# 1. Introduction

In this section we shall look at some simple equations and the methods used to find their solutions.

There are four basic rules:

Rule 1 An equal quantity may be added to both sides of an equation. An equal quantity may be subtracted from both sides of an equation. An equal quantity may multiply both sides of an equation. An equal, non-zero, quantity may divide both sides of an equation.

The application of these rules is illustrated in the following examples.

Example 1: Solve these equations:

(a)3x − 8 = x + 10

Solution: By Rule 1 we may add 8 to both sides:

3x − 8 + 8 = x + 10 + 8, i.e. 3x = x + 18

Then by Rule 2 we may subtract x from both sides:

3xx = x + 18 − x, i.e. 2x = 18

Finally, by Rule 4 we divide both sides by 2 giving x = 9

(b) x / 2 = −6

Solution: By Rule 3 we multiply both sides by 2:

( 2 / 1 ) × ( x / 2 ) = 2 × (−6), so x = −12

It is always a good idea to check that the solution is correct by substituting the value into both sides of the equation. In Example 1(a), by substituting x = 9 into the left hand side of the equation, we see that:

3x − 8 = 3 × 9 − 8 = 19

Substituting x = 9 into the right hand side of the equation gives:

x + 10 = 9 + 10 = 19

Since both sides of the equation are equal when x = 9 , it is a correct solution. In this case it is the only solution to the equation but it is important to note that some equations have more than one solution.

Exercise 1: Solve each of the following equations:

(a) 3x = 18

Solution: Dividing both sides by 3 gives:

3x / 3 = 18 / 3

or:

x = 6
(b) 7x = −14

Solution: Dividing both sides by 7 gives:

7x / 7 = − 14 / 7

or:

x = −2
(c) −2x = −10

Solution: Dividing both sides by −2 gives:

−2x / −2 = − −10 / −2

or:

x = 5
(d) 28x = 35

Solution: Here, 7 is the highest common factor of 28 and 35. First let us divide both sides by this:

28x = 35
28x / 7 = − 35 / 7
4x = 5

Now, divide both sides by 4:

4x / 4 = − 5 / 4
x = − 5 / 4

The solution is thus: x = 5 ⁄ 4

(e) 5x − 3x − 12x = 29 − 2 − 7

Solution: First, simplify both sides. The left hand side is:

5x − 3x − 12x = −10x

The right hand side is:

29 − 2 − 7 = 29 − 9 = 20

The original equation is thus:

−10x = 20

The solution to this is obtained by dividing both sides of the equation by −10:

−10x / −10 = 20 / −10

So that:

x = −2
(f) x / 5 = 3

Solution: In this case we multiply both sides by 5:

x / 5 = 3
5 × x / 5 = 5 × 3
x = 15

- so the solution in this case is: x = −15

Click on questions to reveal solutions

Try the following short quizzes:

Quiz 1: Which of the following is the solution to the equation: 8x + 5x − 3x = 17 − 9 + 22 ?

(a)2 Incorrect - please try again!
(b)−2 Incorrect - please try again!
(c)3 Correct - well done!
(d)−3 Incorrect - please try again!

Solution: Simplify both sides first:

8x + 5x − 3x = 13x − 3x = 10x

and:

17 − 9 + 22 = 8 + 22 = 30

The equation to be solved is thus: 10x = 30, and this clearly has solution: x = 3

Quiz 2: Which of the following is the solution to the equation: x − 13x = 3x − 6 ?

(a) 2 / 5 Correct - well done!
(b) −1 / 5 Incorrect - please try again!
(c) 1 / 3 Incorrect - please try again!
(d) −6 / 17 Incorrect - please try again!

Solution:

x − 13x −12x = 3x − 6 = 3x − 6 = 12x + 3x − 6 = 6 = 2 / 5

# 2. Further Equations

We are now ready to move on to slightly more sophisticated examples.

Example 2: Find the solution to the equation: 5 (x − 3) − 7 (6 − x) = 24 − 3 (8 − x) − 3

Solution: Removing the brackets from both sides first and then simplifying:

 5 (x − 3) − 7 (6 − x) 5x − 15 − 42 + 7x = 24 − 3 (8 − x) − 3 = 24 − 24 + 3x − 3 = 12x + 3x − 6 = 3x − 3 = 3x − 3

12x = 3x − 3 + 57 = 3x + 54

Subtracting 3x from both sides:

12x − 3x = 54 or 9x = 54 giving x = 6

Exercise 2: Find the solution to each of the following equations:

(a)2x + 3 = 16 − (2x − 3)

Solution:

2x + 3 = 16 − (2x − 3) = 16 − 2x + 3 = 19 − 2x

Now add 2x to both sides and subtract 3 from both sides:

4x + 3 4x = 19 = 19 − 3 = 16

and the solution is x = 4. This can be checked by putting x = 4 in both sides of the first equation above and noting that each side will have the value 11.

(b)8 (x − 1) + 17 (x − 3) = 4 (4x − 9) + 4

Solution:

 8x − 8 + 17x − 51 = 16x − 36 + 4 25x − 59 = 16x − 32 25x − 16x − 59 = −32 9x − 59 −32 9x 59 − 32 9x 27 x 3
(c)15 (x − 1) + 4 (x + 3) = 2 (7 + x) = 4 (4x − 9) + 4

Solution:

 15x − 15 + 4x + 12 = 14 + 2x 19x −3 = 2x + 14 19x −2x − 3 = 14 17x −3 14 17x 14 + 3 = 17 x 1

Inserting x = 1 into the equation we can check that both sides have the value 16.

Quiz 3: Which of the following is the solution to the equation: 5x − (4x − 7)(3x − 5) = 6 − 3 (4x − 9)(x − 1) ?

(c)2Correct - well done!

Explanation: First expand the brackets separately using FOIL (see the module on Brackets):

 (4x − 7) (3x − 5) = 12x2 − 20x − 21x + 35 = 12x2 − 41x + 36 (4x − 9) (x −1) = 4x2 − 4x − 9x + 35 = 4x2 − 13x + 9

These can now be substituted, carefully, into the equation:

 5x − [(4x − 7) (3x − 5)] = 6 − 3[(4x − 9) (x −1)] 5x − [12x2 − 41x + 35] = 6 − [4x2 − 13x + 9] 5x − 12x2 + 41x − 35 = 6 − 4x2 + 13x − 9 −12x2 + 46x − 35 = −12x2 + 39x − 21

Notice the extra pair of square brackets in the first equation above. These are to emphasise that the negative sign multiplies all of the parts inside the square brackets. The procedure now follows in an obvious manner. Add 12x2 to both sides, subtract 39x from both sides then add 35 to both sides:

 −12x2 + 46x − 35 = −12x2 + 39x − 21 46x − 35 = 39x − 21 46x − 39x − 35 = −21 46x − 39x = 35 − 21 7x = 14 x = 2

When fractions occur we can sometimes transform the equation to one that does not involve fractions.

Example 3: Find the solution to the equation:

(4x ⁄ 5) − (7 ⁄ 4) = (x ⁄ 5) + (x ⁄ 4)

Solution: The least common multiple of the denominators in the equation is 4 × 5 = 20, and we proceed as follows:

 20(4x5 − 74 ) = 20(x5 + x4 ) 201 × 4x5 − 201 × 74 = 201 × x5 + 201 × x4 16x − 35 = 4x + 5x 16x − 35 = 9x

Adding 35 to both sides and subtracting 9x from both sides leads to:

7x = 35x,  so x = 5 is the solution to the equation.

Exercise 3: Find the solution to each of the following equations:

(a)5x − 6 (x − 5) = 2 (x + 5) + 5 (x − 4)

Solution:

 5x − 6 (x − 5) = 2 (x + 5) + 5 (x − 4) 5x − 6x + 30 = 2x + 10 + 5x − 20 −x + 30 = 7x − 10 30 = x + 7x − 10 30 = 8x − 10 30 + 10 = 8x 8x = 40 x = 5
(b)(x + 15) (x − 3) − (x2 − 6x + 9) = 30 − 15 (x − 1)

Solution: First, using FOIL, we expand:

(x + 15) (x − 3) = x2 − 3x + 15x − 45 = x2 + 12x − 45

Now we have:

 (x + 15) (x − 3) − (x2 − 6x + 9) = 30 − 15 (x − 1) x2 +12x − 45 − x2 + 6x − 9 = 30 − 15x + 15 18x − 54 = 45 − 15x 18x + 15x − 54 = 45 33x − 54 = 45 33x = 45 + 54 33x = 99 x = 3
(c)(x − 2) ⁄ 2 + (x + 10) ⁄ 9 = 5

Solution: This time we multiply both sides by 2 × 9:

 (x − 2)/2 + (x + 10)/9 = 5 2 × 9/1 × (x − 2)/2 + 2 × 9/1 × (x + 10)/9 = 2 × 9 × 5 9 (x − 2) + 2 (x + 10) = 90 9x − 18 + 2x + 20 = 90 11x + 2 = 90 11x = 88 x = 8

Quiz 4: Which of the following is the solution to the equation: (x − 4) ⁄ 7 = (x − 10) ⁄ 5 ?

(d)25Correct - well done!

Explanation:The highest common factor of the denominators is 5 × 7 = 35. Multiplying both sides of the equation by this:

 35/1 × (x − 4)/7 = 35/1 × (x − 10)/5 5 (x − 4) = 7 (x − 10) 5x − 20 = 7x − 70 5x − 20 + 70 = 7x − 70 + 70 5x + 50 = 7x 50 = 7x − 5x = 2x x = 25

so that x = 25 is the solution. This can be checked by putting this value into the original equation and showing that each side will have the value 3.

# 5. Quiz on Equations

In each of the following, solve the equation and choose the solution from the options given:

1.4 (x + 2) ⁄ 5 = 7 + 5x ⁄ 13
(a)5
(b)13
(c)−5
(d)−13
2.(x + 20) ⁄ 9 + 3 x ⁄ 7 = 6
(a)9
(b)7
(c)5
(d)2
3.(x + 35) ⁄ 6 − (x + 7) ⁄ 9 = (x + 21) ⁄ 4
(a)−5
(b)2
(c)4
(d)−1
4.(x + 1) (2x + 1) = (x + 3)(2x + 3) − 14 ?
(a)1
(b)−1
(c)2
(d)−2