Graph the linear equation by finding and plotting its intercepts.
$5x - 3y = 15$
Solve for $x$:
$\begin{aligned}
5x - 3y &= 15 \\
5x - 3(0) &= 15 \quad \text{Let } y = 0. \\
5x &= 15 \quad \text{Simplify.} \\
x &= 3 \quad \text{Divide.}
\end{aligned}$
The $x$-intercept is $(3, 0)$. The graph of $5x - 3y = 15$ passes through the point $(3, 0)$.
A $y$-intercept of a graph is a point where the graph intersects the $y$-axis. To find the $y$-intercept, let $x = 0$ and solve for $y$:
$\begin{aligned}
5x - 3y &= 15 \\
5(0) - 3y &= 15 \quad \text{Let } x = 0. \\
-3y &= 15 \quad \text{Simplify.} \\
y &= \square \quad \text{Divide.}
\end{aligned}$
$5x - 3y = 15$
Solve for $x$:
$\begin{aligned}
5x - 3y &= 15 \\
5x - 3(0) &= 15 \quad \text{Let } y = 0. \\
5x &= 15 \quad \text{Simplify.} \\
x &= 3 \quad \text{Divide.}
\end{aligned}$
The $x$-intercept is $(3, 0)$. The graph of $5x - 3y = 15$ passes through the point $(3, 0)$.
A $y$-intercept of a graph is a point where the graph intersects the $y$-axis. To find the $y$-intercept, let $x = 0$ and solve for $y$:
$\begin{aligned}
5x - 3y &= 15 \\
5(0) - 3y &= 15 \quad \text{Let } x = 0. \\
-3y &= 15 \quad \text{Simplify.} \\
y &= \square \quad \text{Divide.}
\end{aligned}$