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x^{2}+y^{2}=3266.1225
Consider the first equation. Calculate 57.15 to the power of 2 and get 3266.1225.
16x=y\times 10
Consider the second equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 16y, the least common multiple of y,16.
16x-y\times 10=0
Subtract y\times 10 from both sides.
16x-10y=0
Multiply -1 and 10 to get -10.
16x-10y=0,y^{2}+x^{2}=3266.1225
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.
16x-10y=0
Solve 16x-10y=0 for x by isolating x on the left hand side of the equal sign.
16x=10y
Subtract -10y from both sides of the equation.
x=\frac{5}{8}y
Divide both sides by 16.
y^{2}+\left(\frac{5}{8}y\right)^{2}=3266.1225
Substitute \frac{5}{8}y for x in the other equation, y^{2}+x^{2}=3266.1225.
y^{2}+\frac{25}{64}y^{2}=3266.1225
Square \frac{5}{8}y.
\frac{89}{64}y^{2}=3266.1225
Add y^{2} to \frac{25}{64}y^{2}.
\frac{89}{64}y^{2}-3266.1225=0
Subtract 3266.1225 from both sides of the equation.
y=\frac{0±\sqrt{0^{2}-4\times \frac{89}{64}\left(-3266.1225\right)}}{2\times \frac{89}{64}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{5}{8}\right)^{2} for a, 1\times 0\times \frac{5}{8}\times 2 for b, and -3266.1225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{0±\sqrt{-4\times \frac{89}{64}\left(-3266.1225\right)}}{2\times \frac{89}{64}}
Square 1\times 0\times \frac{5}{8}\times 2.
y=\frac{0±\sqrt{-\frac{89}{16}\left(-3266.1225\right)}}{2\times \frac{89}{64}}
Multiply -4 times 1+1\times \left(\frac{5}{8}\right)^{2}.
y=\frac{0±\sqrt{\frac{116273961}{6400}}}{2\times \frac{89}{64}}
Multiply -\frac{89}{16} times -3266.1225 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
y=\frac{0±\frac{1143\sqrt{89}}{80}}{2\times \frac{89}{64}}
Take the square root of \frac{116273961}{6400}.
y=\frac{0±\frac{1143\sqrt{89}}{80}}{\frac{89}{32}}
Multiply 2 times 1+1\times \left(\frac{5}{8}\right)^{2}.
y=\frac{2286\sqrt{89}}{445}
Now solve the equation y=\frac{0±\frac{1143\sqrt{89}}{80}}{\frac{89}{32}} when ± is plus.
y=-\frac{2286\sqrt{89}}{445}
Now solve the equation y=\frac{0±\frac{1143\sqrt{89}}{80}}{\frac{89}{32}} when ± is minus.
x=\frac{5}{8}\times \frac{2286\sqrt{89}}{445}
There are two solutions for y: \frac{2286\sqrt{89}}{445} and -\frac{2286\sqrt{89}}{445}. Substitute \frac{2286\sqrt{89}}{445} for y in the equation x=\frac{5}{8}y to find the corresponding solution for x that satisfies both equations.
x=\frac{5\times \frac{2286\sqrt{89}}{445}}{8}
Multiply \frac{5}{8} times \frac{2286\sqrt{89}}{445}.
x=\frac{5}{8}\left(-\frac{2286\sqrt{89}}{445}\right)
Now substitute -\frac{2286\sqrt{89}}{445} for y in the equation x=\frac{5}{8}y and solve to find the corresponding solution for x that satisfies both equations.
x=\frac{5\left(-\frac{2286\sqrt{89}}{445}\right)}{8}
Multiply \frac{5}{8} times -\frac{2286\sqrt{89}}{445}.
x=\frac{5\times \frac{2286\sqrt{89}}{445}}{8},y=\frac{2286\sqrt{89}}{445}\text{ or }x=\frac{5\left(-\frac{2286\sqrt{89}}{445}\right)}{8},y=-\frac{2286\sqrt{89}}{445}
The system is now solved.