We will solve this system by substitution.
First, isolate the variable x for the second equation to get \(x=15+3y\)
Next, for \(x^2+y^2=25\), substitute x with \(15+3y\) that we obtained from the first question.
We get \((3y+15)^2 +y^2=25\).
We expand the equation to get \(10y^2+90y+225=25\)
Subtract 25 from both sides \(10y^2+90y+200=0\)
Solve with the quadratic formula
Simplify (it seems like a lot, but it actually wasn't so much computation)
After I found y, you should be able to plug it into the equation x-3y = 15, I don't have time to do that since I have to go, but it should be quite easy. :)
y = -5 = \(\frac{-90\pm \sqrt{90^2-4\cdot \:10\cdot \:200}}{2\cdot \:10}\)
y = -4 = \(\frac{-90\pm \sqrt{90^2-4\cdot \:10\cdot \:200}}{2\cdot \:10}\)
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