Solve {l}{y=kx+b}{x^2/4+y^2/3=1}

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

Solve for x, y

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

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

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract kx from both sides.

$$y-kx=b$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 4,3.

$$3x^{2}+4y^{2}=12$$

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.

$$y+\left(-k\right)x=b,3x^{2}+4y^{2}=12$$

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

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

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

$$y=kx+b$$

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

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

Square kx+b.

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

Multiply 4 times k^{2}x^{2}+2bkx+b^{2}.

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

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

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

Subtract 12 from both sides of the equation.

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

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

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

Square 4\times 2kb.

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

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

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

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

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

Add 64k^{2}b^{2} to -16\left(3+4k^{2}\right)\left(b^{2}-3\right).

$$x=\frac{-8bk±\sqrt{144+192k^{2}-48b^{2}}}{2\left(4k^{2}+3\right)}$$

Take the square root of -48b^{2}+192k^{2}+144.

$$x=\frac{-8bk±4\sqrt{9+12k^{2}-3b^{2}}}{2\left(4k^{2}+3\right)}$$

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

$$x=\frac{-8bk±4\sqrt{9+12k^{2}-3b^{2}}}{8k^{2}+6}$$

Now solve the equation x=\frac{-8bk±4\sqrt{9+12k^{2}-3b^{2}}}{8k^{2}+6} when ± is plus. Add -8kb to 4\sqrt{-3b^{2}+12k^{2}+9}.

$$x=\frac{-8bk+4\sqrt{9+12k^{2}-3b^{2}}}{8k^{2}+6}$$

Divide -8bk+4\sqrt{-3b^{2}+12k^{2}+9} by 6+8k^{2}.

$$x=\frac{2\left(-2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}$$

Now solve the equation x=\frac{-8bk±4\sqrt{9+12k^{2}-3b^{2}}}{8k^{2}+6} when ± is minus. Subtract 4\sqrt{-3b^{2}+12k^{2}+9} from -8kb.

$$x=\frac{-8bk-4\sqrt{9+12k^{2}-3b^{2}}}{8k^{2}+6}$$

Divide -8bk-4\sqrt{-3b^{2}+12k^{2}+9} by 6+8k^{2}.

$$x=-\frac{2\left(2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}$$

There are two solutions for x: \frac{2\left(-2bk+\sqrt{-3b^{2}+12k^{2}+9}\right)}{3+4k^{2}} and -\frac{2\left(2kb+\sqrt{9+12k^{2}-3b^{2}}\right)}{3+4k^{2}}. Substitute \frac{2\left(-2bk+\sqrt{-3b^{2}+12k^{2}+9}\right)}{3+4k^{2}} for x in the equation y=kx+b to find the corresponding solution for y that satisfies both equations.

$$y=k\times \frac{2\left(-2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}+b$$

Multiply k times \frac{2\left(-2bk+\sqrt{-3b^{2}+12k^{2}+9}\right)}{3+4k^{2}}.

$$y=\frac{2\left(-2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}k+b$$

Now substitute -\frac{2\left(2kb+\sqrt{9+12k^{2}-3b^{2}}\right)}{3+4k^{2}} for x in the equation y=kx+b and solve to find the corresponding solution for y that satisfies both equations.

$$y=k\left(-\frac{2\left(2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}\right)+b$$

Multiply k times -\frac{2\left(2kb+\sqrt{9+12k^{2}-3b^{2}}\right)}{3+4k^{2}}.

$$y=\left(-\frac{2\left(2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}\right)k+b$$

The system is now solved.

$$y=\frac{2\left(-2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}k+b,x=\frac{2\left(-2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}\text{ or }y=\left(-\frac{2\left(2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}\right)k+b,x=-\frac{2\left(2bk+\sqrt{9+12k^{2}-3b^{2}}\right)}{4k^{2}+3}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{2\left(2bk+\sqrt{3\left(3+4k^{2}-b^{2}\right)}\right)}{4k^{2}+3}\text{, }y=\frac{\sqrt{3}\left(-2k\sqrt{3+4k^{2}-b^{2}}+\sqrt{3}b\right)}{4k^{2}+3}\text{; }x=\frac{2\left(-2bk+\sqrt{3\left(3+4k^{2}-b^{2}\right)}\right)}{4k^{2}+3}\text{, }y=\frac{\sqrt{3}\left(2k\sqrt{3+4k^{2}-b^{2}}+\sqrt{3}b\right)}{4k^{2}+3}\text{, }&k\neq -\frac{\sqrt{3}i}{2}\text{ and }k\neq \frac{\sqrt{3}i}{2}\\x=-\frac{b^{2}-3}{2bk}\text{, }y=\frac{b^{2}+3}{2b}\text{, }&b\neq 0\text{ and }\left(k=-\frac{\sqrt{3}i}{2}\text{ or }k=\frac{\sqrt{3}i}{2}\right)\end{matrix}\right.$$