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x+y=\sqrt{26},y^{2}+x^{2}=16
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=\sqrt{26}
Solve x+y=\sqrt{26} for x by isolating x on the left hand side of the equal sign.
x=-y+\sqrt{26}
Subtract y from both sides of the equation.
y^{2}+\left(-y+\sqrt{26}\right)^{2}=16
Substitute -y+\sqrt{26} for x in the other equation, y^{2}+x^{2}=16.
y^{2}+y^{2}+\left(-2\sqrt{26}\right)y+\left(\sqrt{26}\right)^{2}=16
Square -y+\sqrt{26}.
2y^{2}+\left(-2\sqrt{26}\right)y+\left(\sqrt{26}\right)^{2}=16
Add y^{2} to y^{2}.
2y^{2}+\left(-2\sqrt{26}\right)y+\left(\sqrt{26}\right)^{2}-16=0
Subtract 16 from both sides of the equation.
y=\frac{-\left(-2\sqrt{26}\right)±\sqrt{\left(-2\sqrt{26}\right)^{2}-4\times 2\times 10}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\left(-1\right)\times 2\sqrt{26} for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-\left(-2\sqrt{26}\right)±\sqrt{104-4\times 2\times 10}}{2\times 2}
Square 1\left(-1\right)\times 2\sqrt{26}.
y=\frac{-\left(-2\sqrt{26}\right)±\sqrt{104-8\times 10}}{2\times 2}
Multiply -4 times 1+1\left(-1\right)^{2}.
y=\frac{-\left(-2\sqrt{26}\right)±\sqrt{104-80}}{2\times 2}
Multiply -8 times 10.
y=\frac{-\left(-2\sqrt{26}\right)±\sqrt{24}}{2\times 2}
Add 104 to -80.
y=\frac{-\left(-2\sqrt{26}\right)±2\sqrt{6}}{2\times 2}
Take the square root of 24.
y=\frac{2\sqrt{26}±2\sqrt{6}}{2\times 2}
The opposite of 1\left(-1\right)\times 2\sqrt{26} is 2\sqrt{26}.
y=\frac{2\sqrt{26}±2\sqrt{6}}{4}
Multiply 2 times 1+1\left(-1\right)^{2}.
y=\frac{2\sqrt{6}+2\sqrt{26}}{4}
Now solve the equation y=\frac{2\sqrt{26}±2\sqrt{6}}{4} when ± is plus. Add 2\sqrt{26} to 2\sqrt{6}.
y=\frac{\sqrt{6}+\sqrt{26}}{2}
Divide 2\sqrt{26}+2\sqrt{6} by 4.
y=\frac{2\sqrt{26}-2\sqrt{6}}{4}
Now solve the equation y=\frac{2\sqrt{26}±2\sqrt{6}}{4} when ± is minus. Subtract 2\sqrt{6} from 2\sqrt{26}.
y=\frac{\sqrt{26}-\sqrt{6}}{2}
Divide 2\sqrt{26}-2\sqrt{6} by 4.
x=-\frac{\sqrt{6}+\sqrt{26}}{2}+\sqrt{26}
There are two solutions for y: \frac{\sqrt{26}+\sqrt{6}}{2} and \frac{\sqrt{26}-\sqrt{6}}{2}. Substitute \frac{\sqrt{26}+\sqrt{6}}{2} for y in the equation x=-y+\sqrt{26} to find the corresponding solution for x that satisfies both equations.
x=-\frac{\sqrt{26}-\sqrt{6}}{2}+\sqrt{26}
Now substitute \frac{\sqrt{26}-\sqrt{6}}{2} for y in the equation x=-y+\sqrt{26} and solve to find the corresponding solution for x that satisfies both equations.
x=-\frac{\sqrt{6}+\sqrt{26}}{2}+\sqrt{26},y=\frac{\sqrt{6}+\sqrt{26}}{2}\text{ or }x=-\frac{\sqrt{26}-\sqrt{6}}{2}+\sqrt{26},y=\frac{\sqrt{26}-\sqrt{6}}{2}
The system is now solved.