+569 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 = \(\frac{-90\pm \sqrt{90^2-4\cdot \:10\cdot \:200}}{2\cdot \:10}\)
+118667 Nice answer Iamhappy, (although you did not take 25 from both sides)
You do not need to use the quadratic formula at the end though. It is not necessary.
A quick sketch would help too.
The asker can finish it by him/herself now anyway. 
It is better if they do some of their own thinking.
+569 Thank you Melody for clarifying my slightly hurried and slightly incorrect answer. :) I was about to go to online class. :) I really needed some more reliance since I can't always rely on myself. And yes you are correct that Guest should be able to finish it him/herself. \(:)\)
