Solve {l}{y=kx+2}{x^2-y^2=6}

$$\left\{ \begin{array} { l } { y = k x + 2 } \\ { x ^ { 2 } - y ^ { 2 } = 6 } \end{array} \right.$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}\text{, }y=\frac{\sqrt{2}\left(-k\sqrt{5-3k^{2}}+\sqrt{2}\right)}{1-k^{2}}\text{; }x=\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}\text{, }y=\frac{\sqrt{2}\left(k\sqrt{5-3k^{2}}+\sqrt{2}\right)}{1-k^{2}}\text{, }&|k|\neq 1\text{ and }|k|\leq \frac{\sqrt{15}}{3}\\x=-\frac{5}{2k}\text{, }y=-\frac{1}{2}=-0.5\text{, }&|k|=1\end{matrix}\right.$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract kx from both sides.

$$y-kx=2$$

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.

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

Solve y+\left(-k\right)x=2 for y by isolating y on the left hand side of the equal sign.

$$y+\left(-k\right)x=2$$

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

$$y=kx+2$$

Substitute kx+2 for y in the other equation, x^{2}-y^{2}=6.

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

Square kx+2.

$$x^{2}-\left(k^{2}x^{2}+4kx+4\right)=6$$

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

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

Add x^{2} to \left(-k^{2}\right)x^{2}.

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

Subtract 6 from both sides of the equation.

$$\left(1-k^{2}\right)x^{2}+\left(-4k\right)x-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-k^{2} for a, -2\times 2k for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4k\right)±\sqrt{\left(-4k\right)^{2}-4\left(1-k^{2}\right)\left(-10\right)}}{2\left(1-k^{2}\right)}$$

Square -2\times 2k.

$$x=\frac{-\left(-4k\right)±\sqrt{16k^{2}-4\left(1-k^{2}\right)\left(-10\right)}}{2\left(1-k^{2}\right)}$$

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

$$x=\frac{-\left(-4k\right)±\sqrt{16k^{2}+\left(4k^{2}-4\right)\left(-10\right)}}{2\left(1-k^{2}\right)}$$

Multiply 4k^{2}-4 times -10.

$$x=\frac{-\left(-4k\right)±\sqrt{16k^{2}+40-40k^{2}}}{2\left(1-k^{2}\right)}$$

Add 16k^{2} to 40-40k^{2}.

$$x=\frac{-\left(-4k\right)±\sqrt{40-24k^{2}}}{2\left(1-k^{2}\right)}$$

Take the square root of -24k^{2}+40.

$$x=\frac{-\left(-4k\right)±2\sqrt{10-6k^{2}}}{2\left(1-k^{2}\right)}$$

Multiply 2 times 1-k^{2}.

$$x=\frac{4k±2\sqrt{10-6k^{2}}}{2-2k^{2}}$$

Now solve the equation x=\frac{4k±2\sqrt{10-6k^{2}}}{2-2k^{2}} when ± is plus. Add 4k to 2\sqrt{-6k^{2}+10}.

$$x=\frac{2\sqrt{10-6k^{2}}+4k}{2-2k^{2}}$$

Divide 4k+2\sqrt{-6k^{2}+10} by -2k^{2}+2.

$$x=\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}$$

Now solve the equation x=\frac{4k±2\sqrt{10-6k^{2}}}{2-2k^{2}} when ± is minus. Subtract 2\sqrt{-6k^{2}+10} from 4k.

$$x=\frac{-2\sqrt{10-6k^{2}}+4k}{2-2k^{2}}$$

Divide 4k-2\sqrt{-6k^{2}+10} by -2k^{2}+2.

$$x=\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}$$

There are two solutions for x: \frac{\sqrt{2}\left(\sqrt{2}k+\sqrt{-3k^{2}+5}\right)}{1-k^{2}} and \frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{-3k^{2}+5}\right)}{1-k^{2}}. Substitute \frac{\sqrt{2}\left(\sqrt{2}k+\sqrt{-3k^{2}+5}\right)}{1-k^{2}} for x in the equation y=kx+2 to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}+2$$

Multiply k times \frac{\sqrt{2}\left(\sqrt{2}k+\sqrt{-3k^{2}+5}\right)}{1-k^{2}}.

$$y=\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}k+2$$

Add k\times \frac{\sqrt{2}\left(\sqrt{2}k+\sqrt{-3k^{2}+5}\right)}{1-k^{2}} to 2.

$$y=2+\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}k$$

Now substitute \frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{-3k^{2}+5}\right)}{1-k^{2}} for x in the equation y=kx+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}+2$$

Multiply k times \frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{-3k^{2}+5}\right)}{1-k^{2}}.

$$y=\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}k+2$$

Add k\times \frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{-3k^{2}+5}\right)}{1-k^{2}} to 2.

$$y=2+\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}k$$

The system is now solved.

$$y=2+\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}k,x=\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}\text{ or }y=2+\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}k,x=\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{2}\left(\sqrt{2}k-\sqrt{5-3k^{2}}\right)}{1-k^{2}}\text{, }y=\frac{\sqrt{2}\left(-k\sqrt{5-3k^{2}}+\sqrt{2}\right)}{1-k^{2}}\text{; }x=\frac{\sqrt{2}\left(\sqrt{5-3k^{2}}+\sqrt{2}k\right)}{1-k^{2}}\text{, }y=\frac{\sqrt{2}\left(k\sqrt{5-3k^{2}}+\sqrt{2}\right)}{1-k^{2}}\text{, }&k\neq 1\text{ and }k\neq -1\\x=-\frac{5}{2k}\text{, }y=-\frac{1}{2}=-0.5\text{, }&k=1\text{ or }k=-1\end{matrix}\right.$$