# Linear equations

AS WE pursue exam-type material, I expect that you are making greater effort to understand what is provided and are practising in earnest.

We will begin this week’s lesson by reviewing the answers to last week’s homework. 1. Evaluate: (5n - 2)2

SOLUTION

(5n - 2)2 = (5n - 2)(5n - 2) = (25n2 - 10n -10n + 4) = 25n2 - 20n + 4 2. Expand the following: (a) (k + 4) (k - 3) (b) (t - 1) (t + 4)

SOLUTION

(a) (k + 4) (k - 3) = k2 + 4k -3k - 12 = k2 + k - 12 (b) (t - 1) (t + 4) = t2 - t + 4t - 4 = t2 + 3t - 4 3. Evaluate: ( 1 - 2p)(5 - 3p)

SOLUTION

(1 - 2p)(5 - 3p) = 5 - 10p - 3p + 6p2 = 5 - 13p + 6p2 4. Simplify 2y - 1 /6 - y + 3 /5

SOLUTION

2y - 1/6 - y + 3/5 The LCM of 6 and 5 is 30. 5(2y - 1) - 6(y + 3) /30 = 10y - 5 - 6y - 18 /30 = 4 y - 23 /30

A BRIEF REVIEW OF CHANGE THE SUBJECT OF THE FORMULA

Given the formula, A = xy Then, x = A/y is the formula expressed in terms of x.

Example: Given the formula, V = pi r2h, make r the subject. Solution : Given that V = pi r2h Then r2 = V/ pi h r = √ V/ pi h We will now continue with linear equation.

LINEAR EQUATIONS

The inclusion of the EQUAL sign differentiates an EQUATION from an algebraic expression. This point is commonly missed by students, who sometimes attempt to solve algebraic expressions. Do not make this error. The following points should be noted:

Equations identify either the relationship between variables or the value of a variable.

The value of the variable is maintained by performing identical operations on both sides of the equation.

The methods of clearing brackets and simplifying algebraic expressions are usually required to find solution of linear equations.

In order to solve linear equations, one approach is to simplify each side of the equation and then equate both sides. The above is illustrated by the following example:

EXAMPLE 1

Given that 2(x - 1) - 3x = 6, then x = a) - 8 b) - 4 c) 4 d) 8

Simplifying the left-hand side, 2(x - 1) - 3x = 2x - 2 - 3x = - x - 2 Equating - x - 2 = 6 with - x = 6 + 2 x = - 8, the answer is a)

EXAMPLE 2

Solve 3x/2 + x /4 = 14 Simplify the left-hand side: 3 x /2 + x/4 = 6x/4 + x/4 = 7x/4 Equating both sides: 7x /4 = 14 Multiply both sides by 4 7x/4 x 4 = 14 x 4 = 56 7x = 56 x = 56/7 = 8 Ans.: 8 Example 3 Solve 4x + 5/4 - 9 + 2x/3 = 0

Considering the left-hand side, the LCM of 3 and 4 is 12. 3(4x + 5) - 4(9 + 2x) /12 = 12x + 15 - 36 - 8x /12 = 4x - 21/12 Equating both sides: 4x - 21/12 = 0 Multiply both sides by 12 4x - 21 = 0 or 4x = 21 x = 21/4

ALTERNATIVELY, you may multiply all terms by the LCM of the denominators.

4x + 5/4 - 9 + 2x/3 = 0 Multiply both sides by 12: 4x+5 x 12/4 - 9 +2x x 12 /3 = 0x12 3(4x + 5) - 4(9 + 2x) = 0 12x + 15 -36 - 8x = 0 4x - 21 = 0 x = 21/4

We will now continue algebra with the topic factorization.

Note that an algebraic expression is factorised when it is expressed as the product of its simplest factors. For example, 6b + 15 is expressed as

3( 2b + 5) The usual methods are: (a) Common factor (b) Grouping (c) Factorizing of quadratic expressions (d) Difference of two squares

The methods are adequately explained in the textbooks and you should use them to aid you as you revise for your exams.

It is important that you do the following in all cases:

(a) Bring each factor to its simplest form. For example, a factor 20x + 12 should be expressed as 4(5x + 3) or 2 - 6y as 2( 1 - 3y).

(b) Check your answers, if you have the time, by expanding the factors and comparing the result with the original expression. This week ,we will review the first two methods of Factorization mentioned above.

EXAMPLES OF COMMON FACTOR METHOD

1. Factorise 21ab + 7a 21ab + 7a = 7a ( 3b + 1) NB: 7a is the only factor common to both terms.

Checking, multiplying 7a ( 3b + 1) = 21ab + 7a, the given expression. 2. Factorise 16m + 4 The factors of 16m are 2, 4, 8 and m and 4 has factors 2 and 4. The highest common factor (HCF) is 4. 16m + 4 = 4( 4m + 1 ) Answer : 4( 4m + 1 ) 3. Factorise: 9x2 - 12x The common factor method is used, as 3x is the factor which is common to both terms. Both terms are divided by 3x for us to obtain the second factor.

Answer: 3x(3x - 4) Please note that by expanding the answer, 3x(3x - 4) = 9x2 - 12x, the given expression 4. Factorise: 15x2y -10xy2 Note that the common factor to both terms is 5xy By dividing each term by 5xy, Answer: 5xy(3x - 2y2)

Please factorise each of the following on your own: 6x2 y +12xy 3 - 12mn 2b3 - 6 b2 - 4b

EXAMPLES OF GROUPING METHOD

5. Factorise ax + ay + bx + by Note that a is the common factor of ax + ay and b the common factor of bx + by

ax + ay + bx + by = a(x + y) + b(x + y)

Do you realise that (x + y) is common to both expressions? a(x + y) + b(x + y) = (x + y)(a + b) This method could, therefore, be described as repeated common factor method.

This method could, therefore, be described as repeated common factor method. 4. Factorise 2ax - 6ay + bx - 3by (2ax - 6ay) + (bx - 3by) Factorise each as follows: 2a(x - 3y) + b (x - 3y) = (x - 3y)(2a+ b) As usual, I will close with your homework. 1. Solve 5x + 7 = 3x/2 2. Solve x - 2 /3 + x + 1/4 = 1 3. Factorise: (a) 6x2 - 18x (b) xy2 + x2y 4. Factorise: 2mh - 2nh- 3mk+ 3nk 5. Factorise: (a) 9k2 - 4 (b) 5pq2 - pq + 5rq - r 6. Factorise: 5a2b + 2ab Please identify additional questions to be included in your revision pool.