# How do you solve the system -4x-15y=-17 and -x+5y=-13?

Jun 2, 2015

You can solve it by several methods, let's see them:

Take a variable and separate it, in this case will be easier if we take the $x$ from the second equation:

$x = 13 + 5 y$ (changing signs)
Now we replace the $x$ we found in the other equation in order to see which $y$ accomplishes the equality:

$- 4 \left(13 + 5 y\right) - 15 y = - 17$

So now we have a one variable equation which is easy to solve:
$- 52 - 20 y - 15 y = - 17$
$- 35 y = 35$
$y = - 1$

Now we've got the $y$, let's calculate the $x$ which follows the system with such a $y$ replacing our $y$ in the previous isolated equation:
$x = 13 + 5 y$ with $y = - 1$
$x = 13 - 5 = 7$

You can solve the system by separating the same variable in each equation and matching them, we've seen before:
$x = 13 + 5 y$
And in the other one:
$x = \frac{17 - 15 y}{4}$
We match them and we get:
$13 + 5 y = \frac{17 - 15 y}{4}$
And now we solve the one variable equation:

$52 + 20 y = 17 - 15 y$
$35 y = - 35$
$y = - 1$
Replace that $y$ you found in any equation and you'll find $x = 7$

Instead of those two methods, you can reduce the system, if you don't know how to reduce a matrix by Gauss' method, you just need to multiply one or both equations by a number which, the sum of those will result 0 for one variable, let's see it:

$- 4 \left(- x + 5 y = - 13\right)$
$4 x - 20 y = 52$

If you now sum this equation to the other one, you will notice that:

$0 x - 35 y = 35$

Which gives again $y = - 1$ and replacing this $y$ in any equation, you'll fin $x = 7$.

Any of these methods will give you the same result in case the system has a unique solution, and of course you can apply the method swapping the variables, isolating y, or reducing y or whatever.