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y-x=\frac{5\sqrt{2}}{13},x^{2}+y^{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.
y-x=\frac{5\sqrt{2}}{13}
Solve y-x=\frac{5\sqrt{2}}{13} for y by isolating y on the left hand side of the equal sign.
y=x+\frac{5\sqrt{2}}{13}
Subtract -x from both sides of the equation.
x^{2}+\left(x+\frac{5\sqrt{2}}{13}\right)^{2}=1
Substitute x+\frac{5\sqrt{2}}{13} for y in the other equation, x^{2}+y^{2}=1.
x^{2}+x^{2}+2\times \frac{5\sqrt{2}}{13}x+\left(\frac{5\sqrt{2}}{13}\right)^{2}=1
Square x+\frac{5\sqrt{2}}{13}.
2x^{2}+2\times \frac{5\sqrt{2}}{13}x+\left(\frac{5\sqrt{2}}{13}\right)^{2}=1
Add x^{2} to x^{2}.
2x^{2}+2\times \frac{5\sqrt{2}}{13}x+\left(\frac{5\sqrt{2}}{13}\right)^{2}-1=0
Subtract 1 from both sides of the equation.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\sqrt{\left(2\times \frac{5\sqrt{2}}{13}\right)^{2}-4\times 2\left(-\frac{119}{169}\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 1\times 2\times \frac{5\sqrt{2}}{13} for b, and -\frac{119}{169} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\sqrt{\frac{200}{169}-4\times 2\left(-\frac{119}{169}\right)}}{2\times 2}
Square 1\times 1\times 2\times \frac{5\sqrt{2}}{13}.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\sqrt{\frac{200}{169}-8\left(-\frac{119}{169}\right)}}{2\times 2}
Multiply -4 times 1+1\times 1^{2}.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\sqrt{\frac{200+952}{169}}}{2\times 2}
Multiply -8 times -\frac{119}{169}.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\sqrt{\frac{1152}{169}}}{2\times 2}
Add \frac{200}{169} to \frac{952}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-2\times \frac{5\sqrt{2}}{13}±\frac{24\sqrt{2}}{13}}{2\times 2}
Take the square root of \frac{1152}{169}.
x=\frac{-\frac{10\sqrt{2}}{13}±\frac{24\sqrt{2}}{13}}{4}
Multiply 2 times 1+1\times 1^{2}.
x=\frac{14\sqrt{2}}{4\times 13}
Now solve the equation x=\frac{-\frac{10\sqrt{2}}{13}±\frac{24\sqrt{2}}{13}}{4} when ± is plus. Add -\frac{10\sqrt{2}}{13} to \frac{24\sqrt{2}}{13}.
x=\frac{7\sqrt{2}}{26}
Divide \frac{14\sqrt{2}}{13} by 4.
x=-\frac{\frac{34\sqrt{2}}{13}}{4}
Now solve the equation x=\frac{-\frac{10\sqrt{2}}{13}±\frac{24\sqrt{2}}{13}}{4} when ± is minus. Subtract \frac{24\sqrt{2}}{13} from -\frac{10\sqrt{2}}{13}.
x=-\frac{17\sqrt{2}}{26}
Divide -\frac{34\sqrt{2}}{13} by 4.
y=\frac{7\sqrt{2}}{26}+\frac{5\sqrt{2}}{13}
There are two solutions for x: \frac{7\sqrt{2}}{26} and -\frac{17\sqrt{2}}{26}. Substitute \frac{7\sqrt{2}}{26} for x in the equation y=x+\frac{5\sqrt{2}}{13} to find the corresponding solution for y that satisfies both equations.
y=\frac{5\sqrt{2}}{13}+\frac{7\sqrt{2}}{26}
Add 1\times \frac{7\sqrt{2}}{26} to \frac{5\sqrt{2}}{13}.
y=-\frac{17\sqrt{2}}{26}+\frac{5\sqrt{2}}{13}
Now substitute -\frac{17\sqrt{2}}{26} for x in the equation y=x+\frac{5\sqrt{2}}{13} and solve to find the corresponding solution for y that satisfies both equations.
y=\frac{5\sqrt{2}}{13}-\frac{17\sqrt{2}}{26}
Add 1\left(-\frac{17\sqrt{2}}{26}\right) to \frac{5\sqrt{2}}{13}.
y=\frac{5\sqrt{2}}{13}+\frac{7\sqrt{2}}{26},x=\frac{7\sqrt{2}}{26}\text{ or }y=\frac{5\sqrt{2}}{13}-\frac{17\sqrt{2}}{26},x=-\frac{17\sqrt{2}}{26}
The system is now solved.