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

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

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

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.

$$kx-y+k=0,\frac{1}{3}y^{2}+\frac{1}{4}x^{2}=1$$

Solve kx-y+k=0 for x by isolating x on the left hand side of the equal sign.

$$kx-y+k=0$$

Subtract k from both sides of the equation.

$$kx-y=-k$$

Subtract -y from both sides of the equation.

$$kx=y-k$$

Divide both sides by k.

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

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

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

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

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

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

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

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

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

Subtract 1 from both sides of the equation.

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

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

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

Square \frac{1}{4}\left(-1\right)\times 2\times \frac{1}{k}.

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

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

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

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

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

Add \frac{1}{4k^{2}} to \frac{3}{4k^{2}}+1.

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

Take the square root of \frac{1}{k^{2}}+1.

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

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

$$y=\frac{\frac{1}{2k}±\frac{\sqrt{k^{2}+1}}{|k|}}{\frac{2}{3}+\frac{1}{2k^{2}}}$$

Now solve the equation y=\frac{\frac{1}{2k}±\frac{\sqrt{k^{2}+1}}{|k|}}{\frac{2}{3}+\frac{1}{2k^{2}}} when ± is plus. Add \frac{1}{2k} to \frac{\sqrt{k^{2}+1}}{|k|}.

$$y=\frac{\frac{\sqrt{k^{2}+1}}{|k|}+\frac{1}{2k}}{\frac{2}{3}+\frac{1}{2k^{2}}}$$

Divide \frac{1}{2k}+\frac{\sqrt{k^{2}+1}}{|k|} by \frac{2}{3}+\frac{1}{2k^{2}}.

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

Now solve the equation y=\frac{\frac{1}{2k}±\frac{\sqrt{k^{2}+1}}{|k|}}{\frac{2}{3}+\frac{1}{2k^{2}}} when ± is minus. Subtract \frac{\sqrt{k^{2}+1}}{|k|} from \frac{1}{2k}.

$$y=\frac{-\frac{\sqrt{k^{2}+1}}{|k|}+\frac{1}{2k}}{\frac{2}{3}+\frac{1}{2k^{2}}}$$

Divide \frac{1}{2k}-\frac{\sqrt{k^{2}+1}}{|k|} by \frac{2}{3}+\frac{1}{2k^{2}}.

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

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

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

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

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

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

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

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

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

The system is now solved.

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

Solve for x, y (complex solution)

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