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

1 Answer
Eric Sandin
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+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=17−15y4
We match them and we get:
13+5y=17−15y4
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.

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