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Solve the equations: x^2+y^2=169, x-y-13=0. . . . . ANSWER = {(13,0),(0,-13)}
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x = y + 13
\[(y + 13)^2 = y^2 + 26y +169\]
write x^2+y^2 = (x-y)^2 +2xy
so: \[2y^2 + 26 +169 = 169\]
\[2y(y+13) = 0\]
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forgot a y two posts up
y = 0, or 13
the rest is trivial
negative 13
typos :(
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how did 2y(y+13)=0
you cancel 169, and factor out a 2y
i am not gettin it
2y^2+26+169=169 HOW!!!!!
\[(y+13)^2 + y^2 = 169\] \[(y^2 + 26y + 169) + y^2 = 169\] \[2y^2 + 26 + 169 = 169\]
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what about the values of x? how will i get those
then you just use the second equation
if y = 0: \[x-(0)-13 = 0\]
then x = 13
exactly
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so the ordered pair is (13,0)
ok done@ pretty easy method of solving thanks
Yeah that's right @MrMoose It's correct
You are welcome
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