Cambridge International AS and A Level Mathematics Pure Mathematics 1

Algebra

8

P^

1

EXAMPLE 1.14 Solve the equation 5(x − 3) = 2(x + 6).

SOLUTION

Open the brackets ⇒ 5 x − 15 = 2 x + 12

Subtract 2x from both sides ⇒ 5 x – 2x − 15 = 2 x − 2 x + 12

Tidy up ⇒ 3 x − 15 = 12

Add 15 to both sides ⇒ 3 x − 15 + 15 = 12 + 15

Tidy up ⇒ 3 x = 27

Divide both sides by 3 ⇒ 3

3

27

3

x=

⇒ x = 9

CHECK

When the answer is substituted in the original equation both sides should come

out to be equal. If they are different, you have made a mistake.

Left-hand side Right-hand side

5(x − 3) 2(x + 6)

5(9 − 3) 2(9 + 6)

5 × 6 2 × 15

30 30 (as required).

EXAMPLE 1.15 Solve the equation 12 (x + 6) = x +^13 (2x − 5).

SOLUTION

Start by clearing the fractions. Since the numbers 2 and 3 appear on the bottom

line, multiply through by 6 which cancels both of them.

Multiply both sides by 6 ⇒ 6 × 12 (x + 6) = 6 × x + 6 × 13 (2x − 5)

Tidy up ⇒ 3(x + 6) = 6 x + 2(2x − 5)

Open the brackets ⇒ 3 x + 18 = 6 x + 4 x − 10

Subtract 6x, 4x, and 18

from both sides ⇒ 3 x − 6 x − 4 x = − 10 − 18

Tidy up ⇒ − 7 x = − 28

Divide both sides by (–7) ⇒ –

–

–

7

7

28

7

x=

⇒ x = 4

CHECK

Substituting x = 4 in^12 (x + 6) = x +^13 (2x – 5) gives:

Left-hand side Right-hand side

1

2 (4 + 6)^4 +^

1

3 (8 – 5)

10

2 4 +^

3

3

5 5 (as required).