# How do you solve the system of equations y+2x=3 and -y+4=-x?

Dec 2, 2016

$x = - \frac{1}{3}$ and $y = \frac{11}{3}$ or $3 \frac{2}{3}$

#### Explanation:

$y + 2 x = 3$
$- y + 4 = - x$

In the first equation, we can determine a value for $y$.

$y + 2 x = 3$

Subtract $2 x$ from both sides.

$y = 3 - 2 x$

In the second equation, substitute $y$ with $\textcolor{red}{\left(3 - 2 x\right)}$.

$- \textcolor{red}{\left(3 - 2 x\right)} + 4 = - x$

Open the brackets and simplify. The product of two negatives is a positive, and that of a negative and positive is a negative.

$- 3 + 2 x + 4 = - x$

$2 x + 1 = - x$

Add $x$ to both sides.

$3 x + 1 = 0$

Subtract $1$ from both sides.

$3 x = - 1$

Divide both sides by $3$.

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

In the first equation, substitute $x$ with $- \frac{1}{3}$.

$y + 2 x = 3$

$y + 2 \left(- \frac{1}{3}\right) = 3$

Open the brackets and simplify.

$y - \frac{2}{3} = 3$

Add $\frac{2}{3}$ to both sides.

$y = 3 + \frac{2}{3}$

$y = 3 \frac{2}{3}$ or $y = \frac{11}{3}$