# How do you solve x-y=-2, 2x+y=8 using graphing?

Aug 17, 2017

See a solution process below:

#### Explanation:

First, we can graph the first line by find two points on the line, plotting them and drawing a line through them:

For $x = 0$: $0 - y = - 2$

$- y = - 2$

$\textcolor{red}{- 1} \times - y = \textcolor{red}{- 1} \times - 2$

$y = 2$ or $\left(0 , 2\right)$

For $y = 0$: $x - 0 = - 2$

$x = - 2$ or $\left(- 2 , 0\right)$

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

We can now do the same thing for the second equation:

$2 x + y = 8$

For $x = 0$: $\left(2 \cdot 0\right) + y = 8$

$0 + y = 8$

$y = 8$ or $\left(0 , 8\right)$

For $y = 0$: $2 x + 0 = 8$

$2 x = 8$

$\frac{2 x}{\textcolor{red}{2}} = \frac{8}{\textcolor{red}{2}}$

$x = 4$ or $\left(4 , 0\right)$

graph{(x^2+(y-8)^2-0.125)((x-4)^2+y^2-0.125)(2x+y-8)(x^2+(y-2)^2-0.125)((x+2)^2+y^2-0.125)(x-y+2)=0 [-20,20,-10,10]}

We can now see where the two lines cross:

graph{(2x+y-8)(x-y+2)((x-2)^2+(y-4)^2-0.05)=0 [-5,15,-5,5]}

They cross at $x = 2$ and $y = 4$ or $\left(2 , 4\right)$