Solve {l}{x^2/4+y^2/3=1}{y=k(x+1)}

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

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

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. Use the distributive property to multiply k by x+1.

$$y=kx+k$$

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

$$3x^{2}+4\left(kx+k\right)^{2}=12$$

Square kx+k.

$$3x^{2}+4\left(k^{2}x^{2}+2kkx+k^{2}\right)=12$$

Multiply 4 times k^{2}x^{2}+2kkx+k^{2}.

$$3x^{2}+4k^{2}x^{2}+8k^{2}x+4k^{2}=12$$

Add 3x^{2} to 4k^{2}x^{2}.

$$\left(4k^{2}+3\right)x^{2}+8k^{2}x+4k^{2}=12$$

Subtract 12 from both sides of the equation.

$$\left(4k^{2}+3\right)x^{2}+8k^{2}x+4k^{2}-12=0$$

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

$$x=\frac{-8k^{2}±\sqrt{\left(8k^{2}\right)^{2}-4\left(4k^{2}+3\right)\left(4k^{2}-12\right)}}{2\left(4k^{2}+3\right)}$$

Square 4\times 2kk.

$$x=\frac{-8k^{2}±\sqrt{64k^{4}-4\left(4k^{2}+3\right)\left(4k^{2}-12\right)}}{2\left(4k^{2}+3\right)}$$

Multiply -4 times 3+4k^{2}.

$$x=\frac{-8k^{2}±\sqrt{64k^{4}+\left(-16k^{2}-12\right)\left(4k^{2}-12\right)}}{2\left(4k^{2}+3\right)}$$

Multiply -12-16k^{2} times 4k^{2}-12.

$$x=\frac{-8k^{2}±\sqrt{64k^{4}+144+144k^{2}-64k^{4}}}{2\left(4k^{2}+3\right)}$$

Add 64k^{4} to 144+144k^{2}-64k^{4}.

$$x=\frac{-8k^{2}±\sqrt{144k^{2}+144}}{2\left(4k^{2}+3\right)}$$

Take the square root of 144k^{2}+144.

$$x=\frac{-8k^{2}±12\sqrt{k^{2}+1}}{2\left(4k^{2}+3\right)}$$

Multiply 2 times 3+4k^{2}.

$$x=\frac{-8k^{2}±12\sqrt{k^{2}+1}}{8k^{2}+6}$$

Now solve the equation x=\frac{-8k^{2}±12\sqrt{k^{2}+1}}{8k^{2}+6} when ± is plus. Add -8k^{2} to 12\sqrt{k^{2}+1}.

$$x=\frac{-8k^{2}+12\sqrt{k^{2}+1}}{8k^{2}+6}$$

Divide -8k^{2}+12\sqrt{k^{2}+1} by 6+8k^{2}.

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

Now solve the equation x=\frac{-8k^{2}±12\sqrt{k^{2}+1}}{8k^{2}+6} when ± is minus. Subtract 12\sqrt{k^{2}+1} from -8k^{2}.

$$x=\frac{-8k^{2}-12\sqrt{k^{2}+1}}{8k^{2}+6}$$

Divide -8k^{2}-12\sqrt{k^{2}+1} by 6+8k^{2}.

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

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

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

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

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

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

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

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

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

The system is now solved.

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

Solve for x, y (complex solution)

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