Solve {l}{2x^2-y^2=9}{x+y=0}

$$\left. \begin{array} { l } { 2 x ^ { 2 } - y ^ { 2 } = 9 } \\ { x + y = 0 } \end{array} \right.$$

Solve for x, y

$$x=-3\text{, }y=3$$

$$x=3\text{, }y=-3$$

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.

$$x+y=0,-y^{2}+2x^{2}=9$$

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}=9.

$$-y^{2}+2\left(-y\right)^{2}=9$$

Square -y.

$$-y^{2}+2y^{2}=9$$

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

$$y^{2}=9$$

Subtract 9 from both sides of the equation.

$$y^{2}-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1+2\left(-1\right)^{2} for a, 2\times 0\left(-1\right)\times 2 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 2\times 0\left(-1\right)\times 2.

$$y=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$y=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$y=\frac{0±6}{2}$$

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

$$y=3$$

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

$$y=-3$$

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

$$x=-3$$

Now substitute -3 for y in the equation x=-y and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-3\right)$$

Multiply -1 times -3.

$$x=3$$

The system is now solved.

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