Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Problems - Page 75: 7

Answer

The graph consists of: $y=x$ (in the 1st quadrant) and the entire 3rd quadrant (bounded by $y=0$ and $x=0$).

Work Step by Step

We can explain the graph by tackling the quadrants one by one. Quadrant 1 (the upper right): the only quadrant where both x and y are positive. Therefore, directly, |x|=x and |y|=y. The equation becomes x+x=y+y, 2x=2y, so we get the line x=y (starting from the origin, (0,0)) Quadrant 2 (the upper left): here, x is always negative, while y is always positive. So, |x|=-x, |y|=y. The equation looks like x+(-x)=y+y, so 0=2y and y=0. It coincides with the non-positive part of the x-axis. Quadrant 3 (the lower left): here, x is always negative, and y is also always negative. So, |x|=-x, and |y|=-y. The equation becomes x+(-x)=y+(-y), which is 0=0. This is always true, so the graph includes all points in the 3rd quadrant. Quadrant 4 (the lower right one): here, x is always positive, while y is always negative. So, |x|=x and |y|=-y. So, the equation becomes x+x=y+(-y), which is 2x=0, same as x=0. We get the non-positive part of the y-axis.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.