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

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

Solve for x, y

$$x=-\frac{\sqrt{15}}{5}\approx -0.774596669\text{, }y=-\frac{3\sqrt{5}}{5}\approx -1.341640786$$

$$x=\frac{\sqrt{15}}{5}\approx 0.774596669\text{, }y=\frac{3\sqrt{5}}{5}\approx 1.341640786$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract \sqrt{3}x from both sides.

$$y-\sqrt{3}x=0$$

Reorder the terms.

$$-\sqrt{3}x+y=0$$

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.

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

Solve \left(-\sqrt{3}\right)x+y=0 for x by isolating x on the left hand side of the equal sign.

$$\left(-\sqrt{3}\right)x+y=0$$

Subtract y from both sides of the equation.

$$\left(-\sqrt{3}\right)x=-y$$

Divide both sides by -\sqrt{3}.

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

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

$$3y^{2}+\left(\frac{\sqrt{3}}{3}y\right)^{2}=6$$

Square \frac{\sqrt{3}}{3}y.

$$3y^{2}+\left(\frac{\sqrt{3}}{3}\right)^{2}y^{2}=6$$

Add 3y^{2} to \left(\frac{\sqrt{3}}{3}\right)^{2}y^{2}.

$$\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+3\right)y^{2}=6$$

Subtract 6 from both sides of the equation.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+1\times \left(\frac{\sqrt{3}}{3}\right)^{2} for a, 1\times 0\times 2\times \frac{\sqrt{3}}{3} for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 1\times 0\times 2\times \frac{\sqrt{3}}{3}.

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

Multiply -4 times 3+1\times \left(\frac{\sqrt{3}}{3}\right)^{2}.

$$y=\frac{0±\sqrt{-\frac{40}{3}\left(-6\right)}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+3\right)}$$

Multiply -\frac{40}{3} times -6.

$$y=\frac{0±\sqrt{80}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+3\right)}$$

Take the square root of 80.

$$y=\frac{0±4\sqrt{5}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+3\right)}$$

Multiply 2 times 3+1\times \left(\frac{\sqrt{3}}{3}\right)^{2}.

$$y=\frac{0±4\sqrt{5}}{\frac{20}{3}}$$

Now solve the equation y=\frac{0±4\sqrt{5}}{\frac{20}{3}} when ± is plus.

$$y=\frac{3\sqrt{5}}{5}$$

Now solve the equation y=\frac{0±4\sqrt{5}}{\frac{20}{3}} when ± is minus.

$$y=-\frac{3\sqrt{5}}{5}$$

There are two solutions for y: \frac{3\sqrt{5}}{5} and -\frac{3\sqrt{5}}{5}. Substitute \frac{3\sqrt{5}}{5} for y in the equation x=\frac{\sqrt{3}}{3}y to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}}{3}\times \frac{3\sqrt{5}}{5}$$

Now substitute -\frac{3\sqrt{5}}{5} for y in the equation x=\frac{\sqrt{3}}{3}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}}{3}\left(-\frac{3\sqrt{5}}{5}\right)$$

Multiply \frac{\sqrt{3}}{3} times -\frac{3\sqrt{5}}{5}.

$$x=\left(-\frac{3\sqrt{5}}{5}\right)\times \frac{\sqrt{3}}{3}$$

The system is now solved.

$$x=\frac{\sqrt{3}}{3}\times \frac{3\sqrt{5}}{5},y=\frac{3\sqrt{5}}{5}\text{ or }x=\left(-\frac{3\sqrt{5}}{5}\right)\times \frac{\sqrt{3}}{3},y=-\frac{3\sqrt{5}}{5}$$