Solve {l}{x+y=8}{x^2+y^2=50}

$$\left. \begin{array} { l } { x + y = 8 } \\ { x ^ { 2 } + y ^ { 2 } = 50 } \end{array} \right.$$

Solve for x, y

$$x=7\text{, }y=1$$

$$x=1\text{, }y=7$$

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.

$$x+y=8,y^{2}+x^{2}=50$$

Solve x+y=8 for x by isolating x on the left hand side of the equal sign.

$$x+y=8$$

Subtract y from both sides of the equation.

$$x=-y+8$$

Substitute -y+8 for x in the other equation, y^{2}+x^{2}=50.

$$y^{2}+\left(-y+8\right)^{2}=50$$

Square -y+8.

$$y^{2}+y^{2}-16y+64=50$$

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

$$2y^{2}-16y+64=50$$

Subtract 50 from both sides of the equation.

$$2y^{2}-16y+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 8\left(-1\right)\times 2 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 14}}{2\times 2}$$

Square 1\times 8\left(-1\right)\times 2.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 14}}{2\times 2}$$

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

$$y=\frac{-\left(-16\right)±\sqrt{256-8\times 14}}{2\times 2}$$

Multiply -8 times 14.

$$y=\frac{-\left(-16\right)±\sqrt{256-112}}{2\times 2}$$

Add 256 to -112.

$$y=\frac{-\left(-16\right)±\sqrt{144}}{2\times 2}$$

Take the square root of 144.

$$y=\frac{-\left(-16\right)±12}{2\times 2}$$

The opposite of 1\times 8\left(-1\right)\times 2 is 16.

$$y=\frac{16±12}{2\times 2}$$

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

$$y=\frac{16±12}{4}$$

Now solve the equation y=\frac{16±12}{4} when ± is plus. Add 16 to 12.

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

Divide 28 by 4.

$$y=7$$

Now solve the equation y=\frac{16±12}{4} when ± is minus. Subtract 12 from 16.

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

Divide 4 by 4.

$$y=1$$

There are two solutions for y: 7 and 1. Substitute 7 for y in the equation x=-y+8 to find the corresponding solution for x that satisfies both equations.

$$x=-7+8$$

Add -7 to 8.

$$x=1$$

Now substitute 1 for y in the equation x=-y+8 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-1+8$$

Add -1 to 8.

$$x=7$$

The system is now solved.

$$x=1,y=7\text{ or }x=7,y=1$$