# How do you solve the system of equations -8x + 4y = - 12 and - x - y = - 3?

Jun 8, 2017

$x = 2$
$y = 1$

#### Explanation:

Equation 1: $- 8 x + 4 y = - 12$
Equation 2: $- x - y = - 3$

Because its easier to work with positive values, I'm going to multiple both sides of equation 2 by -1.

$x + y = 3$

I'm going to make $y$ the subject of equation 2 by subtracting $x$ from both sides so I can substitute it into equation 1.

$y = 3 - x$

Substitute this value for $y$ into equation 1.

$- 8 x + 4 \left(3 - x\right) = - 12$

Solve for $x$. Start by expanding the parenthesis.

$- 8 x + 12 - 4 x = - 12$

Then combine the like terms $- 8 x$ and $- 4 x$.

$- 12 x + 12 = - 12$

Subtract 12 from both sides of the equation.

$- 12 x = - 24$

Multiply both sides of the equation by -1 to make them positive.

$12 x = 24$

Divide both sides of the equation by 12 to leave $x$.

$x = 2$

Now we've found $x$ and need to find $y$. Remember earlier how we found that $y = 3 - x$? All we have to do is substitute our value for $x$ into this equation to find $y$.

$y = 3 - \left(2\right)$
$y = 1$