# How do you solve x + y =1 and 2x-y=-2 using substitution?

##### 1 Answer
Jul 10, 2016

By eliminating $x$ in the expression $x + y = 1$, we get

$x = 1 - y$, which we can plug in the second expression:

$2 \left(1 - y\right) - y = - 2$

$2 - 2 y - y = - 2$

$2 - 3 y = - 2$

$- 3 y + 2 = - 2$

$- 3 y = - 4$

$y = \frac{4}{3}$

Knowing $y$, we can now find $x$ by plugging the $y$-value into any of the above equations. For example, by plugging $y = 4$ into the first equation, $x + y = 1$, we get

$x + \frac{4}{3} = 1$

$x = - \frac{1}{3}$

#### Explanation:

We can check whether these values satisfy the equations by plugging them back in:

First: $x + y = 1$

$- \frac{1}{3} + \frac{4}{3} = 1 \to \frac{4}{3} - \frac{1}{3} = 1 \to \frac{12 - 3}{9} = 1 \to \frac{9}{9} = 1$

Second: $2 x - y = - 2$

$2 \left(- \frac{1}{3}\right) - \frac{4}{3} = - 2 \to - \frac{2}{3} - \frac{4}{3} = - 2 \to - \frac{18}{9} = - 2$