Solve {l}{x^2/4+y^2/8=1}{x=y}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 8 } = 1 } \\ { x = y } \end{array} \right.$$

Solve for x, y

$$x=-\frac{2\sqrt{6}}{3}\approx -1.632993162\text{, }y=-\frac{2\sqrt{6}}{3}\approx -1.632993162$$

$$x=\frac{2\sqrt{6}}{3}\approx 1.632993162\text{, }y=\frac{2\sqrt{6}}{3}\approx 1.632993162$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 4,8.

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

Consider the second equation. Subtract y from both sides.

$$x-y=0$$

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.

$$x-y=0,y^{2}+2x^{2}=8$$

Solve x-y=0 for x by isolating x on the left hand side of the equal sign.

$$x-y=0$$

Subtract -y from both sides of the equation.

$$x=y$$

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

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

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

$$3y^{2}=8$$

Subtract 8 from both sides of the equation.

$$3y^{2}-8=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times 3\left(-8\right)}}{2\times 3}$$

Square 2\times 0\times 1\times 2.

$$y=\frac{0±\sqrt{-4\times 3\left(-8\right)}}{2\times 3}$$

Multiply -4 times 1+2\times 1^{2}.

$$y=\frac{0±\sqrt{-12\left(-8\right)}}{2\times 3}$$

Multiply -12 times -8.

$$y=\frac{0±\sqrt{96}}{2\times 3}$$

Take the square root of 96.

$$y=\frac{0±4\sqrt{6}}{2\times 3}$$

Multiply 2 times 1+2\times 1^{2}.

$$y=\frac{0±4\sqrt{6}}{6}$$

Now solve the equation y=\frac{0±4\sqrt{6}}{6} when ± is plus.

$$y=\frac{2\sqrt{6}}{3}$$

Now solve the equation y=\frac{0±4\sqrt{6}}{6} when ± is minus.

$$y=-\frac{2\sqrt{6}}{3}$$

There are two solutions for y: \frac{2\sqrt{6}}{3} and -\frac{2\sqrt{6}}{3}. Substitute \frac{2\sqrt{6}}{3} for y in the equation x=y to find the corresponding solution for x that satisfies both equations.

$$x=\frac{2\sqrt{6}}{3}$$

Now substitute -\frac{2\sqrt{6}}{3} for y in the equation x=y and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{2\sqrt{6}}{3}$$

The system is now solved.

$$x=\frac{2\sqrt{6}}{3},y=\frac{2\sqrt{6}}{3}\text{ or }x=-\frac{2\sqrt{6}}{3},y=-\frac{2\sqrt{6}}{3}$$