# How do you solve the following system: -3x + y = -4, 4x + 2y = 0 ?

Mar 16, 2016

$x = \frac{4}{5}$ and $y = - \frac{8}{5}$

#### Explanation:

we begin with $- 3 x + y = - 4$ and $4 x + 2 y = 0$

Now, I'm going to solve for $y$ in $- 3 x + y = - 4$ which looks like $y = - 4 + 3 x$.

So no we can replace the $y$ in the second equation with $- 4 + 3 x$.

That gives us $4 x + 2 \left(- 4 + 3 x\right) = 0$ which we can simplify to $4 x - 8 + 6 x = 0$. This becomes $10 x = 8$ or $x = \frac{4}{5}$.

No, we can solve for $y$.

$y = - 4 + 3 \left(\frac{4}{5}\right)$, which is $- 4 + \frac{12}{5}$ or $- \frac{8}{5}$.

Now we should double check our math. If we did this right then $4 \left(\frac{4}{5}\right) + 2 \left(- \frac{8}{5}\right)$ should equal $0$. Let's see if it is: $\frac{16}{5} - \frac{16}{5}$ does in fact equal $0$, which means that we were right!

Nice Job!