How do you solve the system #x^2+y^2=81# and #x+y=0#?

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Answer 1 · 2016-08-26T02:44:58

#(9/sqrt2,-9/sqrt2)# or #(-9/sqrt2,9/sqrt2)#

Start with the simpler equation #x+y=0#.

You can rewrite this to be #y=-x# (or #x=-y#).

Now, replace all the #y#s in the more complicated equation with #-x#s.

#x^2+y^2=81rArrx^2+(-x)^2=81#.

#x^2+(-x)^2=81rArrx^2+x^2=81#.

#x^2+x^2=81rArr2x^2=81#.

#2x^2=81rArrx^2=81/2rArrx=+-9/sqrt2#.

Now use the equation from before #y=-x# to see that the solutions are the ordered pairs #(9/sqrt2,-9/sqrt2)# or #(-9/sqrt2,9/sqrt2)#.

You can also see this by graphing both equations and seeing their points of intersection.

graph{(x+y)(x^2+y^2-81)=0 [-40, 40, -20, 20]}