# How do you solve the following system?: -6x -3y =1, -2x +4y = -5

Feb 7, 2016

$x = \frac{11}{30} , y = - \frac{16}{15}$

#### Explanation:

We have the equations:
$- 6 x - 3 y = 1$ and
$- 2 x + 4 y = - 5$

I would say the best way to procede is to eliminate the $x$ terms. Multiply the 2nd equation by 3 so that the coefficient of $x$ on the 2nd equation is the same as the coefficient of $x$ on the first equation giving:

That means the 2nd equation becomes:

$- 6 x + 12 y = - 15$

Now subtract the equations from each other to eliminate the $x$ terms and get:

$\left(- 6 x - 3 y\right) - \left(- 6 x + 12 y\right) = 1 - \left(- 15\right)$
$\to - 15 y = 16 \to y = - \frac{16}{15}$

Now put this value of $y$ back into either of the original equations to find $x$, we'll choose the first equation:

$- 6 x - 3 \left(- \frac{16}{15}\right) = 1$
$- 6 x + \frac{16}{5} = 1$
$\to - 6 x = 1 - \frac{16}{5} = - \frac{11}{5}$
Thus:
$x = \frac{11}{30}$

Hence our final solution. It is good practice to put these values into the other equation to check that they work.