Simple Equations And Variations SS3 Mathematics Lesson Note

 INTRODUCTION TO 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 

8×8 = 5

x = 58

58 is 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 

 CLASSWORK 

1. Solve the following equations

2 – 5t = 20 – 8t 

d = 12 – (11 + 4d)

213d = 28 

CHANGE OF SUBJECT OF FORMULAE

It 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 – axe = 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 both sides. 

x = ± a2 – 4b2

 The general method of Examples 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. 
3. Do not try to place the subject on the left-hand side if it comes more naturally on the right 
4. If more than one term contains the subject, take it outside a bracket (i.e. factorise)
5. Divide both sides by the bracket, then simplify as far as possible. 

 CLASSWORK 

1. Make x the subject of the following equations. 
2. a) x(a – b) = b(c – x) 
3. b) x2 – a2= b 
4. c) (ax – b)(bx + a) = (bx2 + a)a

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

1. Direct Variation 

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

b) The cost of 2 packets at Nx 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 = 14 is often known as a relationship between the variables C and n. 

2. Inverse Variation 

If there are 5 equal sectors in the circle The angle of each sector is 72° in one figure and there is another figure with 12 equal sectors in the circle and the angle of each sector is 30°.

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

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

Similarly, 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, the time taken, T, is inversely proportional to the average speed, S. These statements are written: 

θ∝1n        T ∝ 1S

 Example 1 

If θ∝1n and    = 72 when n = 5, find 

 when n = 12

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

θ= 360n

When n = 12,

θ= 36012= 30

If θ= 360nthen n= 360.

When  = 8, 

n= 3608=45

3. 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 to Z4.

(Alternatively, YZ4 = C)

4. Partial Variation

When a tailor makes a dress, the total cost depends on two things: first the cost of the cloth; and second, the amount of 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 partial variation. The cost is partly constant and partly varies with the amount of time taken. In algebraic form, 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.

1. a) Find the formula which connects R and E. 
2. b) Find R when E = 1300

 Solution:

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. 

 R = 370 + E10

when E = 1300,

    R = 370 + 130010

    = 370 + 130

    = 500

ASSIGNMENT 

1. xyz when y = 7 and z = 3, x = 42. 

Find the relation between x, y and z    

1. 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 gives 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
6. 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.

7. If x – 3 is directly proportional to the square of y and x = 5 when y = 2, find x when y = 6