Solve {l}{y=kx+1}{x^2/5-y^2/4=1}

$$\left. \begin{array} { l } { y = k x + 1 } \\ { \frac { x ^ { 2 } } { 5 } - \frac { y ^ { 2 } } { 4 } = 1 } \end{array} \right.$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}\text{, }y=\frac{2\left(-5k\sqrt{1-k^{2}}+2\right)}{4-5k^{2}}\text{; }x=\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}\text{, }y=\frac{2\left(5k\sqrt{1-k^{2}}+2\right)}{4-5k^{2}}\text{, }&|k|\neq \frac{2\sqrt{5}}{5}\text{ and }|k|\leq 1\\x=-\frac{5}{2k}\text{, }y=-\frac{3}{2}=-1.5\text{, }&|k|=\frac{2\sqrt{5}}{5}\end{matrix}\right.$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract kx from both sides.

$$y-kx=1$$

Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4x^{2}-5y^{2}=20$$

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.

$$y+\left(-k\right)x=1,4x^{2}-5y^{2}=20$$

Solve y+\left(-k\right)x=1 for y by isolating y on the left hand side of the equal sign.

$$y+\left(-k\right)x=1$$

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

$$y=kx+1$$

Substitute kx+1 for y in the other equation, 4x^{2}-5y^{2}=20.

$$4x^{2}-5\left(kx+1\right)^{2}=20$$

Square kx+1.

$$4x^{2}-5\left(k^{2}x^{2}+2kx+1\right)=20$$

Multiply -5 times k^{2}x^{2}+2kx+1.

$$4x^{2}+\left(-5k^{2}\right)x^{2}+\left(-10k\right)x-5=20$$

Add 4x^{2} to \left(-5k^{2}\right)x^{2}.

$$\left(4-5k^{2}\right)x^{2}+\left(-10k\right)x-5=20$$

Subtract 20 from both sides of the equation.

$$\left(4-5k^{2}\right)x^{2}+\left(-10k\right)x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4-5k^{2} for a, -5\times 2k for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10k\right)±\sqrt{\left(-10k\right)^{2}-4\left(4-5k^{2}\right)\left(-25\right)}}{2\left(4-5k^{2}\right)}$$

Square -5\times 2k.

$$x=\frac{-\left(-10k\right)±\sqrt{100k^{2}-4\left(4-5k^{2}\right)\left(-25\right)}}{2\left(4-5k^{2}\right)}$$

Multiply -4 times 4-5k^{2}.

$$x=\frac{-\left(-10k\right)±\sqrt{100k^{2}+\left(20k^{2}-16\right)\left(-25\right)}}{2\left(4-5k^{2}\right)}$$

Multiply -16+20k^{2} times -25.

$$x=\frac{-\left(-10k\right)±\sqrt{100k^{2}+400-500k^{2}}}{2\left(4-5k^{2}\right)}$$

Add 100k^{2} to 400-500k^{2}.

$$x=\frac{-\left(-10k\right)±\sqrt{400-400k^{2}}}{2\left(4-5k^{2}\right)}$$

Take the square root of 400-400k^{2}.

$$x=\frac{-\left(-10k\right)±20\sqrt{1-k^{2}}}{2\left(4-5k^{2}\right)}$$

Multiply 2 times 4-5k^{2}.

$$x=\frac{10k±20\sqrt{1-k^{2}}}{8-10k^{2}}$$

Now solve the equation x=\frac{10k±20\sqrt{1-k^{2}}}{8-10k^{2}} when ± is plus. Add 10k to 20\sqrt{1-k^{2}}.

$$x=\frac{20\sqrt{1-k^{2}}+10k}{8-10k^{2}}$$

Divide 10k+20\sqrt{1-k^{2}} by 8-10k^{2}.

$$x=\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}$$

Now solve the equation x=\frac{10k±20\sqrt{1-k^{2}}}{8-10k^{2}} when ± is minus. Subtract 20\sqrt{1-k^{2}} from 10k.

$$x=\frac{-20\sqrt{1-k^{2}}+10k}{8-10k^{2}}$$

Divide 10k-20\sqrt{1-k^{2}} by 8-10k^{2}.

$$x=\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}$$

There are two solutions for x: \frac{5\left(k+2\sqrt{1-k^{2}}\right)}{4-5k^{2}} and \frac{5\left(k-2\sqrt{1-k^{2}}\right)}{4-5k^{2}}. Substitute \frac{5\left(k+2\sqrt{1-k^{2}}\right)}{4-5k^{2}} for x in the equation y=kx+1 to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}+1$$

Multiply k times \frac{5\left(k+2\sqrt{1-k^{2}}\right)}{4-5k^{2}}.

$$y=\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k+1$$

Add k\times \frac{5\left(k+2\sqrt{1-k^{2}}\right)}{4-5k^{2}} to 1.

$$y=1+\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k$$

Now substitute \frac{5\left(k-2\sqrt{1-k^{2}}\right)}{4-5k^{2}} for x in the equation y=kx+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}+1$$

Multiply k times \frac{5\left(k-2\sqrt{1-k^{2}}\right)}{4-5k^{2}}.

$$y=\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k+1$$

Add k\times \frac{5\left(k-2\sqrt{1-k^{2}}\right)}{4-5k^{2}} to 1.

$$y=1+\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k$$

The system is now solved.

$$y=1+\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k,x=\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}\text{ or }y=1+\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}k,x=\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=\frac{5\left(-2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}\text{, }y=\frac{2\left(-5k\sqrt{1-k^{2}}+2\right)}{4-5k^{2}}\text{; }x=\frac{5\left(2\sqrt{1-k^{2}}+k\right)}{4-5k^{2}}\text{, }y=\frac{2\left(5k\sqrt{1-k^{2}}+2\right)}{4-5k^{2}}\text{, }&k\neq -\frac{2\sqrt{5}}{5}\text{ and }k\neq \frac{2\sqrt{5}}{5}\\x=-\frac{5}{2k}\text{, }y=-\frac{3}{2}=-1.5\text{, }&k=-\frac{2\sqrt{5}}{5}\text{ or }k=\frac{2\sqrt{5}}{5}\end{matrix}\right.$$