Solve {l}{x^2/16+y^2/4=1}{y=2x+9}

$$\left\{ \begin{array} { l } { \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 4 } = 1 } \\ { y = 2 x + 9 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{-2\sqrt{13}i-72}{17}\approx -4.235294118-0.424182503i\text{, }y=\frac{-4\sqrt{13}i+9}{17}\approx 0.529411765-0.848365006i$$

$$x=\frac{-72+2\sqrt{13}i}{17}\approx -4.235294118+0.424182503i\text{, }y=\frac{9+4\sqrt{13}i}{17}\approx 0.529411765+0.848365006i$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 16, the least common multiple of 16,4.

$$x^{2}+4y^{2}=16$$

Consider the second equation. Subtract 2x from both sides.

$$y-2x=9$$

Subtract -2x from both sides of the equation.

$$y=2x+9$$

Substitute 2x+9 for y in the other equation, x^{2}+4y^{2}=16.

$$x^{2}+4\left(2x+9\right)^{2}=16$$

Square 2x+9.

$$x^{2}+4\left(4x^{2}+36x+81\right)=16$$

Multiply 4 times 4x^{2}+36x+81.

$$x^{2}+16x^{2}+144x+324=16$$

Add x^{2} to 16x^{2}.

$$17x^{2}+144x+324=16$$

Subtract 16 from both sides of the equation.

$$17x^{2}+144x+308=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+4\times 2^{2} for a, 4\times 9\times 2\times 2 for b, and 308 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-144±\sqrt{144^{2}-4\times 17\times 308}}{2\times 17}$$

Square 4\times 9\times 2\times 2.

$$x=\frac{-144±\sqrt{20736-4\times 17\times 308}}{2\times 17}$$

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

$$x=\frac{-144±\sqrt{20736-68\times 308}}{2\times 17}$$

Multiply -68 times 308.

$$x=\frac{-144±\sqrt{20736-20944}}{2\times 17}$$

Add 20736 to -20944.

$$x=\frac{-144±\sqrt{-208}}{2\times 17}$$

Take the square root of -208.

$$x=\frac{-144±4\sqrt{13}i}{2\times 17}$$

Multiply 2 times 1+4\times 2^{2}.

$$x=\frac{-144±4\sqrt{13}i}{34}$$

Now solve the equation x=\frac{-144±4\sqrt{13}i}{34} when ± is plus. Add -144 to 4i\sqrt{13}.

$$x=\frac{-144+4\sqrt{13}i}{34}$$

Divide -144+4i\sqrt{13} by 34.

$$x=\frac{-72+2\sqrt{13}i}{17}$$

Now solve the equation x=\frac{-144±4\sqrt{13}i}{34} when ± is minus. Subtract 4i\sqrt{13} from -144.

$$x=\frac{-4\sqrt{13}i-144}{34}$$

Divide -144-4i\sqrt{13} by 34.

$$x=\frac{-2\sqrt{13}i-72}{17}$$

There are two solutions for x: \frac{-72+2i\sqrt{13}}{17} and \frac{-72-2i\sqrt{13}}{17}. Substitute \frac{-72+2i\sqrt{13}}{17} for x in the equation y=2x+9 to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-72+2\sqrt{13}i}{17}+9$$

Now substitute \frac{-72-2i\sqrt{13}}{17} for x in the equation y=2x+9 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-2\sqrt{13}i-72}{17}+9$$

The system is now solved.

$$y=2\times \frac{-72+2\sqrt{13}i}{17}+9,x=\frac{-72+2\sqrt{13}i}{17}\text{ or }y=2\times \frac{-2\sqrt{13}i-72}{17}+9,x=\frac{-2\sqrt{13}i-72}{17}$$