Lesson Notes By Weeks and Term - Senior Secondary School 1

SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 1st TERM

REFERENCE BOOKS

• New General Mathematics SSS 1     M.F. Macrae et al 

 
WEEK TEN

TOPIC: SIMPLE EQUATION AND VARIATIONS

CONTENT

• Change of subject of formulae

• Types of variation such as: direct, inverse, joint and partial

• Application of variations

EQUATIONS 

An equation is a statement of two algebraic expressions which are equal in value. For example, 4 – 4x = 9 – 12x is a linear equation with an unknown x. this equation is only true when x has a particular numerical value. To solve an equation means to find the real number value of the unknown that makes the equation true. 

Example 1 

Solve 4 – 4x = 9 – 12x 

4 – 4x = 9 – 12x 

Add 12x to both sides of the equation. 

4 – 4x + 12x = 9 – 12x + 12x 

4 + 8x = 9 

Subtract 4 from both sides of the equation. 

4 + 8x – 4 = 9 – 4

8x = 5 

Divided both sides of the equation by 8 

8x8 = 5

x = 58

58is the solution or root of the equation. 

Check: when x = 58,

LHS = 4 – 4 x 58 = 4 - 212 = 112

RHS = 9 – 12 x 58 = 9 - 712 = 112 = LHS 

The equation in Example 1 was solved by the balance method. Compare the equation with a pair of sales. If the expressions on opposite sides of the equals sign ‘balance’, they will continue to do so if the same amounts are added to or subtracted from both sides. They will also balance if both sides are multiplied or divided by the same amounts. 

Example 2

Solve 3(4c - 7) – 4(4c - 1) = 0

Remove brackets. 

12c – 21 – 16c + 4 = 0

Collect like terms. 

    -4c – 17  = 0

Add 17 to both sides 

    -4c = 17

Divided both sides by -4.

    C = -174

    C = -414,

Check: When c = -414,

LHS = 3(-17 - 7) – 4(-17 - 1)

    = 3(-24) – 4(-18)

    -72 + 72 = 0 = RHS 

EVALUATION

Solve the following equations

1. 2 – 5t = 20 – 8t 
2. d = 12 – (11 + 4d)
3. 213d = 28 

Change of Subject of Formulae

If is often necessary to change the subject of a formula. To do this, think of the formula as an equation. Solve the equation for the letter which is to become the subject. The following examples show how various formulae can be rearranged to change the subject. 

Example 1 

Make x the subject of the formula a = b(a - x)

Clear brackets.

a = b –bx

rearrange to give terms in x on one side of the equation. 

bx = b – a 

divide both sides by b. 

x = b - a b

Example 2

make x the subject of the formula a = b  +  xb -  x

a = b  +  xb -  x

clear fractions. Multiply both sides by (b – x)

ab – ax = b + x 

collect terms in x on one side of the equation. 

ab – b = ax + x 

take x outside a bracket (factorise).

ab – b = x(a + 1)

divide both sides by (a + 1).

ab -  xa -  1 = x 

    ∴ x = ab -  ba -  1 = b(a - 1)a -  1

Example 3

Make x the subject of the formula 

b = 12a2 - x2

clear fractions. 

    2b = a2 - x2

Square both sides. 

(2b)2 = a2 – x2

4b2 = a2 – x2

Rearrange to give the term in x on one side of the equation.

x2 = a2 – 4b2

Take the square root of the both sides. 

x = ± a2 - 4b2

The general method of Example 1, 2 and 3 is to treat the formula as an equation and the new subject as the unknown of the equation. 

There are many different formulae. Therefore it is not possible to give general rules for changing their subject. However remember the following: 

1. Begin by clearing fractions, brackets and root 
2. Rearrange the formula so that all the terms that contain the new subject are on one side of the equals sign and the rest on the other. Do not try to place the subject on the left-hand side if it comes more naturally on the right 
3. If more than one term contains the subject, take it outside a bracket (i.e. factorise)
4. Divide both sides by the bracket, then simplify as far as possible. 

EVALUATION 

Make x the subject of the following equations. 

1. x(a - b) = b(c - x) 

1. x2 - a2= b 
2. (ax – b)(bx + a) = (bx2 + a)a

Types of Variation such as: Direct, Inverse, Joint and Partial

Direct Variation 

If a person buys some packets of sugar, the total cost is proportional to the number of packets bought 

The cost of 2 packets atNx per packet is N2x

The cost of 3 packets at Nx per packet is N3x.

The cost of n packets at Nx per packet is Nnx. 

Thus, the ratio of total cost to number of packets is the same for any number of packets bought. 

These are both examples of direct variation, or direct proportion. In the first example, the cost, C, varies directly with the number of packets, n.

Example 1 

If 1 packet of sugar costs x naira what will be the cost of 20 packets?

Cost varies directly with the number of packets bought. 

Cost of 1 packet = x naira 

Cost of 20 packets = 20 x n naira 

            = 20x naira

Example 2 

If C n and C = 5 when n = 20, find the formula connecting C and n. 

C n means Cn is constant. 

Let this constant be k. 

Then, Cn = k 

Or C = kn

    C = 5 when n = 20 

Hence 5 = k x 20

    k = 12

thus, C = 14n is the formula which connects C and n. 

a formula such as C = 14nis often known as a relationship between the variables C and n. 

Example 3

If M L and M = 6 when L = 2, find 

1. The relationship between M and L, 
2. The value of L when M = 15. 

1. If M L, then M =kL when k is a constant. 

M = kL

When M = 6, L = 2

Thus, 6 = k x 2

K = 3

Therefore M = 3L is the relationship between M and L. 

1. M = 3L and M = 15,

Thus 15 = 3L

L = 5

EVALUATION 

1. If P Q and P = 4.5 when Q = 12, find 
   1. The relationship between P and Q
   2. P when Q = 16
   3. Q when P = 2.4

Inverse Variation 

1. 5 equal sectors,         b. 12 equal sectors. 

Fig 1                    Fig 2

In fig 1, there ar 5 equal sectors in the circle. The angle of each sector is 72o.

In fig 2, there are 12 equal sectors in the circle. The angle of each sector is 30o.

If there are 18 sectors, the angle of each sector would be 20o.

Therefore, the greater the number of sectors, the smaller the angle of each sector.

Similarity, if a car travels a certain distance, the greater its average speed, the less time it will take. 

These are both examples of inverse variation, or inverse proportion. Sometimes known as indirect variation. In the first example, the size of the angle, , varies inversely with the number of sectors, n. in the second, thetime taken, T, is inversely proportional to the average speed, S. these statement are written: 

θ∝1n        T ∝ 1S

Example 1 

If θ∝1n and    = 72 when n = 5, find 

1. when n = 12
2. n when = 8 

First: find the relation between θ and n

θ ∝ 1nmeansθ= kn where k is a constant. 

θ= Kn

When = 72, n = 5

Thus, 72 = K5

K = 5 x 72 = 360

Thus,θ= 360n

1. θ= 360n

When n = 12,

θ= 36012= 30

1. If θ= 360nthen n= 360.

When = 8, 

n= 3608=45

Joint Variation 

The mass of a sheet of metal is proportional to both the area and the thickness of the metal. Therefore M At (where M, A and t are the mass, area and thickness). This is an example of joint variation. The mass varies jointly with the area and thickness.

Example 1

Y varies inversely as X2 and X varies directly as Z2. Find the relationship between Y and Z, given C is a constant. 

From the first sentence: 

Y ∝ 1X2 and X Z2

Or Y = AX2 and X = BZ2

Where A and B are constants. 

Substituting BZ2 for X in Y = AX2

Y = A(BZ2)2= AB2Z4

Or Y = CZ4 where C is a constant = (AB2)

Thus Y varies inversely as Z4.

(Alternatively, YZ4 = C)

EVALUATION 

1. A rectangle has a constant area, A. its length is l and its breadth is b.
   1. Write a formula for l in terms of A and b
   2. Write a formula for b in terms of A and l.
   3. Does l vary inversely or directly with b?

Partial Variation

When a tailor makes a dress, the total cost depends on two things: first the cost of the cloth; secondly the amount of the time it takes to make the dress. The cost of the cloth is constant, but the time taken to make the dress can vary. A simple dress will take a short time to make; a dress with a difficult pattern will take a long time. This is an example of parital variation. The cost is partly constant and partly varies with the amount of time taken. In algebraic from, C= a + kt, where C is the cost, t is the time taken and a and k are constants. 

Example 1 

R is partly constant and partly varies with E. when R = 350, E = 1,600 and when R = 730, E = 3,600.

• Find the formula which connects R and E. 

• Find R when E = 1300

a.From the first sentence,

    R = c + KE where c and k are constants. Substituting the given values gives two equations. 

    530 = c + 1600k    (1)

    730 = c + 3600k    (2)

These are simultaneous equations. 

Subtract (1) from (2)

200 = 2000k

K = 2002000 = 110

Substituting in (1),

530 = c + 1600 x 110

530 = c + 160

Thus, c = 370

Thus, R = 370 + 110E is the required formula. 

1. R = 370 + E10

when E = 1300,

    R = 370 + 130010

    = 370 + 130

    = 500

EVALUATION 

• The cost of a car service is partly constant and partly varies with time it takes to do the work. It cost N3, 500 for a 512 hour service and N2, 900 for a 4-hour service. 

• Find the formula connecting cost, NC with time, T hours 

• Hence find the cost of a 712 hour service. 

1. X is partly constant and partly varies as y. when y = 2, x = 30, and when y = 6, x = 50.
   1. Find the relationship between x and y.
   2. Find x when y = 3

GENERAL EVALUATION 

1. If a man cycles 15km in 1 hour, how far will he cycle in two hours if he keeps up the same rate? 
2. A piece of string is cut into n pieces of equal length l. 
   1. Does n vary directly or inversely with l?
3. The mass of rice that each woman gets when sharing a sack varies inversely with the number of women. When there are 20 women, each gets 6kg of rice. If there are nine woman, how much does each get? 

READING ASSIGNMENT 

New General Mathematics SSS 1 pages 220 Exercise 18a 11 – 15

WEEKEND ASSIGNMENT 

xyz when y = 7 and z = 3, x = 42. 

1. Find the relation between x, y and z    A. y = xz      B. x = y/z      C. x = 18y/z    D. y = 18xz
2. Find x when y = 5 and z = 9    A. 20     B. 5     C. 2     D. 10 
3. Which of the following give the correct expressions for inverse variation   A. x = ky     B. x = k/y    C. x = kyz      D. x = 1/y 
4. If r 1/T and T = 8 when R = 4, find the relationship between R and T    A. R = 32/T    B. R = 32T     C. R = T/32      D. 32 = R 
5. If D varies inversely as T use the symbol to show a connection between d and t.   A. d t    B. d 1/t    C. d = 1/2t     D. d = 1/t

THEORY 

1. The number of bricks, b, that a man can carry varies inversely with the mass of each brick, m kg, find the relationship between b and m. hence find the number of 312kg bricks that he can carry. 
2. If x – 3 is directly proportional to the square of y and x = 5 when y = 2, find x when y = 6