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 – 72Â Â

Collect like terms

â‡’ 40p – 24p = 8 -16 -72

â‡’ 16p = 8 – 88

â‡’ 16p = – 80

Divide both sides by 16

â‡’ $$\frac {16p}{16} = \frac {-80}{16}$$

âˆ´ p = -5

ii.Â  Â  Â  Â  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

â‡’ 3y + 3y = 34

â‡’ 6y = 34

Divide both sides by 6

â‡’$$\frac {6y}{6} = \frac {34}{6}$$

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

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

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

Expand the brackets

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

Collect like terms

â‡’ 8x – 2x – 16x = 24 – 20 – 6

â‡’ Â  Â  8x – 18x = 24 – 26

â‡’Â  Â  Â  -10x = -2

Multiply both sides by -1

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

â‡’ Â  Â  Â  Â  Â  10x = 2

Divide both sides by 10

â‡’ $$\frac {10x}{10} = \frac {2}{10}$$

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

iv. Â  Â  0 = p + 2.5 – (3.6 – p)

Expand the brackets

i.e. 0 = p + 2.5 – 3.6 + p

Collect like terms

â‡’Â  Â  Â  -2.5 + 3.6 = p + p

â‡’Â  Â  Â  Â  1.1 = 2p

Divide both sides by 2

â‡’ $$\frac {1.1}{2} = \frac {2p}{2}$$

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

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