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

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

#-4(13+5y)-15y=-17#

So now we have a one variable equation which is easy to solve:
#-52-20y-15y=-17#
#-35y=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+5y# 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+5y#
And in the other one:
#x=(17-15y)/4#
We match them and we get:
#13+5y=(17-15y)/4#
And now we solve the one variable equation:

#52+20y=17-15y#
#35y=-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(-x+5y=-13)#
#4x-20y=52#

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

#0x-35y=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.