year 8 maths

434 Chapter 7 Linear relationships 1
Example 14
Using the point of intersection of two lines to solve an equation
Use the graph of y = 4 − x and y = 2x + 1 , shown here, to
answer these questions.
a Write two equations that each have x = −2 as a solution.
b Write four solutions (x, y) for the line with equation
y = 4 − x .
c Write four solutions (x, y) for the line with equation
y = 2x + 1 .
d Write the solution (x, y) that is true for both lines and
show that it satisfies both line equations.
e Solve the equation 4 − x = 2x + 1 .
y = 4 − x
−3 −2
6
5
4
3
2
1
−1
−1
−2
−3
y
O
1
y = 2x + 1
2 3 4 5
x
SOLUTION
a 4 − x = 6
2x + 1 = −3
EXPLANATION
(−2, 6) is on the line y = 4 − x so 4 − x = 6 has solution x = −2 .
(−2, −3) is on the line y = 2x + 1 , so 2x + 1 = −3 has solution
x = −2 .
b (−2, 6)(−1, 5)(1, 3)(4, 0) Many correct answers. Each point on the line y = 4 − x is a
solution to the equation for that line.
c (−2, −3)(0, 1)(1, 3)(2, 5) Many correct answers. Each point on the line y = 2x + 1 is a
solution to the equation for that line.
d (1, 3) (1, 3)
y = 4 − x y = 2x + 1
3 = 4 − 1 3 = 2 × 1 + 1
3 = 3 True 3 = 3 True
The point of intersection (1, 3) is the solution that satisfies both
equations.
Substitute (1, 3) into each equation and show that it makes a true
equation (LHS = RHS).
e x = 1 The solution to 4 − x = 2x + 1 is the x -coordinate at the point
of intersection. The value of both rules is equal for this
x -coordinate.
Exercise 7G
UNDERSTANDING AND FLUENCY
1–8 3–7(½), 8
5–9
1 Use the given rule to complete this table, and then plot and join the points to form a straight line
y = 2x − 1 .
x −2 −1 0 1 2
y
2 Substitute each given y -coordinate into the rule y = 2x − 3 , and then solve the equation
algebraically to find the x -coordinate.
i y = 7 ii y = −5
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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