Solve {l}{x+y=8}{x^2+y^2=64}

$$\left. \begin{array} { l } { x + y = 8 } \\ { x ^ { 2 } + y ^ { 2 } = 64 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=8$$

$$x=8\text{, }y=0$$

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.

$$x+y=8,y^{2}+x^{2}=64$$

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}=64.

$$y^{2}+\left(-y+8\right)^{2}=64$$

Square -y+8.

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

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

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

Subtract 64 from both sides of the equation.

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

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 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-16\right)^{2}.

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

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

$$y=\frac{16±16}{2\times 2}$$

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

$$y=\frac{16±16}{4}$$

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

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

Divide 32 by 4.

$$y=8$$

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

$$y=\frac{0}{4}$$

Divide 0 by 4.

$$y=0$$

There are two solutions for y: 8 and 0. Substitute 8 for y in the equation x=-y+8 to find the corresponding solution for x that satisfies both equations.

$$x=-8+8$$

Add -8 to 8.

$$x=0$$

Now substitute 0 for y in the equation x=-y+8 and solve to find the corresponding solution for x that satisfies both equations.

$$x=8$$

The system is now solved.

$$x=0,y=8\text{ or }x=8,y=0$$