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

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

Solve for x, y

$$x=-\frac{\sqrt{6}}{3}\approx -0.816496581\text{, }y=-\frac{\sqrt{6}}{3}\approx -0.816496581$$

$$x=\frac{\sqrt{6}}{3}\approx 0.816496581\text{, }y=\frac{\sqrt{6}}{3}\approx 0.816496581$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 2.

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

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.

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

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, 2y^{2}+x^{2}=2.

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

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

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

Subtract 2 from both sides of the equation.

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

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

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

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

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

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

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

Multiply -12 times -2.

$$y=\frac{0±\sqrt{24}}{2\times 3}$$

Take the square root of 24.

$$y=\frac{0±2\sqrt{6}}{2\times 3}$$

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

$$y=\frac{0±2\sqrt{6}}{6}$$

Now solve the equation y=\frac{0±2\sqrt{6}}{6} when ± is plus.

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

Now solve the equation y=\frac{0±2\sqrt{6}}{6} when ± is minus.

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

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

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

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

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

The system is now solved.

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