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: 20

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*} 5x-6y&=42\\ 5x -6(0)&=42\\ 5x&=42\\ x&=\frac{42}{5}\\ x&=8.4 \end{align*} Thus, the $x$-intercept is $(8.4, 0)$. \begin{align*} 5x-6y&=42\\ 5(0) -6y&=42\\ -6y&=42\\ y&=\frac{42}{-6}\\ y&=-7 \end{align*} Thus, the $y$-intercept is $(0, -7)$. Plot the points then connect them using a straight line. Refer to the graph above.
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