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

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

Solve for x, y

$$x=-\sqrt{\frac{3}{m^{2}+2m-2}}\left(m+1\right)\text{, }y=-\sqrt{\frac{3}{m^{2}+2m-2}}$$

$$x=\sqrt{\frac{3}{m^{2}+2m-2}}\left(m+1\right)\text{, }y=\sqrt{\frac{3}{m^{2}+2m-2}}\text{, }m<-\sqrt{3}-1\text{ or }m>\sqrt{3}-1$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 3.

$$x^{2}-3y^{2}=3$$

Consider the second equation. Subtract my from both sides.

$$x-my=y$$

Subtract y from both sides.

$$x-my-y=0$$

Combine all terms containing x,y.

$$x+\left(-m-1\right)y=0$$

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.

$$x+\left(-m-1\right)y=0,-3y^{2}+x^{2}=3$$

Solve x+\left(-m-1\right)y=0 for x by isolating x on the left hand side of the equal sign.

$$x+\left(-m-1\right)y=0$$

Subtract \left(-m-1\right)y from both sides of the equation.

$$x=\left(m+1\right)y$$

Substitute \left(m+1\right)y for x in the other equation, -3y^{2}+x^{2}=3.

$$-3y^{2}+\left(\left(m+1\right)y\right)^{2}=3$$

Square \left(m+1\right)y.

$$-3y^{2}+\left(m+1\right)^{2}y^{2}=3$$

Add -3y^{2} to \left(m+1\right)^{2}y^{2}.

$$\left(\left(m+1\right)^{2}-3\right)y^{2}=3$$

Subtract 3 from both sides of the equation.

$$\left(\left(m+1\right)^{2}-3\right)y^{2}-3=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\left(\left(m+1\right)^{2}-3\right)\left(-3\right)}}{2\left(\left(m+1\right)^{2}-3\right)}$$

Square 1\times 0\times 2\left(m+1\right).

$$y=\frac{0±\sqrt{-4\left(\left(m+1\right)^{2}-3\right)\left(-3\right)}}{2\left(\left(m+1\right)^{2}-3\right)}$$

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

$$y=\frac{0±\sqrt{\left(-4\left(m+1\right)^{2}+12\right)\left(-3\right)}}{2\left(\left(m+1\right)^{2}-3\right)}$$

Multiply 12-4\left(m+1\right)^{2} times -3.

$$y=\frac{0±\sqrt{12\left(m+1\right)^{2}-36}}{2\left(\left(m+1\right)^{2}-3\right)}$$

Take the square root of -36+12\left(m+1\right)^{2}.

$$y=\frac{0±2\sqrt{3m^{2}+6m-6}}{2\left(\left(m+1\right)^{2}-3\right)}$$

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

$$y=\frac{0±2\sqrt{3m^{2}+6m-6}}{2\left(m+1\right)^{2}-6}$$

Now solve the equation y=\frac{0±2\sqrt{3m^{2}+6m-6}}{2\left(m+1\right)^{2}-6} when ± is plus.

$$y=\frac{3}{\sqrt{3m^{2}+6m-6}}$$

Now solve the equation y=\frac{0±2\sqrt{3m^{2}+6m-6}}{2\left(m+1\right)^{2}-6} when ± is minus.

$$y=-\frac{3}{\sqrt{3m^{2}+6m-6}}$$

There are two solutions for y: \frac{3}{\sqrt{3m^{2}-6+6m}} and -\frac{3}{\sqrt{3m^{2}-6+6m}}. Substitute \frac{3}{\sqrt{3m^{2}-6+6m}} for y in the equation x=\left(m+1\right)y to find the corresponding solution for x that satisfies both equations.

$$x=\left(m+1\right)\times \frac{3}{\sqrt{3m^{2}+6m-6}}$$

Multiply m+1 times \frac{3}{\sqrt{3m^{2}-6+6m}}.

$$x=\frac{3}{\sqrt{3m^{2}+6m-6}}\left(m+1\right)$$

Now substitute -\frac{3}{\sqrt{3m^{2}-6+6m}} for y in the equation x=\left(m+1\right)y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\left(m+1\right)\left(-\frac{3}{\sqrt{3m^{2}+6m-6}}\right)$$

Multiply m+1 times -\frac{3}{\sqrt{3m^{2}-6+6m}}.

$$x=\left(-\frac{3}{\sqrt{3m^{2}+6m-6}}\right)\left(m+1\right)$$

The system is now solved.

$$x=\frac{3}{\sqrt{3m^{2}+6m-6}}\left(m+1\right),y=\frac{3}{\sqrt{3m^{2}+6m-6}}\text{ or }x=\left(-\frac{3}{\sqrt{3m^{2}+6m-6}}\right)\left(m+1\right),y=-\frac{3}{\sqrt{3m^{2}+6m-6}}$$

Solve for x, y (complex solution)

$$x=\sqrt{3}\left(m^{2}+2m-2\right)^{-\frac{1}{2}}\left(m+1\right)\text{, }y=\sqrt{3}\left(m^{2}+2m-2\right)^{-\frac{1}{2}}$$

$$x=-\sqrt{3}\left(m^{2}+2m-2\right)^{-\frac{1}{2}}\left(m+1\right)\text{, }y=-\sqrt{3}\left(m^{2}+2m-2\right)^{-\frac{1}{2}}\text{, }m\neq \sqrt{3}-1\text{ and }m\neq -\left(\sqrt{3}+1\right)$$