How do you solve the system x^2+y^2=81x2+y2=81 and x+y=0x+y=0?

1 Answer
mason m
Aug 26, 2016

(9/sqrt2,-9/sqrt2)(9√2,−9√2) or (-9/sqrt2,9/sqrt2)(−9√2,9√2)

Explanation:

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

You can rewrite this to be y=-xy=−x (or x=-yx=−y).

Now, replace all the yys in the more complicated equation with -x−xs.

x^2+y^2=81rArrx^2+(-x)^2=81x2+y2=81⇒x2+(−x)2=81.

x^2+(-x)^2=81rArrx^2+x^2=81x2+(−x)2=81⇒x2+x2=81.

x^2+x^2=81rArr2x^2=81x2+x2=81⇒2x2=81.

2x^2=81rArrx^2=81/2rArrx=+-9/sqrt22x2=81⇒x2=812⇒x=±9√2.

Now use the equation from before y=-xy=−x to see that the solutions are the ordered pairs (9/sqrt2,-9/sqrt2)(9√2,−9√2) or (-9/sqrt2,9/sqrt2)(−9√2,9√2).

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]}

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