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

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

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Answer 1 · 2016-08-26T02:44:58

$(\frac{9}{\sqrt{2}}, - \frac{9}{\sqrt{2}})$ $(9/sqrt2,-9/sqrt2)$ or $(- \frac{9}{\sqrt{2}}, \frac{9}{\sqrt{2}})$ $(-9/sqrt2,9/sqrt2)$

Explanation:

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

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

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

$x^{2} + y^{2} = 81 \Rightarrow x^{2} + {(- x)}^{2} = 81$ $x^2+y^2=81rArrx^2+(-x)^2=81$ .

$x^{2} + {(- x)}^{2} = 81 \Rightarrow x^{2} + x^{2} = 81$ $x^2+(-x)^2=81rArrx^2+x^2=81$ .

$x^{2} + x^{2} = 81 \Rightarrow 2 x^{2} = 81$ $x^2+x^2=81rArr2x^2=81$ .

$2 x^{2} = 81 \Rightarrow x^{2} = \frac{81}{2} \Rightarrow x = \pm \frac{9}{\sqrt{2}}$ $2x^2=81rArrx^2=81/2rArrx=+-9/sqrt2$ .

Now use the equation from before $y = - x$ $y=-x$ to see that the solutions are the ordered pairs $(\frac{9}{\sqrt{2}}, - \frac{9}{\sqrt{2}})$ $(9/sqrt2,-9/sqrt2)$ or $(- \frac{9}{\sqrt{2}}, \frac{9}{\sqrt{2}})$ $(-9/sqrt2,9/sqrt2)$ .

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]}

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

1 Answer

Explanation:

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