Solve {l}{x^2/3+y^2=1}{y=k(x-1)}

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

Solve for x, y

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

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

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. Use the distributive property to multiply k by x-1.

$$y=kx-k$$

Substitute kx-k for y in the other equation, x^{2}+3y^{2}=3.

$$x^{2}+3\left(kx-k\right)^{2}=3$$

Square kx-k.

$$x^{2}+3\left(k^{2}x^{2}+2\left(-k\right)kx+\left(-k\right)^{2}\right)=3$$

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

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

Add x^{2} to 3k^{2}x^{2}.

$$\left(3k^{2}+1\right)x^{2}+6\left(-k\right)kx+3\left(-k\right)^{2}=3$$

Subtract 3 from both sides of the equation.

$$\left(3k^{2}+1\right)x^{2}+6\left(-k\right)kx+3\left(-k\right)^{2}-3=0$$

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

$$x=\frac{-6\left(-k\right)k±\sqrt{\left(6\left(-k\right)k\right)^{2}-4\left(3k^{2}+1\right)\left(3k^{2}-3\right)}}{2\left(3k^{2}+1\right)}$$

Square 3\times 2k\left(-k\right).

$$x=\frac{-6\left(-k\right)k±\sqrt{36k^{4}-4\left(3k^{2}+1\right)\left(3k^{2}-3\right)}}{2\left(3k^{2}+1\right)}$$

Multiply -4 times 1+3k^{2}.

$$x=\frac{-6\left(-k\right)k±\sqrt{36k^{4}+\left(-12k^{2}-4\right)\left(3k^{2}-3\right)}}{2\left(3k^{2}+1\right)}$$

Multiply -4-12k^{2} times -3+3k^{2}.

$$x=\frac{-6\left(-k\right)k±\sqrt{36k^{4}+12+24k^{2}-36k^{4}}}{2\left(3k^{2}+1\right)}$$

Add 36k^{4} to 24k^{2}+12-36k^{4}.

$$x=\frac{-6\left(-k\right)k±\sqrt{24k^{2}+12}}{2\left(3k^{2}+1\right)}$$

Take the square root of 12+24k^{2}.

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

Multiply 2 times 1+3k^{2}.

$$x=\frac{6k^{2}±2\sqrt{6k^{2}+3}}{6k^{2}+2}$$

Now solve the equation x=\frac{6k^{2}±2\sqrt{6k^{2}+3}}{6k^{2}+2} when ± is plus. Add 6k^{2} to 2\sqrt{3+6k^{2}}.

$$x=\frac{6k^{2}+2\sqrt{6k^{2}+3}}{6k^{2}+2}$$

Divide 6k^{2}+2\sqrt{3+6k^{2}} by 2+6k^{2}.

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

Now solve the equation x=\frac{6k^{2}±2\sqrt{6k^{2}+3}}{6k^{2}+2} when ± is minus. Subtract 2\sqrt{3+6k^{2}} from 6k^{2}.

$$x=\frac{6k^{2}-2\sqrt{6k^{2}+3}}{6k^{2}+2}$$

Divide 6k^{2}-2\sqrt{3+6k^{2}} by 2+6k^{2}.

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

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

$$y=k\times \frac{\sqrt{3}\left(\sqrt{3}k^{2}+\sqrt{2k^{2}+1}\right)}{3k^{2}+1}-k$$

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

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

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

$$y=k\times \frac{\sqrt{3}\left(\sqrt{3}k^{2}-\sqrt{2k^{2}+1}\right)}{3k^{2}+1}-k$$

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

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

The system is now solved.

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

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{3}\left(\sqrt{3}k^{2}-\sqrt{2k^{2}+1}\right)}{3k^{2}+1}\text{, }y=-\frac{k\left(\sqrt{3\left(2k^{2}+1\right)}+1\right)}{3k^{2}+1}\text{; }x=\frac{\sqrt{3}\left(\sqrt{3}k^{2}+\sqrt{2k^{2}+1}\right)}{3k^{2}+1}\text{, }y=\frac{k\left(\sqrt{3\left(2k^{2}+1\right)}-1\right)}{3k^{2}+1}\text{, }&k\neq -\frac{\sqrt{3}i}{3}\text{ and }k\neq \frac{\sqrt{3}i}{3}\\x=\frac{k^{2}-1}{2k^{2}}\text{, }y=\frac{-k^{2}-1}{2k}\text{, }&k=-\frac{\sqrt{3}i}{3}\text{ or }k=\frac{\sqrt{3}i}{3}\end{matrix}\right.$$