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Solve for x, y
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x-y=0
Consider the second equation. Subtract y from both sides.
x-y=0,y^{2}+x^{2}=1
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
x-y=0
Solve x-y=0 for x by isolating x on the left hand side of the equal sign.
x=y
Subtract -y from both sides of the equation.
y^{2}+y^{2}=1
Substitute y for x in the other equation, y^{2}+x^{2}=1.
2y^{2}=1
Add y^{2} to y^{2}.
2y^{2}-1=0
Subtract 1 from both sides of the equation.
y=\frac{0±\sqrt{0^{2}-4\times 2\left(-1\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 0\times 1\times 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\times 2\left(-1\right)}}{2\times 2}
Square 1\times 0\times 1\times 2.
y=\frac{0±\sqrt{-8\left(-1\right)}}{2\times 2}
Multiply -4 times 1+1\times 1^{2}.
y=\frac{0±\sqrt{8}}{2\times 2}
Multiply -8 times -1.
y=\frac{0±2\sqrt{2}}{2\times 2}
Take the square root of 8.
y=\frac{0±2\sqrt{2}}{4}
Multiply 2 times 1+1\times 1^{2}.
y=\frac{\sqrt{2}}{2}
Now solve the equation y=\frac{0±2\sqrt{2}}{4} when ± is plus.
y=-\frac{\sqrt{2}}{2}
Now solve the equation y=\frac{0±2\sqrt{2}}{4} when ± is minus.
x=\frac{\sqrt{2}}{2}
There are two solutions for y: \frac{\sqrt{2}}{2} and -\frac{\sqrt{2}}{2}. Substitute \frac{\sqrt{2}}{2} for y in the equation x=y to find the corresponding solution for x that satisfies both equations.
x=-\frac{\sqrt{2}}{2}
Now substitute -\frac{\sqrt{2}}{2} for y in the equation x=y and solve to find the corresponding solution for x that satisfies both equations.
x=\frac{\sqrt{2}}{2},y=\frac{\sqrt{2}}{2}\text{ or }x=-\frac{\sqrt{2}}{2},y=-\frac{\sqrt{2}}{2}
The system is now solved.