# How do you solve x=8-2y and x=-3y+12?

Jun 12, 2018

First solve for $y$ using substitution, then solve for $x$, you will find that $y = 4$ and $x = 0$

#### Explanation:

We're very lucky for this problem, as both expressions leave $x$ by itself on one side. If you wanted to write this as one long expression:

$8 - 2 y = x = - 3 y + 12$

Now, we can ignore $x$ and write the expression as follows (BECAUSE both expressions equal $x$):

$8 - 2 y = - 3 y + 12$

Since $x$ is not in either equation, we can solve for $y$ easily:

$\cancel{8} - 2 y \textcolor{red}{+ 3 y} \textcolor{b l u e}{\cancel{- 8}} = \cancel{- 3 y} + 12 \textcolor{red}{\cancel{+ 3 y}} \textcolor{b l u e}{- 8}$

color(green)(y=4

Finally, we can solve for $x$:

$x = 8 - 2 y$

$x = 8 - 2 \left(4\right)$

$x = 8 - 8$

$\textcolor{g r e e n}{x = 0}$

Jun 12, 2018

$\left(x , y\right) \to \left(0 , 4\right)$

#### Explanation:

$x = 8 - 2 y \to \left(1\right)$

$x = - 3 y + 12 \to \left(2\right)$

$\text{since both equations express x in terms of y , equate}$
$\text{the right sides}$

$8 - 2 y = - 3 y + 12$

$\text{add "3y" to both sides}$

$8 - 2 y + 3 y = 12$

$\text{subtract 8 from both sides}$

$y = 12 - 8 = 4$

$\text{substitute "y=4" into either of the 2 equations}$

$\left(1\right) \to x = 8 - 8 = 0$

$\text{the point of intersection } = \left(0 , 4\right)$
graph{(y+1/2x-4)(y+1/3x-4)=0 [-10, 10, -5, 5]}