year 8 maths

7G
Solving linear equations using graphical techniques
433
■ A point (x, y) is a solution to the equation for a line if its coordinates make the equation true.
• An equation is true when LHS = RHS after the coordinates are substituted.
• Every point on a straight line is a solution to the equation for that line.
• Every point that is not on a straight line is not a solution to the equation for that line.
■ The point of intersection of two straight lines is the only solution that
satisfies both equations.
• The point of intersection is the shared point where two straight lines
cross each other.
• This is the only point with coordinates that makes both equations
true.
For example, (1, 3) is the only point that makes both y = 6 − 3x
and y = 2x + 1 true.
Substituting (1, 3) :
y = 6 − 3x y = 2x + 1
3 = 6 − 3 × 1 3 = 2 × 1 + 1
3 = 3 (True) 3 = 3 (True)
5
4
3
2
1
O
y
1
(1, 3)
Point of
intersection
2 3
x
Example 13
Using a linear graph to solve an equation
Use the graph of y = 2x + 1 , shown here, to solve each of the
following equations.
a 2x + 1 = 5
b 2x + 1 = 0
c 2x + 1 = −4
−4 −3 −2 −1
5
4
3
2
1
−1
−2
−3
−4
−5
y
y = 2x + 1
O 1 2 3 4
x
SOLUTION EXPLANATION
a x = 2 Locate the point on the line with y -coordinate 5 . The x -coordinate of this point
is 2 , so x = 2 is the solution to 2x + 1 = 5 .
b x = −0.5 Locate the point on the line with y -coordinate 0 . The x -coordinate of this point is
−0.5 , so x = −0.5 is the solution to 2x + 1 = 0 .
c x = −2.5 Locate the point on the line with y -coordinate −4 . The x -coordinate of this point
is −2.5 , so x = −2.5 is the solution to 2x + 1 = −4 .
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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