TOPIC: SIMPLE EQUATIONS AND VARIATIONS
CONTENT:
Change of subject of formulae
Subject of formula and substitution
Types of variations – direct and inverse
Joint and partial
Application of variation
Simple equation is an algebraic equation involving one variable and the power or index of the variable is one.
CHANGE OF SUBJECT OF FORMULAE
A literal equation is a simple equation that involves more than one variable (unknown). The procedure of solving such an equation is usually to find one of the variables (unknowns) in terms of the other(s).
Examples: (1) Given A=1/2 h(a+b), make a the subject of the formula
Solution: A=1/2 h(a+b)
Cross-multiply
2A=h(a+b)
2A=ah+bh
2A-bh=ah
Also, ah=2A-bh
a=(2A-bh)/h
a=2A/h-bh/h
∴a=2A/h-b
https://youtu.be/cbKc_qilgzA
(2) Make v the subject of the formula;
H=(m(v^2-u^2))/2gx
Solution: H=(m(v^2-u^2))/2gx
Cross-multiply
2gxH= m(v^2-u^2)
2gxH= mv^2-mu^2
2gxH+mu^2=mv^2
(2gxH+mu^2)/m=v^2
Take square root of both sides;
v=±√((2gxH+mu^2)/m) or ±√(2gxH/m+u^2 )
https://youtu.be/TTa9fqmrCjM
(3) The period of a compound pendulum is given by T=2π√(((h^2+k^2)/gh) ) express k in terms of T,h and g, taking π^2 as 10
Solution: T=2π√(((h^2+k^2)/gh) )
Take the square of both side
T^2=(2π)^2 ((h^2+k^2)/gh)
T^2=〖4π〗^2 ((h^2+k^2)/gh)
Cross-multiply
T^2 gh=〖4π〗^2 (h^2+k^2 )
Divided both sides by 〖4π〗^2
(T^2 gh)/〖4π〗^2 =h^2+k^2
k^2=(T^2 gh)/〖4π〗^2 -h^2
Take square root of both sides
k=√((T^2 gh)/〖4π〗^2 -h^2 )
Using the value of 10 for π^2,
∴k=√((T^2 gh)/40-h^2 )
SUBJECT OF FORMULA AND SUBSTITUTION
A formula is an equation in which letters represent quantities. The value of one variable in a formula or algebraic equation may be found by substituting (i.e replacing) known values in the same formula
Examples: (1) The sum of the squares of the first n integers is given by s_n=(n(n+1)(2n+1))/6
Calculate (a) s_20 (b) the sum of the squares from 21 to 40 inclusive.
Solution: (a) s_20 means the values of s_n when n=20
s_20=(20(20+1)(2×20+1))/6
s_20=(20(21)(41))/6
=2870
(b) Sum of squares from 21 to 40 means = s_40-s_20
But, s_20 is already known to be 2870,
s_40=(40(40+1)(2×40+1))/6
=(40(41)(81))/6
=22140
∴s_40-s_20=22140-2870
=19270
(2) Find the value of 2π√(l/g) , when π=3 1/7 ,l=98 and g=32
Solution: 2π√(l/g) = 2×22/7 √(98/32)
= 44/7 √(49/16)
= 44/7×7/4
= 11
https://youtu.be/t8Z9SUFcAR4
EVALUATION:
(1) Make the letters that appear as the subject of the formula
I=nE/(R+nr) , (n,R)
T=E/√(R^2-W^2 L^2 ) , (R,L)
V=πh^2 (r+h/3) , (r)
(2) (a) If v=kl(D^2-d^2), find d when v=1250,D=1000,k=5 and l=4
(b) Simplify p(q-2p)-(p+q)(q-2p) find the value of the expression when p=-2 and q=1
(c) Evaluate √((s^2-(b+c)s+bc)/bc) when b=25,c=7 and s=28
Assignment: NGM, page 82, Ex 6f nos 15,31 and 35; page 83, Ex 6g nos 10,13 and 16
TOPIC :
VARIATIONS
Variation is a connection of sets of numerical values by an equation which indicates some kind of proportionality. We have four major types of variations which are direct, inverse, joint and partial variations.
DIRECT VARIATIONS:
Considering two quantities x and y. If when y increases, x also increases and when y decreases, x also decreases in a constant proportion, then x and y are said to be in direct variation. i.e. x varies directly as y or x is directly proportional to y: Written as xy x = ky where k is the constant of proportionality or variation.
Other examples are as follows:
(i) A varies directly as the square of B
i.e. A=B2
* A = KB2 (where K is the constant of variation)
(ii) P varies directly as the square root of q.
i.e. P-q
* P = K-q (where k is the constant of variation).
Example 1: (A) (i) if P varies directly with q and q = 2, p = 10. Find p when q is 5.
Solution: p∝q
p=kq
10=2k
k=5
Formula connecting p and q is p = 5q,
Then, p = 5 x 5 = 25
(ii) if x-3 is directly proportional to the square of y and x=5 when y=2,find x when y=6
Solution; x-3∝y2
x-3=ky2
5-3=k×22
2=4k
k=1/2
Formula connecting x-3 and y is x-3=1/2 y2.
Then, x-3=1/2×6×6
x-3=18
∴x=21
https://youtu.be/gSOW6zyM-58
Example 2:
The area of a box varies directly as the square of its length. When the area is 175cm2, the length is 5cm. Find the;
Equation connecting the quantities.
Area when the length is 8cm
Length when the Area is 100cm2
Solution:
Let A represent Area
L represent Length
A = L2
A = KL2 –- (1) (Where k is Constant)
If A = 175 and L = 5 then
175 * k(5)2
k = 175/25
k = 7
Substitute for k in equation (1) to get the law of variation.
A = 7L2
A = 7L2 is the equation connecting the quantities.
(ii) To find A when L = 8
put values in the law of variation A = 7L2 to have
A = 7(8)2
A =7 * 64
A = 448cm2.
(iii) To find L when A = 100cm2
put values in the law of variation A = 7L2 to have
100 = 7L2
100 = L2
7
L = 100
7
L = 10 or 3.78cm
7
https://youtu.be/wdoBR9JUxv8
EVALUATION
x is directly proportional to y. If x=5 when y=3,find y when x=2/7
The wages of a labourer varies directly as the number of hours worked by the labourer. The labourer earned N500 when he worked for 2 hours. Find
(i) The amount he would earn if he works for 7 hours.
(ii) The number of hours he would work if he is paid N800.
y varies directly as the square of x. If y = 98 when x = 7, calculate y when x = 5.
[WAEC]
If D∝S∝and D=140 when S=35,find
The relationship between D and S
The value of S when D = 176
INVERSE VARIATIONS.
Considering two quantities x and y. If when y increases, x decreases and when x increases y decreases in a constant ratio, then we say x and y varies inversely.
It is normally written as x*1/y or y*1/x.
Note that if xµ1/y, then x = k/y where K is the constant of variation or proportionality.
Other examples are as follows:
(i) A varies inversely as the cube root of B
i.e. A µ ¬1
3*B
A = K x 1
3*B (where k is constant)
(ii) P varies inversely as the square of q
I.e. P µ 1
q2
P = K 1
q2 where k is constant
Example
(i) t∝1/d and t=0.15 when d=120
(a) find t when d = 45
(b) find d when t = 0.12
Solution: t∝1/d
t=k/d
0.15×120=k
k=18
Formula connecting t and d, t=18/d
t=18/45
=0.4
t=18/d
0.12d=18
d=18/0.12
∴d=150
Example
If P is inversely proportional to the square root of q, when P = 3, q = 25. Find
(a) P when q = 49
(b) q when p = 13
Solution:
P µ 1
q
P = k
q -------------------(1) (where k is constant)
When P = 3, q = 25
3 = k
25
3 x 5 = k
k = 15
Substitute in (1) to get the law of variation
P = 15
q
(a) To find p when q = 49
Substitute in the law of variation to have
P = 15
49
p = 15
7
p = 21/7
(b) To find q when p = 13, substitute in the law of variation to have
13 = 15
p
13*p = 15
*p = 15
13
p = 15 2
13
p = 225
169
p = 156/169
https://youtu.be/awp2vxqd-l4
https://youtu.be/blubR7QeUaA
EVALUATION:
if x varies inversely as the square of y and x=8 when y=4,find y when x=32
(2) If y varies inversely as x and y = 6 when x = 2. Find
(i) The law of variation
(ii) x when y = 10
(iii) y when x = 7
(3)If y varies inversely as the square of x and y = 8 when x = 3. Find
(i) The relationship between x and y.
(ii) x when y = 5
(iii) y when x = 9
(4) If P varies inversely as the square root of q and p = 12 when q = 4. Find
(i) p when q = 25
(ii) q when p = 8.
(5)Exercise 18c, page 217 of New General Mathematics for senior Secondary School, Book 1.
JOINT VARIATION
This type of variation involves three or more quantities joined together with a combination of two direct variations or one direct and one inverse or two inverse variations.
Examples:
(i) x varies directly as y and jointly as z
I.e. x*yz
x = kyz (where k is constant)
(ii) x varies directly as y and inversely as the square of z.
i.e. x*y
z2
* x = ky (where k is constant)
z2
(iii) P varies inversely as q and inversely as the square root of r.
I.e. p * 1
q*r
* p = k (where k is constant)
q*r
Example :
If x varies directly as y and inversely as the square of z. When y = 5 and z = 3, x = 20. Find (a) z when x = 213/5 and y = 15
(b) x when y = 6 and z = 4.
Solution:
X * y
z2
x = ky -------- (1) (where k is constant)
z2
when y = 5 and z = 3, x = 20
20 = k5
32
20 x 9 = k5
k = 20 x 9
5
k = 4 x 9
* k = 36.
Substitute in (1) above to get the law of variation.
X = 36y
z2
(a) To find z when x = 213/5 and y = 15
108 = 36 x 15
5 z2
108z2 = 36 x 15 x 5
z2 = 36 x 15 x 5
108
z = *25
* z = 5.
(b) To find x when y = 6 and z = 4.
x = 36 x 6
42
x = 36 x 6
16
*x = 13.5
Examples; (1) x varies directly as the product of u and v and inversely as their sum,if x=3 when u=3 and v=1,what is the value of x if u=3 and v=3?
Solution: x∝uv/(u+v)
x=kuv/(u+v)
3= (k×3×1)/(3+1)
3×4=3k
k=4
Law connecting x,u and v is x=4uv/(u+v)
Then, x=(4×3×3)/(3+3)
∴x=6
(2) A∝BC,when B=4 and C=9,A=6
(a) find the formula that connects A,B & C
(b) find A when B = 3 and C = 10
(c) find C if A = 20 and B = 15
Solution: A∝BC
A=kBC
6=k×4×9
k=1/6
A=1/6 BC
A=1/6 BC
A=1/6×3×10
A = 5
A=1/6 BC
20=1/6×15×C ∴C=8
https://youtu.be/73-H7OyNLUM
https://youtu.be/lbVJWrqC6Gw
EVALUATION
(2) A varies directly as B and inversely as C. When B = 3, C = 5, A = 40. Find
(i) The law of variation
(ii) A when B = 8 and C = 12.
(3) xµy. when y = 36 and z = 16, x = 81.
Öz
Find (i) The law of variation
(ii) y when x = 56 and z = 25
(iii) x when y = 27 and z = 9
(4) U varies directly as V and inversely as the square of W. When V = 3 and W = 4, U = 24. Find (i) U when V = 5 and W = 8
(ii) W when U = 30 and V = 8
(5) The electrical resistance of a copper wire varies directly as its length and inversely as the square of its radius. If a copper wire 500 meters long and radius 0.2cm has a resistance of 30 ohms, calculate the resistance of the same type of copper wire 750 meters long and radius 0.25cm.
(WAEC).
(6) If PµQÖR. When R = 16 and Q = 3, P = 48. Find (i) The law of variation
(ii) P when R = 25 and Q = 7
(iii) R when P = 36 and Q = 9
PARTIAL VARIATION
This consist of two or more parts or quantities added together. Both parts may be made of variables or one part may be constant, while the other can either vary directly or inversely. The values of the constants of variation are usually found out by solving simultaneous equations.
Examples:
(i) x is partly constant, and partly varies as y
x = a+by (where a and b are constant)
(ii) p varies partly as q and partly as r.
p = aq + br (where a and b are constants)
(iii) v varies partly as u and partly as the reciprocal of w2.
Examples: (1) P is partly constant and partly varies as Q, when Q is 5, P is 20 and when Q is 8, P is 26. Find P when Q is 4.
Solution: P = c + kQ
Note: c & k are constants which must be obtained through simultaneous equation.
20=c+5k ………(i)
-26=c+8k……….(ii)
-6 = -3k
K = 2
substituting the value of k in equation (i)
20=c+5(2)
20-10=c
c=10
Thus, P = 10 + 2Q (formula connecting P & Q)
Then, P = 10 + 2(4)
P = 10 + 8
∴ P = 18
Example :
X is partly constant and partly varies as y. when y = 7, x = 15; and when y = 5, x = 7. Find
(a) The law of variation
(b) x when y = 2
(c) y when x = 11
Solution
(a) x = a + by --- (1) where a and b are constants of variation.
When y = 7, x = 15 and when y = 5, x = 7
15 = a + 7y ----------------- (2)
7 = a + 5y ----------------- (3)
Solve equation (2) and (3) simultaneously
Eqn (2): 15 = a + 7b _
Eqn (3): 7 = a + 5b
8 = 2b
8 = b/2
b = 4
Put in eqn (2) to have
15 = a + 7 x 4
15 = a + 28
a = 15 – 28
a = -13
Substitute a = -13 and b = 4 in eqn (1) to get the law of variation.
x = -13 + 4y -------------(law of variation)
(b) To find x when y = 2, put in the law of variation
x = -13 + 4 x 2
x = -13 + 8
x = -5
(c) To find y when x = 11, put in the law of variation
11 = -13 + 4y
11 + 13 = 4y
24 = 4y
y = 24/4
y = 6.
https://youtu.be/zeKZ0daCE58
EVALUATION:
M is partly constant and partly varies with N, when N = 40, M = 150 and when N = 54, M = 192
(a) Find the formula connecting M and N, (b) Hence find M when N = 73
Two quantities P and Q are connected by a linear relation of the form P = aQ + b, where a & b are constants. If Q = 80 when P = 12 and Q = 300 when P = 50, find the equation connecting P and Q
The cost of producing a wooden frame varies directly as the width of the frame and partly as the square root of its length. When the width is 10cm and the length is 25cm, the cost is N115.00 and when the width is 18cm and the length is 36cm, the cost is N240. Find the
(a) Law of variation
(b) Cost of a frame of width 12cm and the length 49cm.
(2) The cost of producing a textbook is partly constant and partly varies as the number of books produced. It cost N4000 to produce 20 books and N6000 to produce 70 books. Find the
(a) Cost of producing 120 books
(b) Number of books produced at N10, 000.
(3) The cost of sinking a well varies partly as the depth of the well and partly as the number of laborers used for the job. If it cost N1500 to sink a well of 6 metres deep with 2 laborers and N2500 to sink a well of 9 metres deep with 5 laborers. Find the
(a) Law of variation
(b) Cost of sinking a well 15 metres deep with 6 laborers
(c) Number of laborers that would sink a well of 12 metres deep at N3000.
APPLICATION OF VARIATION
Examples; (1) A particle moves in such a way that its displacement S metres, at time t seconds is given by the relation S = at^2, where ‘a’ is constant. Calculate ‘a’ if S = 32 when t = 4
Solution; S = at2
32 = a x 42
32/16=a
a = 2
(2) The cost of feeding a number of students of Deeper Life High School in a regional excursion is partly constant and partly varies directly as the number of students. The cost of feeding 75 students during the excursion is $875 and the cost of feeding 100 students during the same period of time is $1000. Find the cost of feeding 220 students over the same period of time.
Solution: The equation is given as; C = a + kN where C is the cost and N is the number of students
875 = a + 75k ……….(i)
1000 = a + 100k ………. (ii)
Solving both equations simultaneously, we obtain k = 5 & a = 500.
Formula connecting C and N implies; C = 500 + 5N
Then, C = 500 + 5(220)
C = 500 + 1100
∴ C = $1600
GENERAL EVALUATION:
The electrical resistance R of a wire varies inversely as the square of the radius r. use a constant k to show the law between R and r
Given that the energy E, varies directly as the resistance R and inversely as the square if the distance d, (i) obtain an equation connecting E, R and d given that E = 32/25 when R = 8 and d = 5. (ii) calculate (a) the value of R when E = 16 and d = 3 (b) the value of d when R = 5 and d = 5/6 (c) find the percentage increase in the value of R when each of E and d increases by 3%
READING ASSIGNMENT:
NGM for SSS book 1, pages 80-84, 213-219.
Mathematical Association of Nigeria (MAN) pages 63- 77
ASSIGNMENT:
Make W the subject of the formula; R-d=√(R^2-W^2 ) . Given that R = 1.25 and d = 0.25. calculate W
Given that T=2π√((l^2+k^2)/2gl)
Make k the subject of the formula
Find the value of k when l=1 and T^2=〖4π〗^2 g
The resistance R to the motion of a car is partly constant and partly proportional to the square of the speed v. when the speed is 30km/h, the resistance is 190
O and when the speed is 50km/h, the resistance is 350
O. find for what speed the resistance is 302.5
O
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