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

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

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

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

Consider the second equation. Subtract kx from both sides.

$$y-kx=k$$

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.

$$y+\left(-k\right)x=k,x^{2}+2y^{2}=4$$

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

$$y+\left(-k\right)x=k$$

Subtract \left(-k\right)x from both sides of the equation.

$$y=kx+k$$

Substitute kx+k for y in the other equation, x^{2}+2y^{2}=4.

$$x^{2}+2\left(kx+k\right)^{2}=4$$

Square kx+k.

$$x^{2}+2\left(k^{2}x^{2}+2kkx+k^{2}\right)=4$$

Multiply 2 times k^{2}x^{2}+2kkx+k^{2}.

$$x^{2}+2k^{2}x^{2}+4k^{2}x+2k^{2}=4$$

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

$$\left(2k^{2}+1\right)x^{2}+4k^{2}x+2k^{2}=4$$

Subtract 4 from both sides of the equation.

$$\left(2k^{2}+1\right)x^{2}+4k^{2}x+2k^{2}-4=0$$

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

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

Square 2\times 2kk.

$$x=\frac{-4k^{2}±\sqrt{16k^{4}-4\left(2k^{2}+1\right)\left(2k^{2}-4\right)}}{2\left(2k^{2}+1\right)}$$

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

$$x=\frac{-4k^{2}±\sqrt{16k^{4}+\left(-8k^{2}-4\right)\left(2k^{2}-4\right)}}{2\left(2k^{2}+1\right)}$$

Multiply -4-8k^{2} times 2k^{2}-4.

$$x=\frac{-4k^{2}±\sqrt{16k^{4}+16+24k^{2}-16k^{4}}}{2\left(2k^{2}+1\right)}$$

Add 16k^{4} to 16+24k^{2}-16k^{4}.

$$x=\frac{-4k^{2}±\sqrt{24k^{2}+16}}{2\left(2k^{2}+1\right)}$$

Take the square root of 24k^{2}+16.

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

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

$$x=\frac{-4k^{2}±2\sqrt{6k^{2}+4}}{4k^{2}+2}$$

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

$$x=\frac{-4k^{2}+2\sqrt{6k^{2}+4}}{4k^{2}+2}$$

Divide -4k^{2}+2\sqrt{6k^{2}+4} by 2+4k^{2}.

$$x=\frac{-2k^{2}+\sqrt{6k^{2}+4}}{2k^{2}+1}$$

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

$$x=\frac{-4k^{2}-2\sqrt{6k^{2}+4}}{4k^{2}+2}$$

Divide -4k^{2}-2\sqrt{6k^{2}+4} by 2+4k^{2}.

$$x=-\frac{2k^{2}+\sqrt{6k^{2}+4}}{2k^{2}+1}$$

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

$$y=k\times \frac{-2k^{2}+\sqrt{6k^{2}+4}}{2k^{2}+1}+k$$

Multiply k times \frac{-2k^{2}+\sqrt{6k^{2}+4}}{1+2k^{2}}.

$$y=\frac{-2k^{2}+\sqrt{6k^{2}+4}}{2k^{2}+1}k+k$$

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

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

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

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

The system is now solved.

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

Solve for x, y (complex solution)

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