Intermediate Algebra, Chapter 3, 3.1, Section 3.1, Problem 42

Determine the $x$- and $y$-intercepts. Then graph the equation $\displaystyle \frac{5}{7}x + \frac{6}{7}y = -2$.

To find $x$-intercept, we set $y = 0$


$
\begin{equation}
\begin{aligned}

\frac{5}{7}x + \frac{6}{7}(0) =& -2
\\
\\
\frac{5}{7}x =& -2
\\
\\
x =& -2 \left( \frac{7}{5} \right)
\\
\\
x =& - \frac{14}{5}

\end{aligned}
\end{equation}
$


To find $y$-intercept, we set $x = 0$


$
\begin{equation}
\begin{aligned}

\frac{5}{7}(0) + \frac{6}{7} y
\\
\\
\frac{6}{7}y =& -2
\\
\\
y =& -2 \left( \frac{7}{6} \right)
\\
\\
y =& - \frac{7}{3}

\end{aligned}
\end{equation}
$


The line $\displaystyle \frac{5}{7}x + \frac{6}{7}y = -2$ passes through the points $\displaystyle \left( - \frac{14}{5}, 0 \right)$ and $\displaystyle \left( 0, - \frac{7}{3} \right)$. So the graph is