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

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

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 8,4.

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

Consider the second equation. Subtract x from both sides.

$$y-x=0$$

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.

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

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

$$y-x=0$$

Subtract -x from both sides of the equation.

$$y=x$$

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

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

Add x^{2} to 2x^{2}.

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

Subtract 8 from both sides of the equation.

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

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}.

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

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

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

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

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

Multiply -12 times -8.

$$x=\frac{0±\sqrt{96}}{2\times 3}$$

Take the square root of 96.

$$x=\frac{0±4\sqrt{6}}{2\times 3}$$

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

$$x=\frac{0±4\sqrt{6}}{6}$$

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

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

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

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

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

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

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

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

The system is now solved.

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