Solve {l}{x+y=31}{x^2+y^2=25^2}

$$\left\{ \begin{array} { l } { x + y = 31 } \\ { x ^ { 2 } + y ^ { 2 } = 25 ^ { 2 } } \end{array} \right.$$

Solve for x, y

$$x=24\text{, }y=7$$

$$x=7\text{, }y=24$$

Steps Using Substitution and Quadratic Formula

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.

$$x+y=31,y^{2}+x^{2}=625$$

Solve x+y=31 for x by isolating x on the left hand side of the equal sign.

$$x+y=31$$

Subtract y from both sides of the equation.

$$x=-y+31$$

Substitute -y+31 for x in the other equation, y^{2}+x^{2}=625.

$$y^{2}+\left(-y+31\right)^{2}=625$$

Square -y+31.

$$y^{2}+y^{2}-62y+961=625$$

Add y^{2} to y^{2}.

$$2y^{2}-62y+961=625$$

Subtract 625 from both sides of the equation.

$$2y^{2}-62y+336=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 31\left(-1\right)\times 2 for b, and 336 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-62\right)±\sqrt{\left(-62\right)^{2}-4\times 2\times 336}}{2\times 2}$$

Square 1\times 31\left(-1\right)\times 2.

$$y=\frac{-\left(-62\right)±\sqrt{3844-4\times 2\times 336}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-62\right)±\sqrt{3844-8\times 336}}{2\times 2}$$

Multiply -8 times 336.

$$y=\frac{-\left(-62\right)±\sqrt{3844-2688}}{2\times 2}$$

Add 3844 to -2688.

$$y=\frac{-\left(-62\right)±\sqrt{1156}}{2\times 2}$$

Take the square root of 1156.

$$y=\frac{-\left(-62\right)±34}{2\times 2}$$

The opposite of 1\times 31\left(-1\right)\times 2 is 62.

$$y=\frac{62±34}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$y=\frac{62±34}{4}$$

Now solve the equation y=\frac{62±34}{4} when ± is plus. Add 62 to 34.

$$y=\frac{96}{4}$$

Divide 96 by 4.

$$y=24$$

Now solve the equation y=\frac{62±34}{4} when ± is minus. Subtract 34 from 62.

$$y=\frac{28}{4}$$

Divide 28 by 4.

$$y=7$$

There are two solutions for y: 24 and 7. Substitute 24 for y in the equation x=-y+31 to find the corresponding solution for x that satisfies both equations.

$$x=-24+31$$

Add -24 to 31.

$$x=7$$

Now substitute 7 for y in the equation x=-y+31 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-7+31$$

Add -7 to 31.

$$x=24$$

The system is now solved.

$$x=7,y=24\text{ or }x=24,y=7$$