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

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

Solve for x, y

$$x=2\sqrt{6}-2\approx 2.898979486\text{, }y=2-\sqrt{6}\approx -0.449489743$$

$$x=-2\sqrt{6}-2\approx -6.898979486\text{, }y=\sqrt{6}+2\approx 4.449489743$$

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.

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

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

$$x+2y=2$$

Subtract 2y from both sides of the equation.

$$x=-2y+2$$

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

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

Square -2y+2.

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

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

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

Subtract 8 from both sides of the equation.

$$2y^{2}-8y-4=0$$

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

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-4\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-8\right)±\sqrt{64-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$y=\frac{-\left(-8\right)±\sqrt{64+32}}{2\times 2}$$

Add 64 to 32.

$$y=\frac{-\left(-8\right)±\sqrt{96}}{2\times 2}$$

Take the square root of 96.

$$y=\frac{-\left(-8\right)±4\sqrt{6}}{2\times 2}$$

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

$$y=\frac{8±4\sqrt{6}}{2\times 2}$$

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

$$y=\frac{8±4\sqrt{6}}{4}$$

Now solve the equation y=\frac{8±4\sqrt{6}}{4} when ± is plus. Add 8 to 4\sqrt{6}.

$$y=\frac{4\sqrt{6}+8}{4}$$

Divide 8+4\sqrt{6} by 4.

$$y=\sqrt{6}+2$$

Now solve the equation y=\frac{8±4\sqrt{6}}{4} when ± is minus. Subtract 4\sqrt{6} from 8.

$$y=\frac{8-4\sqrt{6}}{4}$$

Divide 8-4\sqrt{6} by 4.

$$y=2-\sqrt{6}$$

There are two solutions for y: 2+\sqrt{6} and 2-\sqrt{6}. Substitute 2+\sqrt{6} for y in the equation x=-2y+2 to find the corresponding solution for x that satisfies both equations.

$$x=-2\left(\sqrt{6}+2\right)+2$$

Now substitute 2-\sqrt{6} for y in the equation x=-2y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\left(2-\sqrt{6}\right)+2$$

The system is now solved.

$$x=-2\left(\sqrt{6}+2\right)+2,y=\sqrt{6}+2\text{ or }x=-2\left(2-\sqrt{6}\right)+2,y=2-\sqrt{6}$$