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

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

Solve for x, y

$$x=4\text{, }y=2$$

$$x=-20\text{, }y=14$$

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.

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

Solve x+2y=8 for x by isolating x on the left hand side of the equal sign.

$$x+2y=8$$

Subtract 2y from both sides of the equation.

$$x=-2y+8$$

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

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

Square -2y+8.

$$-2y^{2}+4y^{2}-32y+64=8$$

Add -2y^{2} to 4y^{2}.

$$2y^{2}-32y+64=8$$

Subtract 8 from both sides of the equation.

$$2y^{2}-32y+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2+1\left(-2\right)^{2} for a, 1\times 8\left(-2\right)\times 2 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 2\times 56}}{2\times 2}$$

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

$$y=\frac{-\left(-32\right)±\sqrt{1024-4\times 2\times 56}}{2\times 2}$$

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

$$y=\frac{-\left(-32\right)±\sqrt{1024-8\times 56}}{2\times 2}$$

Multiply -8 times 56.

$$y=\frac{-\left(-32\right)±\sqrt{1024-448}}{2\times 2}$$

Add 1024 to -448.

$$y=\frac{-\left(-32\right)±\sqrt{576}}{2\times 2}$$

Take the square root of 576.

$$y=\frac{-\left(-32\right)±24}{2\times 2}$$

The opposite of 1\times 8\left(-2\right)\times 2 is 32.

$$y=\frac{32±24}{2\times 2}$$

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

$$y=\frac{32±24}{4}$$

Now solve the equation y=\frac{32±24}{4} when ± is plus. Add 32 to 24.

$$y=\frac{56}{4}$$

Divide 56 by 4.

$$y=14$$

Now solve the equation y=\frac{32±24}{4} when ± is minus. Subtract 24 from 32.

$$y=\frac{8}{4}$$

Divide 8 by 4.

$$y=2$$

There are two solutions for y: 14 and 2. Substitute 14 for y in the equation x=-2y+8 to find the corresponding solution for x that satisfies both equations.

$$x=-2\times 14+8$$

Multiply -2 times 14.

$$x=-28+8$$

Add -2\times 14 to 8.

$$x=-20$$

Now substitute 2 for y in the equation x=-2y+8 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\times 2+8$$

Multiply -2 times 2.

$$x=-4+8$$

Add -2\times 2 to 8.

$$x=4$$

The system is now solved.

$$x=-20,y=14\text{ or }x=4,y=2$$