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

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

Solve for x, y

$$x=\frac{4\sqrt{3}}{3}\approx 2.309401077\text{, }y=-\frac{\sqrt{3}}{3}\approx -0.577350269$$

$$x=0\text{, }y=\sqrt{3}\approx 1.732050808$$

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.

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

Consider the second equation. Add x to both sides.

$$y+x=\sqrt{3}$$

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.

$$y+x=\sqrt{3},x^{2}+2y^{2}=6$$

Solve y+x=\sqrt{3} for y by isolating y on the left hand side of the equal sign.

$$y+x=\sqrt{3}$$

Subtract x from both sides of the equation.

$$y=-x+\sqrt{3}$$

Substitute -x+\sqrt{3} for y in the other equation, x^{2}+2y^{2}=6.

$$x^{2}+2\left(-x+\sqrt{3}\right)^{2}=6$$

Square -x+\sqrt{3}.

$$x^{2}+2\left(x^{2}+\left(-2\sqrt{3}\right)x+\left(\sqrt{3}\right)^{2}\right)=6$$

Multiply 2 times x^{2}+\left(-2\sqrt{3}\right)x+\left(\sqrt{3}\right)^{2}.

$$x^{2}+2x^{2}+\left(-4\sqrt{3}\right)x+2\left(\sqrt{3}\right)^{2}=6$$

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

$$3x^{2}+\left(-4\sqrt{3}\right)x+2\left(\sqrt{3}\right)^{2}=6$$

Subtract 6 from both sides of the equation.

$$3x^{2}+\left(-4\sqrt{3}\right)x+2\left(\sqrt{3}\right)^{2}-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+2\left(-1\right)^{2} for a, 2\left(-1\right)\times 2\sqrt{3} for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

The square root of b^{2} is |b|. Substitute 2\left(-1\right)\times 2\sqrt{3} for b.

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

The opposite of 2\left(-1\right)\times 2\sqrt{3} is 4\sqrt{3}.

$$x=\frac{4\sqrt{3}±4\sqrt{3}}{2\times 3}$$

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

$$x=\frac{4\sqrt{3}±4\sqrt{3}}{6}$$

Now solve the equation x=\frac{4\sqrt{3}±4\sqrt{3}}{6} when ± is plus. Add 4\sqrt{3} to 4\sqrt{3}.

$$x=\frac{8\sqrt{3}}{6}$$

Divide 8\sqrt{3} by 6.

$$x=\frac{4\sqrt{3}}{3}$$

Now solve the equation x=\frac{4\sqrt{3}±4\sqrt{3}}{6} when ± is minus. Subtract 4\sqrt{3} from 4\sqrt{3}.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

There are two solutions for x: \frac{4\sqrt{3}}{3} and 0. Substitute \frac{4\sqrt{3}}{3} for x in the equation y=-x+\sqrt{3} to find the corresponding solution for y that satisfies both equations.

$$y=-\frac{4\sqrt{3}}{3}+\sqrt{3}$$

Now substitute 0 for x in the equation y=-x+\sqrt{3} and solve to find the corresponding solution for y that satisfies both equations.

$$y=\sqrt{3}$$

The system is now solved.

$$y=-\frac{4\sqrt{3}}{3}+\sqrt{3},x=\frac{4\sqrt{3}}{3}\text{ or }y=\sqrt{3},x=0$$