1 – 5y/2
x = 1 – 5y /2 ………… eqn 3
Substitute eq3 into eqn 2
3 (1-5y) /2 - 2y = 30
3 – 15y - 4y/2 = 30
Multiple through by 2 or find the LCM and cross multiply.
3 – 15y – 4y = 60
3 – 19y = 60
-19y = 60 – 3
-10y = 57 3
-10 -10
y = -3
Substitute y = -3 into eq 3
x = 1 – 5y/2
x = 1 – 5 (-3) /2 = 1 + 15 /2 = 16/2
x = 8
x = 8, y = -3
Example:
Solve the following equations simultaneously
2/x - 1/y = 3
4/x + 3/y = 16
Solution
2/x - 1/y = 3
4 /x + 3/y = 16
Instead of using x and y as the unknown, let the unknown be (1/x) an (1/y).
2(1/x) - (1/y) = 3 ……………. eq1
4 (1/x) - 3 (1/y) = 16 …………… eq 2
Using elimination method, multiply equation 1 by 2 to eliminate x.
4(1/x) – 2(1/y) = 6 …………….. eq3
4 (1/x) + 3(1/y) = 16 ……………. eq4
-5 (1/y)/-5 = -10/-5
1/y = 2
y = ½
Substitute (1/y) = 2 into eqn 1
2 (1/x) – (1/y) = 3
2 (1/x) – (2) = 3
2(1/x) = 3 + 2
2 (1/x) = 5
1/x = 5/2
x = 2/5
y = ½, x = 2/5
https://youtu.be/ZSJ32Bq9sbQ
Evaluation
New General Mathematics for SSII
Exercise 7b no. 15 – 19.
FURTHER EXAMPLES
Example:
Solve for x and y simultaneously: 2x – 3y + 2 = x + 2y – 5 = 3x + y.
Solutions.
2x – 3y + 2 = x + 2y – 5 = 3x + y
Form two equations out of the question
2x – 3y + 2 = 3x + y
x + 2y – 5 = 3x + y
OR
2x – 3y + 2 = x + 2y – 5 ------------- eq 1
x + 2y – 5 = 3x + y -------------- eq 2
Rearrange the equations to put the unknown on one side and the constant at the other side.
2x – 3y – x – 2y = - 5 – 2
2x – x – 3y – 2y = -7
x – 5y = -7 ---------------- eq 3
From eq 2.
x – 3x + 2y – y – 5
- 2x + y = 5 ------------- eq 4
Using substitution method solve eq 3 & 4
x – 5y = -7 ---------------- eq 3
-2x + y = 5 --------------- eq 4
Make y the subject in eq 4.
y = 5 + 2x --------------- eq 5
Substitute eq 5 into eq 3.
x – 5 (5 + 2x) = -7
x – 25 – 10x = -7
-9x – 25 = -7
-9x = -7 + 25
-9x = 18
x = 18
-9
x= -2
Substitute x = -2 into eq 5
y = 5 + 2x
y = 5 + 2(-2)
y = 5 – 4
y = 1
x = -2, y = 1
Example
Solve the equations
By comparison
4x – 2 = 9x – 3y
4x – 9x + 3y = 2
- 5x + 3y =2 --------- eq 4
Solve equation 3 and 4 simultaneously
2x – y = 0 --------- eq 3
-5x + 3y =2 ---------- eq 4
Using elimination method: multiply equation 3 by 3
6x – 3y = 0 -------- eq 3
-5x + 3y = 2 ---------- eq 4
eq 3 + eq 4
x = 2
Substitute x = 2 into eq 3
2x – y = 0
2 (2) – y = 0
4 – y = 0
4 = 0+y
4 = y
x = 2, y=4 Word problem
https://youtu.be/WJbeyscO6Is
Example
.Seven cups and eight plates cost N1750, eight cups and seven plates cost N1700. Calculate the cost of a cup and a plate
solution
Let cups be x and plates be y
7x + 8y = 1750 -------------- eq 1
8x + 7y = 1700 -------------- eq 2
Multiply eq 1 by 8 and eq 2 by 7 to eliminate x (cups).
56x + 64y = 14000 ---------- eq 3
56x + 49y = 11900 ---------- eq 4
Subtracting eq 4 from eq 3
15y = 2100
y = 2100/15
Substitute y = 140 into eq 2
8x + 7y = 1700
8x + 7 (140) = 1700
8x + 980 = 1700
8x = 1700 – 980
x = 720/8
x = 90
Each cup cost N90 and each plate cost N140
https://youtu.be/zS6iXOZFkG4
Example
Find a two digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.
Solution
Let the two digit number be ab, where a is the tens digit and b is the unit digit
From the first statement,
2a + 3 = 3b
2a – 3b = -3 ………….eq1
From the second statement,
4(10a + b) – 99 = 10b + a
40a + 4b – 99 = 10b + a
40a – a + 4b – 10b = 99
39a – 6b = 99
Dividing through by 3
13a – 2b = 33 ………….eq2
Solving both equations simultaneously,
a = 3 , b = 3
Hence, the two digit number is 33
https://youtu.be/9tqlzouVQfs
Evaluation
Reading assignment : Essential Mathematics, Bk2 Chapter 9
New General Mathematics, Bk2
ASSIGNMENT
OBJECTIVE
1. If (x-y) log10 6 = log10 216 and 2 x+y =32 , calculate the values of x and y
a. x=1 , y=4 b. x= 4 , y =1 c. x=-4 , y= 1 d. x=4, y= -1
2. The point of intersection of the lines 3x- 2y =-12 and x + 2y = 4 is …
a. (5, 0) b. (3, 4) c. (-2, 5) d. (-2, 3)
3. Find the value of (x - y) , if 2x + 2y =16 and 8x – 2y = 44
a. 2 b. 4 c. 5 d. 6
4. If 5 (p +2q) =5 and 4 (p+3q) =16 , the value of 3 (p+q) is …..
a.0 b. -1 c.2 d. 1
5. Given 4x – 3y = 11 , evaluate y2 – 3x
7x – 4y 23 3
a-2 b. 3 c. -3 d. 2
THEORY
1. Given that 2 1- x/y = 1/32, find x in terms of y, and hence solve the simultaneous
equation 2x + 3y – 30 = 0 ; 2 1- x/y = 1/32 .(WAEC)
2. A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the
original number. (WAEC)
WEEK 7
TOPIC: SIMULTANEOUS EQUATION (One linear, One quadratic)
Examples.
Solve simultaneously for x and y (i.e. the points of their intersection)
3x + y = 10
2x2 +y2 = 19
Note: One linear, One quadratic is only possible analytically using substitution method.
Solution.
3x + y = 10 ----------- eq 1
2x2 + y2 = 19 --------- eq 2
Make y the subject in eq 1 (linear equation)
y = 10 – 3x ---------- eq 3
Substitute eq 3 into eq 2
2x2 + (10-3x) 2 = 19
2x2 + (10 – 3x) (10 – 3x) = 19
2x2 + 100 – 30x – 30x + 9x2 = 19
2x2 + 9x2 - 30x – 30x + 100 – 19 = 0
11x2 - 60x + 81 = 0
11x2 - 33x – 27x + 81= 0
11x (x-3) – 27 (x – 3) = 0
(11x – 27) (x – 3) = 0
11x – 27 = 0 or x-3 = 0
11x = 27 or x = 3
x = 27/11 or 3
Substitute the values of x into eq 3.
When x = 3
y = 10 – 3(x)
y = 10 - 3(3)
y = 10 – 9 = 1
When x =27/11
y = 10 – 3(27/11)
y = 10 - 51/11
y = 110 - 51/11
y = 59/11
w hen x = 3, y = 1
x = 27/11 , y = 59/11
https://youtu.be/ZdD9Rde3B08
EXAMPLE
Solve the equations simultaneously
3x + 4y = 11
xy = 2
solution
3x + 4y = 11 -------- eq 1
x y = 2 -------- eq 2
Make y the subject in eq 1
4y = 11 – 3x
y = 11 – 3x / 4 ………… eq3
substitute eq 3 into eq 2
x y = 2
x ( 11- 3x/4 ) = 2
x (11-3x) = 2x4
11x – 3x2 = 8
-3x2 + 11x – 8 = 0
-3x2 + 3x + 8x – 8 = 0
-3x (x-1) +8 (x-1) = 0
(-3x + 8) (x-1) = 0
-3x + 8 = 0 or x – 1 = 0
3x = 8 or x = 1
x = 8/3 or 1
Substitute the values of x into eq 3
y = 11- 3x
4
when x = 1
y = 11 – 3(1) = 11-3 = 8
4 4 2
y = 4
when x = 8/3
y = 11 – 3(8/3)
4
y = 33 – 24 = 9 = 3
12 12 4
x = 1, y = 2
x = 8/3, y = 3/4.
https://youtu.be/p15_GD_Eks0
Evaluation
Solve for x and y
I. 3x 2 - 4y = -1 II. 4x2 + 9y2 = 20
2x – y = 1 2x – 9y =-2
Reading Assignment New General Maths for SS II Ex 7d, 1 b, e, 2 b, c.
Simultaneous Equation - (More Examples)
Solve simultaneously for x and y.
3x – y = 3 -------- eq 1
9x2 - y 2 = 45 --------- eq 2
Solution
From eq 2
(3x)2 - y 2 = 45
(3x-y) (3x+y) = 45 ---------- eq 3
Substitute eq 1 into eq 3
3 (3x + y) = 45
3x + y = 15 ……………..eq4
Solve eq 1 and eq 4 simultaneously.
3x – y = 3 --------- eq 1
3x + y = 15 -------- eq 4
eq 1 + eq 4
6x = 18
x = 18/ 6
x = 3
Substitute x = 3 into eq 4.
3x + y = 15
3 (3) + y = 15
9 + y = 15
y = 15 – 9
y = 6
x = 3, y = 6
https://youtu.be/3jS9dQoJcjI
Evaluation
1. a. Given that :
4x2 – y2 = 15 b. Given that : 3x2 +5xy –y2 =3
2x – y = 5 x - y = 4
Solve for x and y. solve for x and y.
WORD PROBLEM
Example
The product of two numbers is 12. The sum of the larger number and twice the smaller number is 11. Find the two numbers.
Solution
Let x = the larger number
y = the smaller number
Product, x y = 12 …………….eq1
From the last statement,
x + 2y = 11 ………….. eq2
From eq2, x = 11 – 2y …………...eq3
Sub. Into eq1
y(11 – 2y) = 12
11y – 2y2 = 12
2y2 -11y + 12 = 0
2y2 – 8y – 3y + 12 = 0
2y(y-4) – 3(y-4) = 0
(2y-3)(y-4) =0
2y-3 =0 or y-4 =0
2y = 3 or y = 4
y= 3/2 or 4
when y = 3/2 when y=4
x = 11 – 2y x = 11- 2y
x = 11 – 2(3/2) x = 11 – 2(4)
x = 11 – 3 x = 11 – 8
x = 8 x = 3
Therefore, (8 , 3/2)(3 , 4)
https://youtu.be/EjfLPMJGU70
https://youtu.be/9tqlzouVQfs
https://youtu.be/MynlGmwhm78
Assignment
Objective
In each of the following pairs of equations solve simultaneously,
1. xy = -12 ; x – y = 7 a. (3 , -4)(4 ,-3)
b. (-2 ,4)(-3, -4)
c. (-4, 5)(-2 , 3)
d(3 ,-3)(4,-4)
2. x – 5y = 5 ; x2 – 25y2 = 55 a (-8, 0)(3/5 , 0) b(0, 0)(-8 , 3/5) c (8 , 3/5) d (0, 8)(0, 3/5)
3. y = x2 and y = x + 6
(a).(0,6) (3,9)
(b)(-3,0) (2,4)
(c)(-2,4) (3,9) (d).(-2, 3), (-3,2)
4. x – y = -3/2 ; 4x2 + 2xy – y2 = 11/4 :
a(-1, 1/2)(1, 5/2).
b.(3, 2/5)(1, 1/2)
c.(3/2 , -1)(4,2)
d.(-1 , -1/2)(-1 , 5/2)
5. m2 + n2 = 25 ; 2m + n – 5 = 0 :
a. (0,5)(4, -3)
b.(5,0)(-3,4)
c.(4,0)(-3,5)
d(-5,3)(0,4)
Theory
1 a. Find the coordinate of the points where the line 2x – y = 5 meets the curve 3x2 – xy -4 =10
b. Solve the simultaneous equations: x + 2y = 5, 7(x2 +1) = y(x + 3y)
2. A woman is q years old while her son is p years old. The sum of their ages is equal to twice the difference of their ages.
The product of their ages is 675.
Write down the equations connecting their ages and solve the equations in order to find the ages of the woman and her son. (WAEC)
Reading Assignment
1. New General Mathematics for SS 2 By J B Channon & Co page 73 – 78
2. Past SSCE Questions
3. Exam Focus (Mathematics).
Examples.
Solve simultaneously for x and y (i.e. the points of their intersection)
3x + y = 10
2x2 +y2 = 19
Note: One linear, One quadratic is only possible analytically using substitution method.
Solution.
3x + y = 10 ----------- eq 1
2x2 + y2 = 19 --------- eq 2
Make y the subject in eq 1 (linear equation)
y = 10 – 3x ---------- eq 3
Substitute eq 3 into eq 2
2x2 + (10-3x) 2 = 19
2x2 + (10 – 3x) (10 – 3x) = 19
2x2 + 100 – 30x – 30x + 9x2 = 19
2x2 + 9x2 - 30x – 30x + 100 – 19 = 0
11x2 - 60x + 81 = 0
11x2 - 33x – 27x + 81= 0
11x (x-3) – 27 (x – 3) = 0
(11x – 27) (x – 3) = 0
11x – 27 = 0 or x-3 = 0
11x = 27 or x = 3
x = 27/11 or 3
Substitute the values of x into eq 3.
When x = 3
y = 10 – 3(x)
y = 10 - 3(3)
y = 10 – 9 = 1
When x =27/11
y = 10 – 3(27/11)
y = 10 - 51/11
y = 110 - 51/11
y = 59/11
w hen x = 3, y = 1
x = 27/11 , y = 59/11
https://youtu.be/ZdD9Rde3B08
EXAMPLE
Solve the equations simultaneously
3x + 4y = 11
xy = 2
solution
3x + 4y = 11 -------- eq 1
x y = 2 -------- eq 2
Make y the subject in eq 1
4y = 11 – 3x
y = 11 – 3x / 4 ………… eq3
substitute eq 3 into eq 2
x y = 2
x ( 11- 3x/4 ) = 2
x (11-3x) = 2x4
11x – 3x2 = 8
-3x2 + 11x – 8 = 0
-3x2 + 3x + 8x – 8 = 0
-3x (x-1) +8 (x-1) = 0
(-3x + 8) (x-1) = 0
-3x + 8 = 0 or x – 1 = 0
3x = 8 or x = 1
x = 8/3 or 1
Substitute the values of x into eq 3
y = 11- 3x
4
when x = 1
y = 11 – 3(1) = 11-3 = 8
4 4 2
y = 4
when x = 8/3
y = 11 – 3(8/3)
4
y = 33 – 24 = 9 = 3
12 12 4
x = 1, y = 2
x = 8/3, y = 3/4.
https://youtu.be/p15_GD_Eks0
Evaluation
Solve for x and y
I. 3x 2 - 4y = -1 II. 4x2 + 9y2 = 20
2x – y = 1 2x – 9y =-2
Reading Assignment New General Maths for SS II Ex 7d, 1 b, e, 2 b, c.
Simultaneous Equation - (More Examples)
Solve simultaneously for x and y.
3x – y = 3 -------- eq 1
9x2 - y 2 = 45 --------- eq 2
Solution
From eq 2
(3x)2 - y 2 = 45
(3x-y) (3x+y) = 45 ---------- eq 3
Substitute eq 1 into eq 3
3 (3x + y) = 45
3x + y = 15 ……………..eq4
Solve eq 1 and eq 4 simultaneously.
3x – y = 3 --------- eq 1
3x + y = 15 -------- eq 4
eq 1 + eq 4
6x = 18
x = 18/ 6
x = 3
Substitute x = 3 into eq 4.
3x + y = 15
3 (3) + y = 15
9 + y = 15
y = 15 – 9
y = 6
x = 3, y = 6
https://youtu.be/3jS9dQoJcjI
Evaluation
1. a. Given that :
4x2 – y2 = 15 b. Given that : 3x2 +5xy –y2 =3
2x – y = 5 x - y = 4
Solve for x and y. solve for x and y.
WORD PROBLEM
Example
The product of two numbers is 12. The sum of the larger number and twice the smaller number is 11. Find the two numbers.
Solution
Let x = the larger number
y = the smaller number
Product, x y = 12 …………….eq1
From the last statement,
x + 2y = 11 ………….. eq2
From eq2, x = 11 – 2y …………...eq3
Sub. Into eq1
y(11 – 2y) = 12
11y – 2y2 = 12
2y2 -11y + 12 = 0
2y2 – 8y – 3y + 12 = 0
2y(y-4) – 3(y-4) = 0
(2y-3)(y-4) =0
2y-3 =0 or y-4 =0
2y = 3 or y = 4
y= 3/2 or 4
when y = 3/2 when y=4
x = 11 – 2y x = 11- 2y
x = 11 – 2(3/2) x = 11 – 2(4)
x = 11 – 3 x = 11 – 8
x = 8 x = 3
Therefore, (8 , 3/2)(3 , 4)
https://youtu.be/EjfLPMJGU70
https://youtu.be/9tqlzouVQfs
https://youtu.be/MynlGmwhm78
Assignment
Objective
In each of the following pairs of equations solve simultaneously,
1. xy = -12 ; x – y = 7 a. (3 , -4)(4 ,-3)
b. (-2 ,4)(-3, -4)
c. (-4, 5)(-2 , 3)
d(3 ,-3)(4,-4)
2. x – 5y = 5 ; x2 – 25y2 = 55 a (-8, 0)(3/5 , 0) b(0, 0)(-8 , 3/5) c (8 , 3/5) d (0, 8)(0, 3/5)
3. y = x2 and y = x + 6
(a).(0,6) (3,9)
(b)(-3,0) (2,4)
(c)(-2,4) (3,9) (d).(-2, 3), (-3,2)
4. x – y = -3/2 ; 4x2 + 2xy – y2 = 11/4 :
a(-1, 1/2)(1, 5/2).
b.(3, 2/5)(1, 1/2)
c.(3/2 , -1)(4,2)
d.(-1 , -1/2)(-1 , 5/2)
5. m2 + n2 = 25 ; 2m + n – 5 = 0 :
a. (0,5)(4, -3)
b.(5,0)(-3,4)
c.(4,0)(-3,5)
d(-5,3)(0,4)
Theory
1 a. Find the coordinate of the points where the line 2x – y = 5 meets the curve 3x2 – xy -4 =10
b. Solve the simultaneous equations: x + 2y = 5, 7(x2 +1) = y(x + 3y)
2. A woman is q years old while her son is p years old. The sum of their ages is equal to twice the difference of their ages.
The product of their ages is 675.
Write down the equations connecting their ages and solve the equations in order to find the ages of the woman and her son. (WAEC)
Reading Assignment
1. New General Mathematics for SS 2 By J B Channon & Co page 73 – 78
2. Past SSCE Questions
3. Exam Focus (Mathematics).
WEEK 8
SPECIFIC TOPIC: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
OBJECTIVE: At the end of the lesson, the students should be able to: Solve problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
TOPIC: GRAPHICAL SOLUTION OF SIMULTANEOUS EQUATION: ONE LINEAR, ONE QUADRATIC AND GRADIENT OF CURVE AT A GIVEN POINT.
CONTENT:
• Drawing of simultaneous graph, (one linear, one quadratic)
• Solving problems on graph
• Gradient of a curve at a given point
TOPIC: Straight line graph
SUBTOPIC: Sketching graphs of straight lines (b) Gradient of a straight line
REFERENCES BOOK
a. NGM for WA SS2 by JB Channon et al
b. MAN mathematics for SS2
c. Effective mathematics for SS2 by popoola et al
BEHAVIOURAL OBJECTIVES: At the end of the lesson, the students should be able to: Sketch graphs
SUBTOPIC: Gradient of a straight line
BEHAVIOURAL OBJECTIVES: At the end of the end of lesson, students should be able to:
a. Locate points on Cartesian graph
b. Find the gradient of a straight line
https://youtu.be/s7NKLWXkEEE
https://youtu.be/9Uc62CuQjc4
CONTENT: Find the gradients of the line joining the following pairs of points.
a. (1, 4) (-1, 5)
b. (2, 3) (6, -5)
c. (-4, -4) (-1, 5)
d. (3, 2) (-3, -5)
https://youtu.be/ScZIIAENy50
EVALUATION:
a. Locate the following point on the graph and
b. Find the gradients of the line joining pair of points
(i) (9, 7) (2, 5)
(ii) (-3, 2) (4, 4)
(iii) (-4, -4) (-1, 5)
(iv) (5, 3) (0, 0)
SUBTOPIC: Sketching graphs of straight line
BEHAVIOURAL OBJECTIVES: At the end of the lesson, students should be able to: (i) draw the graph of straight lines (ii) find the gradient of the lines
CONTENT
a. Sketch the graph of the line whose equation is
(i) 2x – y = -1 (ii) y = 3x -2
b. Write down the gradients of the lines represented by:
(i) Y = 2x + 3 (ii) 3x + 7y = 5
https://youtu.be/edrwgW-lcMY
EVALUATION:
Draw the graph of y = x + 6
(i) find the gradient of the lines whose equations are 5x – 2y = 6 and y = 3x + 5
ASSIGNMENT
a. Draw the graph
b. Find the gradient of the lines whose equations are 4y+ 3x + 15 = 0 and 4x – 5=12y
Drawing of simultaneous graph (one linear one quadratic)
Example:
Using the scale 2cm to 1 units on x-axis and 2cm to 2 unit on y-axis, draw the graph of y=x2 – x – 1 and y=2x – 1 (on the same scale and axis for values of x: - 3< 4)
Solution
Table of values for y=x2 – x – 1
x -3 -2 -1 0 1 2 3 4
x2 9 4 1 0 1 4 9 16
-x +3 +2 +1 0 -1 -2 -3 -4
-1 -1 -1 -1 -1 -1 -1 -1 -1
y 11 5 +1 -1 -1 1 5 11
x -3 -2 -1 0 1 2 3 4
y 11 5 1 -1 -1 1 5 11
https://youtu.be/cMa02EEAFC8
https://youtu.be/NPzlCNDEJqA
Table of values for y = 2x – 1
x -3 -2 -1 0 1 2
2x -6 -4 -2 0 2 4
-1 -1 -1 -1 -1 -1 -1
y -7 -5 -3 -1 1 3
y = 2x – 1
x -3 -2 -1 0 1 2 3
y -7 -5 -3 -1 1 3 5
https://youtu.be/m2syMt1tWN0
Evaluation
a. Copy and complete the table for the Fn y = 2x2 – 3x – 7
x| = |-2| -1|0 |1 |2|3 |4 |5 |
y| = |--| -1 |--|--|--|--|--|-- |
b. Using a scale of 2cm to 1 unit on x-axis and 2cm to 5 unit on y-axis, draw the graph of the relation y=2x2-3x-7 for -3 < 5
c. Using the same scale and axis, draw the graph of y=2x-1
SOLUTIONS ON SIMULTANEOUS GRAPH (ONE LINEAR ONE QUADRATIC) & GRADIENT OF CURVE AT A GIVEN POINT.
1. Using the example (graph) of last period of y=x2 – x – 1 and y = 2x - 1
Estimate from your graph
a. x2 – 3x = 0
b. x2 – x – 3 = 0
c. The minimum value of y from the graph of y = x2 – x – 1
d. What is the gradient of the graph y = 2x – 1
e. What is the gradient of the graph y = x2 – x – 1 at x = - 2
Solution
(a) At the point of intersections
x2-x -1 = 2x -1
x2 – 3x = 0
The solution of x2 -3x = 0 are x = 0 or 3
(b) x2 – x – 3 =0
x2 – x – 3 + 2 = 0 + 2
x2 – x – 1 = 2
i.e: The value of x when x2 – x – 1 = 2
i.e when y = 2 on the curve
= when y = 2, x = -1.3 or 2.3
c. The minimum value of y = -1.3 (1dp)
d. The gradient of y = 2x – 1
e. The gradient of the curve at x = -2
i. Draw a tangent on the curve at x = - 2
ii. Pick two points on that tangent
iii. Trace each point to x and y axis and obtain the values y2, y1, x2, x1
iv. Find the gradient (i.e the gradient of the drawn tangent)
y2 – y1/x2 – x1 =
https://youtu.be/JAe5-CB-TCY
Evaluation
Using the graph of the previous evaluation.
Estimate from the graph y = 2x2-3x-7 and y=2x-1 correct to 1 d.p
(i) The minimum value of y in the graph 2x2 -3x-7
(ii) The roots of the equation 2x2-3x-7=0
(iii) Solution of the simultaneous equations y= 2x2-3x-7 and y=2x-1
(iv) Solution of the equation 2x2-3x-7=2x-1
(v) Values of X in 2x2-3x-7 =5
(vi) Roots of the equation 2x2-3x-4=0
(vii) The gradient of the curve y = 2x2-3x-1 at x= 2
(viii) The gradient of the linear graph y=2x-1
EXERCISE
(a) Draw on the same axes the graph of the functions y = x2 -3x -3 and y = 2x -4 for -2 ≤ x ≤ 5
(b) From your graphs
(i) Solve the equation x2-5x + 1 = 0
(ii) Find the minimum value of the function y = x2-3x-3
(iii) Find the roots of the equation x2-3x-3x-3=0
Hint choose any convenient scale.
Solutions
Table of values
Y= x2 -3x -3
X -2 -1 0 1 2 3 4 5
X2 4 1 0 1 4 9 16 25
-3x 6 3 0 -3 -6 -9 -12 -15
y 7 1 -3 -5 -5 -3 1 7
X -2 -1 0 1 2 3 4 5
Y 7 1 -3 -5 -5 -3 1 7
y = 2x – 4
X -2 -1 0 1 2 3 4 5
2X -4 -2 0 2 4 6 8 10
-4 -4 -4 -4 -4 -4 -4 -4 -4
Y -8 -6 -4 -2 0 2 4 6
X -2 -1 0 1 2 3 4 5
Y -8 -6 -4 -2 0 2 4 6
(a) The equation : x2 – 5x + 1 = 0 is x2 – 3x – 3 = 2x – 4 (the point of intersection of the two graph).
= x = 4.8 or 0.2
(b) The minimum value of y = x2 – 3x – 3 is -5.3
(c) The roots of the equation x2 – 3x – 3 = 0 is x = -0.8 or 3.8.
https://youtu.be/K7POmMh47RA
ASSIGNMENT
Theory
1. Solve the simultaneous equation analytically
2 + x + x2 = y
1 – x = y
2a. Copy and complete the table of value for the relation y = 2 + x – x2.
x -2 -1 0 1 2 3
y 0 -4
(b) Using a scale of 2m to 1 unit on x-axis and 4cm to 1 unit on y axis, draw the graph of the relation y = 2 + x – x2 for the interval – 2 < x < 3.
(c) From your graph, find the maximum value of y and the corresponding value of x for which this occurs.
(d) Using the same scale and axis, draw the graph of y = 1 – x
(e) Use your graphs to solve the equation 1 + 2x – x2 = 0
(f) Find the gradient of the curve at x = 1.5
Objective
Use the sketch below to answer the question.
OBJECTIVE: At the end of the lesson, the students should be able to: Solve problems using graphical solution
CONTENT: GRAPHICAL SOLUTION TO SIMULTANEOUS EQUATIONS, ONE LINEAR AND ONE QUADRATIC.
TOPIC: GRAPHICAL SOLUTION OF SIMULTANEOUS EQUATION: ONE LINEAR, ONE QUADRATIC AND GRADIENT OF CURVE AT A GIVEN POINT.
CONTENT:
• Drawing of simultaneous graph, (one linear, one quadratic)
• Solving problems on graph
• Gradient of a curve at a given point
TOPIC: Straight line graph
SUBTOPIC: Sketching graphs of straight lines (b) Gradient of a straight line
REFERENCES BOOK
a. NGM for WA SS2 by JB Channon et al
b. MAN mathematics for SS2
c. Effective mathematics for SS2 by popoola et al
BEHAVIOURAL OBJECTIVES: At the end of the lesson, the students should be able to: Sketch graphs
SUBTOPIC: Gradient of a straight line
BEHAVIOURAL OBJECTIVES: At the end of the end of lesson, students should be able to:
a. Locate points on Cartesian graph
b. Find the gradient of a straight line
https://youtu.be/s7NKLWXkEEE
https://youtu.be/9Uc62CuQjc4
CONTENT: Find the gradients of the line joining the following pairs of points.
a. (1, 4) (-1, 5)
b. (2, 3) (6, -5)
c. (-4, -4) (-1, 5)
d. (3, 2) (-3, -5)
https://youtu.be/ScZIIAENy50
EVALUATION:
a. Locate the following point on the graph and
b. Find the gradients of the line joining pair of points
(i) (9, 7) (2, 5)
(ii) (-3, 2) (4, 4)
(iii) (-4, -4) (-1, 5)
(iv) (5, 3) (0, 0)
SUBTOPIC: Sketching graphs of straight line
BEHAVIOURAL OBJECTIVES: At the end of the lesson, students should be able to: (i) draw the graph of straight lines (ii) find the gradient of the lines
CONTENT
a. Sketch the graph of the line whose equation is
(i) 2x – y = -1 (ii) y = 3x -2
b. Write down the gradients of the lines represented by:
(i) Y = 2x + 3 (ii) 3x + 7y = 5
https://youtu.be/edrwgW-lcMY
EVALUATION:
Draw the graph of y = x + 6
(i) find the gradient of the lines whose equations are 5x – 2y = 6 and y = 3x + 5
ASSIGNMENT
a. Draw the graph
b. Find the gradient of the lines whose equations are 4y+ 3x + 15 = 0 and 4x – 5=12y
Drawing of simultaneous graph (one linear one quadratic)
Example:
Using the scale 2cm to 1 units on x-axis and 2cm to 2 unit on y-axis, draw the graph of y=x2 – x – 1 and y=2x – 1 (on the same scale and axis for values of x: - 3< 4)
Solution
Table of values for y=x2 – x – 1
x -3 -2 -1 0 1 2 3 4
x2 9 4 1 0 1 4 9 16
-x +3 +2 +1 0 -1 -2 -3 -4
-1 -1 -1 -1 -1 -1 -1 -1 -1
y 11 5 +1 -1 -1 1 5 11
x -3 -2 -1 0 1 2 3 4
y 11 5 1 -1 -1 1 5 11
https://youtu.be/cMa02EEAFC8
https://youtu.be/NPzlCNDEJqA
Table of values for y = 2x – 1
x -3 -2 -1 0 1 2
2x -6 -4 -2 0 2 4
-1 -1 -1 -1 -1 -1 -1
y -7 -5 -3 -1 1 3
y = 2x – 1
x -3 -2 -1 0 1 2 3
y -7 -5 -3 -1 1 3 5
https://youtu.be/m2syMt1tWN0
Evaluation
a. Copy and complete the table for the Fn y = 2x2 – 3x – 7
x| = |-2| -1|0 |1 |2|3 |4 |5 |
y| = |--| -1 |--|--|--|--|--|-- |
b. Using a scale of 2cm to 1 unit on x-axis and 2cm to 5 unit on y-axis, draw the graph of the relation y=2x2-3x-7 for -3 < 5
c. Using the same scale and axis, draw the graph of y=2x-1
SOLUTIONS ON SIMULTANEOUS GRAPH (ONE LINEAR ONE QUADRATIC) & GRADIENT OF CURVE AT A GIVEN POINT.
1. Using the example (graph) of last period of y=x2 – x – 1 and y = 2x - 1
Estimate from your graph
a. x2 – 3x = 0
b. x2 – x – 3 = 0
c. The minimum value of y from the graph of y = x2 – x – 1
d. What is the gradient of the graph y = 2x – 1
e. What is the gradient of the graph y = x2 – x – 1 at x = - 2
Solution
(a) At the point of intersections
x2-x -1 = 2x -1
x2 – 3x = 0
The solution of x2 -3x = 0 are x = 0 or 3
(b) x2 – x – 3 =0
x2 – x – 3 + 2 = 0 + 2
x2 – x – 1 = 2
i.e: The value of x when x2 – x – 1 = 2
i.e when y = 2 on the curve
= when y = 2, x = -1.3 or 2.3
c. The minimum value of y = -1.3 (1dp)
d. The gradient of y = 2x – 1
e. The gradient of the curve at x = -2
i. Draw a tangent on the curve at x = - 2
ii. Pick two points on that tangent
iii. Trace each point to x and y axis and obtain the values y2, y1, x2, x1
iv. Find the gradient (i.e the gradient of the drawn tangent)
y2 – y1/x2 – x1 =
https://youtu.be/JAe5-CB-TCY
Evaluation
Using the graph of the previous evaluation.
Estimate from the graph y = 2x2-3x-7 and y=2x-1 correct to 1 d.p
(i) The minimum value of y in the graph 2x2 -3x-7
(ii) The roots of the equation 2x2-3x-7=0
(iii) Solution of the simultaneous equations y= 2x2-3x-7 and y=2x-1
(iv) Solution of the equation 2x2-3x-7=2x-1
(v) Values of X in 2x2-3x-7 =5
(vi) Roots of the equation 2x2-3x-4=0
(vii) The gradient of the curve y = 2x2-3x-1 at x= 2
(viii) The gradient of the linear graph y=2x-1
EXERCISE
(a) Draw on the same axes the graph of the functions y = x2 -3x -3 and y = 2x -4 for -2 ≤ x ≤ 5
(b) From your graphs
(i) Solve the equation x2-5x + 1 = 0
(ii) Find the minimum value of the function y = x2-3x-3
(iii) Find the roots of the equation x2-3x-3x-3=0
Hint choose any convenient scale.
Solutions
Table of values
Y= x2 -3x -3
X -2 -1 0 1 2 3 4 5
X2 4 1 0 1 4 9 16 25
-3x 6 3 0 -3 -6 -9 -12 -15
y 7 1 -3 -5 -5 -3 1 7
X -2 -1 0 1 2 3 4 5
Y 7 1 -3 -5 -5 -3 1 7
y = 2x – 4
X -2 -1 0 1 2 3 4 5
2X -4 -2 0 2 4 6 8 10
-4 -4 -4 -4 -4 -4 -4 -4 -4
Y -8 -6 -4 -2 0 2 4 6
X -2 -1 0 1 2 3 4 5
Y -8 -6 -4 -2 0 2 4 6
(a) The equation : x2 – 5x + 1 = 0 is x2 – 3x – 3 = 2x – 4 (the point of intersection of the two graph).
= x = 4.8 or 0.2
(b) The minimum value of y = x2 – 3x – 3 is -5.3
(c) The roots of the equation x2 – 3x – 3 = 0 is x = -0.8 or 3.8.
https://youtu.be/K7POmMh47RA
ASSIGNMENT
Theory
1. Solve the simultaneous equation analytically
2 + x + x2 = y
1 – x = y
2a. Copy and complete the table of value for the relation y = 2 + x – x2.
x -2 -1 0 1 2 3
y 0 -4
(b) Using a scale of 2m to 1 unit on x-axis and 4cm to 1 unit on y axis, draw the graph of the relation y = 2 + x – x2 for the interval – 2 < x < 3.
(c) From your graph, find the maximum value of y and the corresponding value of x for which this occurs.
(d) Using the same scale and axis, draw the graph of y = 1 – x
(e) Use your graphs to solve the equation 1 + 2x – x2 = 0
(f) Find the gradient of the curve at x = 1.5
Objective
Use the sketch below to answer the question.
WEEK 9
TOPIC: ADDITION SUBTRACTION AND MULTIPLICATION OF SURD
CONTENTS:
1. Definition of surd
2. Like surds
3. Simplification of surd
4. Addition and subtraction of surd
5. multiplication of surds
DEFINITION OF SURDS
A number which can be expressed as a quotient m/n of two integers, (n≠ 0) is called a rational number. Any real number which is not rational is called IRRATIONAL. Irrational numbers which are in the form of roots are called SURDS.
Examples.
https://youtu.be/WPbA5peq6FQ
SIMPLIFICATION OF SURD
Illustration
By putting m = 16 and n = 9, find which of the following pairs of expression are equal.
Solutions to question 1, 3, and 6 are equal.
Note:- The illustration demonstrates the fact that
To simplify surd where the square root sign as a product of the factor one of which should be a perfect square, then simplify the surd by taking the square root of the perfect square.
https://youtu.be/yiW2rJeupx4
Example 1
EVALUATION
ADDITION AND SUBTRACTION OF SURD
Two or more surds can be added together or subtracted from one another if they are like surds. Before addition or subtraction, the surds should first be simplified if possible.
https://youtu.be/oNA0kOdpb3M
Examples
https://youtu.be/dOw4h2pc19A
EVALUATION:
Simplify the following
MULTIPLICATION OF SURDS
When two or more surds are multiplied together, they should first be simplified if possible, then multiply whole number with whole number and surd with surd.
Examples:
https://youtu.be/tC5kYXcFhsg
EVALUATION
https://youtu.be/0mAQjd2eUrE
GENERAL EVALUATION
READING ASSIGNMENT:
1. New General Mathematics for SS 2 3rd Edition by J B Channon & Co page 138 - 140
2. Exam Focus (mathematics) by S.A. Omeni page 19 – 20
3. SSCE past Questions (mathematics)
EVALUATION
REFERENCE BOOKS:
1. New General Mathematics for SS 2 by J. B. Channon and co 3rd Edition..
2. Distinction in mathematics. A comprehensive revision text by A. A.
Adelodun 3rd Edition
3. Exam Focus (Mathematics) for WASSCE & SSCE by S. A, lori, I. U. Jahun and B. A. Omoni
4. Intermediate Pure Mathematics S I unit by J Blakey 5th Edition
5. Further Mathematics Project I revised Edition by Tuttuh – Adegun, M.R., Sivasubramanian,
S. & Adegoke, R.
6. Additional mathematics
7. S. S.C.E mathematics questions.
DIVISION OF SURD (RATIONALIZATION)
1. Division of Surd (Rationalization)
2. Multiplication of Surd involving and conjugate of binomial surd.
3. Expression with Surds
DIVISION OF SURD (RATIONALIZATION)
Examples
Note: When a fraction has Surd as the denomination, the Surd is rationalized.
To rationalize the denominator of a rational surd, means to rationalize denominator remove the square root sign from the denominator.
Two types are considered
https://youtu.be/IKrt-P1rxdY
https://youtu.be/Zq5sbARFEvA
Multiplication of Surd involving brackets and conjugate of binominal Surds
When multiplying surds in brackets use the ordinary algebraic method i.e.
(a + b) (c + d) = ac + ad + bc + bd
Examples
BINOMIAL DENOMINATOR:
Conjugate of a binominal surd
https://youtu.be/WAw7JpZWaxE
Example:
Simplify
https://youtu.be/aU9B6X7uak0
EVALUATION
Simplify
EVALUATION OF EXPRESSION WITH SURDS:
When evaluating a fraction containing a surd, it is advisable to rationalize its denominator.
https://youtu.be/yKsshxT4tq8
Example:
https://youtu.be/ApPCgZWxzro
READING ASSIGNMENT
1. New General mathematics for SS2 by J.B. Channon & Co. 3rd Edition Page 140 - 142
2. Exam Focus (Mathematics) by S. A. Ilori, I.U. Jahun and B.A. Omeni Page 19 - 20
ASSIGNMENT
Reading Assignment:
1. New General mathematics for SS 2 - by J. B. Channon & Co.
page 78 – 80.
2. Exam focus (Mathematics) By Ilori & Co.
CONTENTS:
1. Definition of surd
2. Like surds
3. Simplification of surd
4. Addition and subtraction of surd
5. multiplication of surds
DEFINITION OF SURDS
A number which can be expressed as a quotient m/n of two integers, (n≠ 0) is called a rational number. Any real number which is not rational is called IRRATIONAL. Irrational numbers which are in the form of roots are called SURDS.
Examples.
https://youtu.be/WPbA5peq6FQ
SIMPLIFICATION OF SURD
Illustration
By putting m = 16 and n = 9, find which of the following pairs of expression are equal.
Solutions to question 1, 3, and 6 are equal.
Note:- The illustration demonstrates the fact that
To simplify surd where the square root sign as a product of the factor one of which should be a perfect square, then simplify the surd by taking the square root of the perfect square.
https://youtu.be/yiW2rJeupx4
Example 1
EVALUATION
ADDITION AND SUBTRACTION OF SURD
Two or more surds can be added together or subtracted from one another if they are like surds. Before addition or subtraction, the surds should first be simplified if possible.
https://youtu.be/oNA0kOdpb3M
Examples
https://youtu.be/dOw4h2pc19A
EVALUATION:
Simplify the following
MULTIPLICATION OF SURDS
When two or more surds are multiplied together, they should first be simplified if possible, then multiply whole number with whole number and surd with surd.
Examples:
https://youtu.be/tC5kYXcFhsg
EVALUATION
https://youtu.be/0mAQjd2eUrE
GENERAL EVALUATION
READING ASSIGNMENT:
1. New General Mathematics for SS 2 3rd Edition by J B Channon & Co page 138 - 140
2. Exam Focus (mathematics) by S.A. Omeni page 19 – 20
3. SSCE past Questions (mathematics)
EVALUATION
REFERENCE BOOKS:
1. New General Mathematics for SS 2 by J. B. Channon and co 3rd Edition..
2. Distinction in mathematics. A comprehensive revision text by A. A.
Adelodun 3rd Edition
3. Exam Focus (Mathematics) for WASSCE & SSCE by S. A, lori, I. U. Jahun and B. A. Omoni
4. Intermediate Pure Mathematics S I unit by J Blakey 5th Edition
5. Further Mathematics Project I revised Edition by Tuttuh – Adegun, M.R., Sivasubramanian,
S. & Adegoke, R.
6. Additional mathematics
7. S. S.C.E mathematics questions.
DIVISION OF SURD (RATIONALIZATION)
1. Division of Surd (Rationalization)
2. Multiplication of Surd involving and conjugate of binomial surd.
3. Expression with Surds
DIVISION OF SURD (RATIONALIZATION)
Examples
Note: When a fraction has Surd as the denomination, the Surd is rationalized.
To rationalize the denominator of a rational surd, means to rationalize denominator remove the square root sign from the denominator.
Two types are considered
https://youtu.be/IKrt-P1rxdY
https://youtu.be/Zq5sbARFEvA
Multiplication of Surd involving brackets and conjugate of binominal Surds
When multiplying surds in brackets use the ordinary algebraic method i.e.
(a + b) (c + d) = ac + ad + bc + bd
Examples
BINOMIAL DENOMINATOR:
Conjugate of a binominal surd
https://youtu.be/WAw7JpZWaxE
Example:
Simplify
https://youtu.be/aU9B6X7uak0
EVALUATION
Simplify
EVALUATION OF EXPRESSION WITH SURDS:
When evaluating a fraction containing a surd, it is advisable to rationalize its denominator.
https://youtu.be/yKsshxT4tq8
Example:
https://youtu.be/ApPCgZWxzro
READING ASSIGNMENT
1. New General mathematics for SS2 by J.B. Channon & Co. 3rd Edition Page 140 - 142
2. Exam Focus (Mathematics) by S. A. Ilori, I.U. Jahun and B.A. Omeni Page 19 - 20
ASSIGNMENT
Reading Assignment:
1. New General mathematics for SS 2 - by J. B. Channon & Co.
page 78 – 80.
2. Exam focus (Mathematics) By Ilori & Co.