year 8 maths

72 Chapter 2 Equations 2
15 a Prove that 7x + 4 = 39 and −2x + 13 = 3 are equivalent by filling in the missing steps.
7x + 4 = 39
−4
−4
7x = 35
÷7 ÷7
––– = –––
× −2 × −2
––– = –––
+13 +13
−2x + 13 = 3
b Prove that 10k + 4 = 24 and 3k − 1 = 5 are equivalent.
16 a Prove that 4x + 3 = 11 and 2x = 4 are equivalent. Try to use just two steps to get from one
equation to the other.
b Are the equations 5x + 2 = 17 and x = 5 equivalent?
c Prove that 10 − 2x = 13 and 14x + 7 = 20 are not equivalent, no matter how many steps
are used.
17 A student has taken the equation x = 5 and performed some operations to both sides:
x = 5
×4 ×4
4x = 20
+3 +3
4x + 3 = 23
×2 ×2
(4x + 3) × 2 = 46
a Solve (4x + 3) × 2 = 46 systematically.
b Describe how the steps you used in your solution compare with the steps used by the student.
c Give an example of another equation that has x = 5 as its solution.
d Explain why there are infinitely many different equations with the solution x = 5 .
ENRICHMENT
– –
18
Dividing whole expressions
18 It is possible to solve 2x + 4 = 20 by first dividing both sides by 2 , as long as every term is divided
by 2 . So you could solve it in either of these fashions.
2x + 4 = 20
−4
−4
2x = 16
÷2 ÷2
x = 8
2x + 4 = 20
÷2 ÷2
x + 2 = 10
−2
−2
x = 8
Note that 2x + 4 divided by 2 is x + 2 , not x + 4 . Use this method of dividing first to solve the
following equations, and then check that you get the same answer as if you subtracted first.
a 2x + 6 = 12 b 4x + 12 = 16
c 10x + 30 = 50 d 2x + 5 = 13
e 5x + 4 = 19 f 3 + 2x = 5
g 7 = 2x + 4 h 10 = 4x + 10
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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