Solve {l}{x^2/4+y^2/3=1}{kx-1-y=0}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 4 } + \frac { y ^ { 2 } } { 3 } = 1 } \\ { k x - 1 - y = 0 } \end{array} \right.$$

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y^{2}=12$$

Consider the second equation. Add 1 to both sides. Anything plus zero gives itself.

$$kx-y=1$$

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.

$$kx-y=1,4y^{2}+3x^{2}=12$$

Solve kx-y=1 for x by isolating x on the left hand side of the equal sign.

$$kx-y=1$$

Subtract -y from both sides of the equation.

$$kx=y+1$$

Divide both sides by k.

$$x=\frac{1}{k}y+\frac{1}{k}$$

Substitute \frac{1}{k}y+\frac{1}{k} for x in the other equation, 4y^{2}+3x^{2}=12.

$$4y^{2}+3\left(\frac{1}{k}y+\frac{1}{k}\right)^{2}=12$$

Square \frac{1}{k}y+\frac{1}{k}.

$$4y^{2}+3\left(\left(\frac{1}{k}\right)^{2}y^{2}+2\times \frac{1}{k}\times \frac{1}{k}y+\left(\frac{1}{k}\right)^{2}\right)=12$$

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

$$4y^{2}+3\times \left(\frac{1}{k}\right)^{2}y^{2}+6\times \left(\frac{1}{k}\right)^{2}y+3\times \left(\frac{1}{k}\right)^{2}=12$$

Add 4y^{2} to 3\times \left(\frac{1}{k}\right)^{2}y^{2}.

$$\left(4+3\times \left(\frac{1}{k}\right)^{2}\right)y^{2}+6\times \left(\frac{1}{k}\right)^{2}y+3\times \left(\frac{1}{k}\right)^{2}=12$$

Subtract 12 from both sides of the equation.

$$\left(4+3\times \left(\frac{1}{k}\right)^{2}\right)y^{2}+6\times \left(\frac{1}{k}\right)^{2}y+3\times \left(\frac{1}{k}\right)^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4+3\times \left(\frac{1}{k}\right)^{2} for a, 3\times 2\times \frac{1}{k}\times \frac{1}{k} for b, and \frac{3}{k^{2}}-12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3\times 2\times \frac{1}{k}\times \frac{1}{k}.

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

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

$$y=\frac{-6\times \left(\frac{1}{k}\right)^{2}±\sqrt{\frac{36}{k^{4}}+\left(-16-\frac{12}{k^{2}}\right)\left(-12+\frac{3}{k^{2}}\right)}}{2\left(4+3\times \left(\frac{1}{k}\right)^{2}\right)}$$

Multiply -16-\frac{12}{k^{2}} times \frac{3}{k^{2}}-12.

$$y=\frac{-6\times \left(\frac{1}{k}\right)^{2}±\sqrt{\frac{36}{k^{4}}+192+\frac{96}{k^{2}}-\frac{36}{k^{4}}}}{2\left(4+3\times \left(\frac{1}{k}\right)^{2}\right)}$$

Add \frac{36}{k^{4}} to 192+\frac{96}{k^{2}}-\frac{36}{k^{4}}.

$$y=\frac{-6\times \left(\frac{1}{k}\right)^{2}±\sqrt{192+\frac{96}{k^{2}}}}{2\left(4+3\times \left(\frac{1}{k}\right)^{2}\right)}$$

Take the square root of 192+\frac{96}{k^{2}}.

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

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

$$y=\frac{-\frac{6}{k^{2}}±\frac{4\sqrt{12k^{2}+6}}{|k|}}{8+\frac{6}{k^{2}}}$$

Now solve the equation y=\frac{-\frac{6}{k^{2}}±\frac{4\sqrt{12k^{2}+6}}{|k|}}{8+\frac{6}{k^{2}}} when ± is plus. Add -\frac{6}{k^{2}} to \frac{4\sqrt{12k^{2}+6}}{|k|}.

$$y=\frac{\frac{4\sqrt{12k^{2}+6}}{|k|}-\frac{6}{k^{2}}}{8+\frac{6}{k^{2}}}$$

Divide -\frac{6}{k^{2}}+\frac{4\sqrt{12k^{2}+6}}{|k|} by 8+\frac{6}{k^{2}}.

$$y=\frac{2\sqrt{12k^{2}+6}k^{2}-3|k|}{|k|\left(4k^{2}+3\right)}$$

Now solve the equation y=\frac{-\frac{6}{k^{2}}±\frac{4\sqrt{12k^{2}+6}}{|k|}}{8+\frac{6}{k^{2}}} when ± is minus. Subtract \frac{4\sqrt{12k^{2}+6}}{|k|} from -\frac{6}{k^{2}}.

$$y=\frac{-\frac{4\sqrt{12k^{2}+6}}{|k|}-\frac{6}{k^{2}}}{8+\frac{6}{k^{2}}}$$

Divide -\frac{6}{k^{2}}-\frac{4\sqrt{12k^{2}+6}}{|k|} by 8+\frac{6}{k^{2}}.

$$y=-\frac{2\sqrt{12k^{2}+6}k^{2}+3|k|}{|k|\left(4k^{2}+3\right)}$$

There are two solutions for y: \frac{-3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|} and -\frac{3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|}. Substitute \frac{-3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|} for y in the equation x=\frac{1}{k}y+\frac{1}{k} to find the corresponding solution for x that satisfies both equations.

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

Add \frac{1}{k}\times \frac{-3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|} to \frac{1}{k}.

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

Now substitute -\frac{3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|} for y in the equation x=\frac{1}{k}y+\frac{1}{k} and solve to find the corresponding solution for x that satisfies both equations.

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

Add \frac{1}{k}\left(-\frac{3|k|+2k^{2}\sqrt{12k^{2}+6}}{\left(3+4k^{2}\right)|k|}\right) to \frac{1}{k}.

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

The system is now solved.

$$x=\frac{1}{k}+\frac{1}{k}\times \frac{2\sqrt{12k^{2}+6}k^{2}-3|k|}{|k|\left(4k^{2}+3\right)},y=\frac{2\sqrt{12k^{2}+6}k^{2}-3|k|}{|k|\left(4k^{2}+3\right)}\text{ or }x=\frac{1}{k}+\frac{1}{k}\left(-\frac{2\sqrt{12k^{2}+6}k^{2}+3|k|}{|k|\left(4k^{2}+3\right)}\right),y=-\frac{2\sqrt{12k^{2}+6}k^{2}+3|k|}{|k|\left(4k^{2}+3\right)}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{2\left(-\sqrt{6\left(2k^{2}+1\right)}+2k\right)}{4k^{2}+3}\text{, }y=-\frac{2k\sqrt{6\left(2k^{2}+1\right)}+3}{4k^{2}+3}\text{; }x=\frac{2\left(\sqrt{6\left(2k^{2}+1\right)}+2k\right)}{4k^{2}+3}\text{, }y=\frac{2k\sqrt{6\left(2k^{2}+1\right)}-3}{4k^{2}+3}\text{, }&k\neq -\frac{\sqrt{3}i}{2}\text{ and }k\neq \frac{\sqrt{3}i}{2}\\x=-\frac{1}{k}\text{, }y=-2\text{, }&k=-\frac{\sqrt{3}i}{2}\text{ or }k=\frac{\sqrt{3}i}{2}\end{matrix}\right.$$