Beginning Algebra With Applications, Chapter 5, 5.3, Section 5.3, Problem 54

Graph $\displaystyle 5x-2y = 10$ by using the slope and $y$-intercept.

$y$-intercept:


$
\begin{equation}
\begin{aligned}

5x-2y =& 10
&& \text{Given equation}
\\
5(0)-2y =& 10
&& \text{To find the $y$-intercept, let } x = 0
\\
-2y =& 10
&& \text{Simplify}
\\
y =& -5
&&

\end{aligned}
\end{equation}
$


The $y$-intercept is $(0,-5)$

Writing the equation in slope form, $y = mx+b$


$
\begin{equation}
\begin{aligned}

-2y =& 10-5x
\\
\\
y =& \frac{10-5x}{-2}
\\
\\
y =& \frac{5}{2}x - 5

\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}

m =& \frac{\text{change in } y}{\text{change in } x}
\\
\\
m =& \frac{5}{2}

\end{aligned}
\end{equation}
$


Beginning at the $y$-intercept, move to the right 2 units and then up 5 units.







$(2, 0)$ are the coordinates of a second point on the graph.

Draw a line through $(0,-5)$ and $(2, 0)$