# Solve equations y=2x-13 and y=-2x+23 by elimination method?

Feb 2, 2017

$x = 9$ and $y = 5$

#### Explanation:

As $y = 2 x - 13$ and

$y = - 2 x + 23$

Just adding them gives $2 y = 2 x - 13 - 2 x + 23 = 10$

Hence $y = \frac{10}{2} = 5$ and putting this in first equation, we get

$5 = 2 x - 13$

or $2 x = 5 + 13 = 18$

Hence $x = \frac{18}{2} = 9$

Feb 2, 2017

$x = 9 \mathmr{and} y = 5$

#### Explanation:

$y = 2 x - 13 \mathmr{and} y = - 2 x + 23$

This is the best scenario you can get if you want to use the elimination method.

Notice that the two $x =$terms are additive inverses.

$y = \textcolor{b l u e}{2 x} - 13 \mathmr{and} y = \textcolor{b l u e}{- 2 x} + 23$

If you add the two equations together, the $x -$terms will add up to 0 and thereby be eliminated.

$\textcolor{w h i t e}{\ldots \ldots \ldots . .} y = \text{ } 2 x - 13 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} y = - 2 x + 23 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . B$

A+B:" :2y = color(blue)(0x) +10
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} y = 5$

Substitute to find $x$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} 5 = 2 x - 13 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . A$
$\textcolor{w h i t e}{\ldots . .} 5 + 13 = 2 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} 18 = 2 x$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots} 9 = x$

Check in B

$\textcolor{w h i t e}{\ldots \ldots \ldots . .} y = - 2 \left(9\right) + 23 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . B$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} y = - 18 + 23$
$\textcolor{w h i t e}{\ldots \ldots \ldots . .} y = 5$