Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.4 - Linear Functions and Slope - Exercise Set - Page 213: 70

Answer

$x=-5$, $y=-3$, see graph

Work Step by Step

Consider the given linear equation $3x+5y+15=0$ Use intercepts to graph the given equation as follows: Step 1: Find the $x$ intercept. To find the x intercept of the line, substitute $y=0$ in the equation and calculate the value of x. $\begin{align} & 3x+5\times 0+15=0 \\ & 3x=-15 \\ & x=\frac{\left( -15 \right)}{3} \\ & =-5 \end{align}$ So, the $x$ intercept of equation $3x+5y+15=0$ is $-5$. Hence, the line passes through the point $\left( -5,0 \right)$. Step 2: Find the $y$ intercept. To calculate the y intercept of the line, substitute $x=0$ in the equation and calculate the value of y. $\begin{align} & 3\times 0+5y+15=0 \\ & 5y=-15 \\ & y=\frac{\left( -15 \right)}{5} \\ & =-3 \end{align}$ So, the $y$ intercept of the given equation $3x+5y+15=0$ is $-3$. Hence, the line passes through $\left( 0,-3 \right)$. Step 3: Graph the equation of the straight line. Locate the two intercepts $\left( -5,0 \right)$ and $\left( 0,-3 \right)$. Connect them by a straight line. The graph of the given equation $3x+5y+15=0$ is as follows:
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