# How do you graph the equation 2x-8y=-15?

Aug 2, 2017

See a solution below:

#### Explanation:

First, we can determine two points on the line by finding the $x$ and $y$ intercepts by setting one variable to $0$ and then solve for the other variable:

x-intercept

$2 x - \left(8 \cdot 0\right) = - 15$

$2 x - 0 = - 15$

$2 x = - 15$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{15}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = - \frac{15}{2}$

$x = - \frac{15}{2}$ or $\left(- \frac{15}{2} , 0\right)$

y-intercept

$\left(2 \cdot 0\right) - 8 y = - 15$

$0 - 8 y = - 15$

$- 8 y = - 15$

$\frac{- 8 y}{\textcolor{red}{- 8}} = \frac{- 15}{\textcolor{red}{- 8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 8}}} y}{\cancel{\textcolor{red}{- 8}}} = \frac{15}{8}$

$y = \frac{15}{8}$ or $\left(0 , \frac{15}{8}\right)$

We can next plot these two points:

graph{((x+ (15/2))^2+(y)^2-0.125)((x)^2+(y-(15/8))^2-0.125)=0 [-20, 20, -10, 10]}

Now, we can draw a line through the two points giving:

graph{(2x - 8y + 15)((x+ (15/2))^2+(y)^2-0.125)((x)^2+(y-(15/8))^2-0.125)=0 [-20, 20, -10, 10]}