year 8 maths

2B
Equivalent equations
69
Example 4 Finding equivalent equations
Show the result of applying the given operation to both sides of these equations.
a 8y = 40 [÷8] b 10 + 2x = 36 [−10] c 5a − 3 = 12 [+3]
SOLUTION
a 8y = 40
÷8 ÷8
y = 5
b 10 + 2x = 36
−10
2x = 26
c 5a − 3 = 12
+3 +3
5a = 15
−10
EXPLANATION
Write out the equation and then divide both sides
by 8 .
40 ÷ 8 is 5 and 8y ÷ 8 is y .
Write out the equation and then subtract 10 from
both sides.
36 − 10 is 26.
10 + 2x − 10 is 2x .
Write out the equation and then add 3 to
both sides.
12 + 3 is 15 .
5a − 3 + 3 is 5a.
Example 5 Solving equations systematically
Solve the following equations and check the solution by substituting.
a x − 4 = 16 b 2u + 7 = 17 c 40 − 3x = 22
SOLUTION
a x − 4 = 16
+4 +4
x = 20
So the solution is x = 20 .
b 2u + 7 = 17
−7
−7
2u = 10
÷2 ÷2
u = 5
So the solution is u = 5 .
c 40 − 3x = 22
−40
−40
−3x = −18
÷−3 ÷−3
x = 6
So the solution is x = 6 .
EXPLANATION
By adding 4 to both sides of the equation, we get
an equivalent equation.
Check: 20 − 4 = 16
✓
To remove the +7 , we subtract 7 from both sides.
Finally, we divide by 2 to reverse the 2u .
Remember that 2u means 2 × u .
Check: 2(5) + 7 = 10 + 7 = 17
We subtract 40 from both sides to remove the
40 at the start of the LHS.
Since −3 × x = −18 , we divide by −3 to get the
final solution.
Check: 40 − 3(6) = 40 − 18 = 22
✓
✓
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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