year 8 maths

74 Chapter 2 Equations 2
Continued from previous page
4y + 15
b
= 3
9
×9 ×9
4y + 15 = 27
−15
−15
4y = 12
÷4 ÷4
y = 3
Multiplying both sides by 9 removes the denominator
of 9 .
The equation 4y + 15 = 27 is solved in the usual
fashion (subtract 15 , divide by 4 ).
c
d
4 + 5x
2 = 29
−4 −4
5x
2 = 25
×2 ×2
5x = 50
÷5
÷5
x = 10
7 − 2x
2 = 5
−7
−7
− 2x
3 = −2
×−1 ×−1
2x
3 = 2
×3 ×3
2x = 6
÷2 ÷2
x = 3
We must subtract 4 first because we do not have a
fraction by itself on the left-hand side. Once there
is a fraction by itself, multiply by the denominator (2) .
Subtract 7 first to get a fraction. When both sides are
negative, multiplying (or dividing) by −1 makes them
both positive.
Exercise 2C
UNDERSTANDING AND FLUENCY
1–3, 4–5(½) 2, 3, 4–5(½)
4–5(½)
1 a If x = 4 , find the value of x 2 + 6 .
b If x = 4 , find the value of x + 6
2 .
x
c Are
2 + 6 and x + 6
equivalent expressions?
2
2 Fill in the missing steps to solve these equations.
a
x
3 = 10
×3 ×3
x = __
Cambridge Maths NSW
Stage 4 Year 8 Second edition
b
m
5 = 2
×5 ×5
m = __
c 11 = q 2
□
__ = q
□
d
□
p
10 = 7
p = __
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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