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x-y=2,y^{2}+x^{2}=100
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=2
Solve x-y=2 for x by isolating x on the left hand side of the equal sign.
x=y+2
Subtract -y from both sides of the equation.
y^{2}+\left(y+2\right)^{2}=100
Substitute y+2 for x in the other equation, y^{2}+x^{2}=100.
y^{2}+y^{2}+4y+4=100
Square y+2.
2y^{2}+4y+4=100
Add y^{2} to y^{2}.
2y^{2}+4y-96=0
Subtract 100 from both sides of the equation.
y=\frac{-4±\sqrt{4^{2}-4\times 2\left(-96\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 2\times 1\times 2 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-4±\sqrt{16-4\times 2\left(-96\right)}}{2\times 2}
Square 1\times 2\times 1\times 2.
y=\frac{-4±\sqrt{16-8\left(-96\right)}}{2\times 2}
Multiply -4 times 1+1\times 1^{2}.
y=\frac{-4±\sqrt{16+768}}{2\times 2}
Multiply -8 times -96.
y=\frac{-4±\sqrt{784}}{2\times 2}
Add 16 to 768.
y=\frac{-4±28}{2\times 2}
Take the square root of 784.
y=\frac{-4±28}{4}
Multiply 2 times 1+1\times 1^{2}.
y=\frac{24}{4}
Now solve the equation y=\frac{-4±28}{4} when ± is plus. Add -4 to 28.
y=6
Divide 24 by 4.
y=-\frac{32}{4}
Now solve the equation y=\frac{-4±28}{4} when ± is minus. Subtract 28 from -4.
y=-8
Divide -32 by 4.
x=6+2
There are two solutions for y: 6 and -8. Substitute 6 for y in the equation x=y+2 to find the corresponding solution for x that satisfies both equations.
x=8
Add 1\times 6 to 2.
x=-8+2
Now substitute -8 for y in the equation x=y+2 and solve to find the corresponding solution for x that satisfies both equations.
x=-6
Add -8 to 2.
x=8,y=6\text{ or }x=-6,y=-8
The system is now solved.