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

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

Solve for y, x

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract x from both sides.

$$y-x=0$$

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.

$$2x^{2}-y^{2}=4$$

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.

$$y-x=0,2x^{2}-y^{2}=4$$

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, 2x^{2}-y^{2}=4.

$$2x^{2}-x^{2}=4$$

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

$$x^{2}=4$$

Subtract 4 from both sides of the equation.

$$x^{2}-4=0$$

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

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

Square -0\times 2.

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

Multiply -4 times -4.

$$x=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{0±4}{2}$$

Now solve the equation x=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$x=2$$

Now solve the equation x=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$x=-2$$

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

$$y=2$$

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

$$y=-2$$

The system is now solved.

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