Lesson 4, Topic 2
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# Equations with Brackets

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Hint: To solve an equation containing brackets eliminate the brackets first, collect like terms on separate sides of the equation and then solve the equation.

Example 2

Solve the following equations:

i. 8(5p – 1) + 16 = 8(3p – 9)

ii. 2(2y – 5) – (y + 3) = 3(7 – y)

iii. 4(2x + 5) – 2(x – 3) – 8(2x + 4) = 0

iv. 0 = p + 2.5 – (3.6 – p)

Solution:

i. 8(5p – 1) + 16 = 8(3p – 9)

Expand the brackets

40p – 8 + 16 = 24p – 72

Collect like terms

40p – 24p = 8 -16 -72

16p = 8 – 88

16p = – 80

Divide both sides by 16

$$\frac {16p}{16} = \frac {-80}{16}$$

p = -5

ii.        2(2y – 5) – (y + 3) = 3(7 – y)

Expand the brackets

4y – 10 – y – 3 = 21 – 3y

Collect like terms

i.e. 4y – y + 3y = 21 + 10 + 3

3y + 3y = 34

6y = 34

Divide both sides by 6

$$\frac {6y}{6} = \frac {34}{6}$$

y = $$\frac{17}{3}$$

y = $$\scriptsize 5 \normalsize \frac{2}{3}$$

iii.     4(2x + 5) − 2(x – 3) – 8(2x + 4) = 0

Expand the brackets

i.e.    8x + 20 – 2x + 6 – 16x – 24 = 0

Collect like terms

8x – 2x – 16x = 24 – 20 – 6

8x – 18x = 24 – 26

-10x = -2

Multiply both sides by -1

i.e.       $$\scriptsize -10x \; \times \; -1 \; = \; -2 \; \times \; -1$$

10x = 2

Divide both sides by 10

$$\frac {10x}{10} = \frac {2}{10}$$

$$\therefore \scriptsize x =\normalsize \frac {1}{5}$$

iv.     0 = p + 2.5 – (3.6 – p)

Expand the brackets

i.e. 0 = p + 2.5 – 3.6 + p

Collect like terms

-2.5 + 3.6 = p + p

1.1 = 2p

Divide both sides by 2

$$\frac {1.1}{2} = \frac {2p}{2}$$

$$\therefore \scriptsize p = \normalsize \frac {1.1}{2} \scriptsize \: or \: p = \normalsize \frac {11}{20}$$

error: