Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1 - Linear Functions - Chapter Review Exercises - Page 121: 19

Answer

Refer to the graph below.

Work Step by Step

Recall: (1) The $x$-intercept of a line is the point where the line crosses the $x$-axis and can be found by setting $y=0$ then solving for $x$. (1) The $y$-intercept of a line is the point where the line crosses the $y$-axis and can be found by setting $x=0$ then solving for $y$. Solve for the $x$ and $y$ intercepts to obtain: \begin{align*} 2x+3y&=24\\ 2x + 3(0)&=24\\ 2x&=24\\ x&=\frac{24}{2}\\ x&=12 \end{align*} Thus, the $x$-intercept is $(12, 0)$. \begin{align*} 2x+3y&=24\\ 2(0) + 3y&=24\\ 3y&=24\\ y&=\frac{24}{3}\\ y&=8 \end{align*} Thus, the $y$-intercept is $(0, 8)$. Plot the points then connect them using a straight line. Refer to the graph above.
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