## Precalculus (6th Edition) Blitzer First we will replace the inequality sign with an equals sign. Therefore, \begin{align} & x+y=7 \\ & x+4y=-8 \end{align} Now we will graph these lines: Consider $x+y=7$ For the $x$ -intercept set $y=0$ , Thus, \begin{align} & x+0=7 \\ & x=7 \end{align} The line passes through $\left( 7,0 \right)$. For the $y$ - intercept, put $x=0$ , Thus, \begin{align} & 0+y=7 \\ & y=7 \end{align} The line passes through $\left( 0,7 \right)$. Test that the statement is true or not by substituting $\left( 0,0 \right)$ in $x+y\le 7$ , $0+0\le 7$ Which is true. Shade the half plane containing the points. Consider $x+4y=-8$ For the $x$-intercept, put $y=0$ , \begin{align} & x+4\times 0=-8 \\ & x=-8 \end{align} The line passes through $\left( -8,0 \right)$. For the $y$ - intercept, put $x=0$ , Thus, \begin{align} & 0+4y=-8 \\ & 4y=-8 \\ & y=\frac{-8}{4} \\ & =-2 \end{align} The line passes through $\left( 0,-2 \right)$. Test that the statement is true or not by substituting $\left( 0,0 \right)$ in $x+4y>-8$ , $0+0>-8\text{ false}$ Shade the half plane containing the points. Plot the points $\left( -8,0 \right)$, $\left( 0,-2 \right)$ for the equation $x+y=7$ and the points $\left( 7,0 \right)$, $\left( 0,7 \right)$ for $x+4y=-8$ The graph of a solution of inequalities is plotted above.