Solve {l}{x+y=20}{x^2+y^2=208}

$$\left\{ \begin{array} { l } { x + y = 20 } \\ { x ^ { 2 } + y ^ { 2 } = 208 } \end{array} \right.$$

Solve for x, y

$$x=12\text{, }y=8$$

$$x=8\text{, }y=12$$

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.

$$x+y=20,y^{2}+x^{2}=208$$

Solve x+y=20 for x by isolating x on the left hand side of the equal sign.

$$x+y=20$$

Subtract y from both sides of the equation.

$$x=-y+20$$

Substitute -y+20 for x in the other equation, y^{2}+x^{2}=208.

$$y^{2}+\left(-y+20\right)^{2}=208$$

Square -y+20.

$$y^{2}+y^{2}-40y+400=208$$

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

$$2y^{2}-40y+400=208$$

Subtract 208 from both sides of the equation.

$$2y^{2}-40y+192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 20\left(-1\right)\times 2 for b, and 192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 2\times 192}}{2\times 2}$$

Square 1\times 20\left(-1\right)\times 2.

$$y=\frac{-\left(-40\right)±\sqrt{1600-4\times 2\times 192}}{2\times 2}$$

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

$$y=\frac{-\left(-40\right)±\sqrt{1600-8\times 192}}{2\times 2}$$

Multiply -8 times 192.

$$y=\frac{-\left(-40\right)±\sqrt{1600-1536}}{2\times 2}$$

Add 1600 to -1536.

$$y=\frac{-\left(-40\right)±\sqrt{64}}{2\times 2}$$

Take the square root of 64.

$$y=\frac{-\left(-40\right)±8}{2\times 2}$$

The opposite of 1\times 20\left(-1\right)\times 2 is 40.

$$y=\frac{40±8}{2\times 2}$$

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

$$y=\frac{40±8}{4}$$

Now solve the equation y=\frac{40±8}{4} when ± is plus. Add 40 to 8.

$$y=\frac{48}{4}$$

Divide 48 by 4.

$$y=12$$

Now solve the equation y=\frac{40±8}{4} when ± is minus. Subtract 8 from 40.

$$y=\frac{32}{4}$$

Divide 32 by 4.

$$y=8$$

There are two solutions for y: 12 and 8. Substitute 12 for y in the equation x=-y+20 to find the corresponding solution for x that satisfies both equations.

$$x=-12+20$$

Add -12 to 20.

$$x=8$$

Now substitute 8 for y in the equation x=-y+20 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-8+20$$

Add -8 to 20.

$$x=12$$

The system is now solved.

$$x=8,y=12\text{ or }x=12,y=8$$