Precalculus (6th Edition) Blitzer

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

Chapter 7 - Review Exercises - Page 878: 39

Answer

The graph is shown below:

Work Step by Step

Let us consider the inequality $3x-4y\gt12$; substitute the equal symbol in place of the inequality and rewrite the equation as given below: $3x-4y=12$ To find the value of the x-intercept, put y = 0 as given below: $\begin{align} & 3x-4y=12 \\ & 3x-\left( 4\times 0 \right)=12 \\ & 3x=12 \\ & x=4 \end{align}$ To find the value of the y-intercept, substitute x = 0 as given below: $\begin{align} & 3x-4y=12 \\ & \left( 3\times 0 \right)-4y=12 \\ & -4y=12 \\ & y=-3 \end{align}$ Then, take the origin $\left( 0,0 \right)$ as a test point and check the region in the graph to shade: $\begin{align} & 3x-4y>12 \\ & \left( 3\times 0 \right)-\left( 4\times 0 \right)>12 \\ & 0>12 \end{align}$ So, the above expression is incorrect, which means the shaded region will not contain the test point; that is, the region away from the origin will be shaded. Therefore, the line passes through points $\left( 4,0 \right)$ and $\left( 0,-3 \right)$. Plot the graph using the intercepts as given below. Thus, the graph of the provided inequality is plotted and the line passes through points $\left( 4,0 \right)$ and $\left( 0,-3 \right)$. Also, we get that the shaded region does not contain the origin.
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