Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 3 - Section 3.2 - Graphing Linear Equations - Exercise Set: 5

Answer

Point A: $y=\frac{1}{3}x$ $y=\frac{1}{3}\times(0)$ $y=0$ Point B: $y=\frac{1}{3}x$ $y=\frac{1}{3}\times(6)$ $y=2$ Point C: $y=\frac{1}{3}x$ $y=\frac{1}{3}\times(-3)$ $y=-1$
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Work Step by Step

To get the coordinates of point A, substitute the given x-value of 0 into the given equation $y=\frac{1}{3}x$ to get the y-value of y=0. Hence, we know that when x is 0, y is 0. i.e (0,0) is a point on the line. To get the coordinates of point B, substitute the given x-value of 6 into the given equation $y=\frac{1}{3}x$ to get the y-value of y=2. Hence, we know that when x is 6, y is 2. i.e (6,2) is a point on the line. To get the coordinates of point C, substitute the given x-value of -3 into the given equation $y=\frac{1}{3}x$ to get the y-value of y=-1. Hence, we know that when x is -3, y is -1. i.e (-3,-1) is a point on the line. Graph points A, B and C using their respective x and y values.. Using a ruler, draw a straight line passing through all three points to graph the equation of $y=\frac{1}{3}x$.
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