Solve {l}{x^2/6+y^2/3=1}{x+2y=0}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 6 } + \frac { y ^ { 2 } } { 3 } = 1 } \\ { x + 2 y = 0 } \end{array} \right.$$

Solve for x, y

$$x=-2\text{, }y=1$$

$$x=2\text{, }y=-1$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 6,3.

$$x^{2}+2y^{2}=6$$

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.

$$x+2y=0,2y^{2}+x^{2}=6$$

Solve x+2y=0 for x by isolating x on the left hand side of the equal sign.

$$x+2y=0$$

Subtract 2y from both sides of the equation.

$$x=-2y$$

Substitute -2y for x in the other equation, 2y^{2}+x^{2}=6.

$$2y^{2}+\left(-2y\right)^{2}=6$$

Square -2y.

$$2y^{2}+4y^{2}=6$$

Add 2y^{2} to 4y^{2}.

$$6y^{2}=6$$

Subtract 6 from both sides of the equation.

$$6y^{2}-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2+1\left(-2\right)^{2} for a, 1\times 0\left(-2\right)\times 2 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 6\left(-6\right)}}{2\times 6}$$

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

$$y=\frac{0±\sqrt{-4\times 6\left(-6\right)}}{2\times 6}$$

Multiply -4 times 2+1\left(-2\right)^{2}.

$$y=\frac{0±\sqrt{-24\left(-6\right)}}{2\times 6}$$

Multiply -24 times -6.

$$y=\frac{0±\sqrt{144}}{2\times 6}$$

Take the square root of 144.

$$y=\frac{0±12}{2\times 6}$$

Multiply 2 times 2+1\left(-2\right)^{2}.

$$y=\frac{0±12}{12}$$

Now solve the equation y=\frac{0±12}{12} when ± is plus. Divide 12 by 12.

$$y=1$$

Now solve the equation y=\frac{0±12}{12} when ± is minus. Divide -12 by 12.

$$y=-1$$

There are two solutions for y: 1 and -1. Substitute 1 for y in the equation x=-2y to find the corresponding solution for x that satisfies both equations.

$$x=-2$$

Now substitute -1 for y in the equation x=-2y and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\left(-1\right)$$

Multiply -2 times -1.

$$x=2$$

The system is now solved.

$$x=-2,y=1\text{ or }x=2,y=-1$$