Question

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

Answer 1

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.