Lesson Progress
0% Complete

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

\( \frac {1.1}{2} = \frac {2p}{2} \)

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

Responses

Your email address will not be published. Required fields are marked *

error: