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

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

Solve for x, y (complex solution)

$$x=-\sqrt{2}i-2\approx -2-1.414213562i\text{, }y=-\frac{\sqrt{2}i}{2}+1\approx 1-0.707106781i$$

$$x=-2+\sqrt{2}i\approx -2+1.414213562i\text{, }y=\frac{\sqrt{2}i}{2}+1\approx 1+0.707106781i$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4.

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

Consider the second equation. Subtract \frac{1}{2}x from both sides.

$$y-2-\frac{1}{2}x=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$y-\frac{1}{2}x=2$$

Subtract -\frac{1}{2}x from both sides of the equation.

$$y=\frac{1}{2}x+2$$

Substitute \frac{1}{2}x+2 for y in the other equation, x^{2}+4y^{2}=4.

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

Square \frac{1}{2}x+2.

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

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

$$x^{2}+x^{2}+8x+16=4$$

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

$$2x^{2}+8x+16=4$$

Subtract 4 from both sides of the equation.

$$2x^{2}+8x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+4\times \left(\frac{1}{2}\right)^{2} for a, 4\times 2\times \frac{1}{2}\times 2 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\times 12}}{2\times 2}$$

Square 4\times 2\times \frac{1}{2}\times 2.

$$x=\frac{-8±\sqrt{64-4\times 2\times 12}}{2\times 2}$$

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

$$x=\frac{-8±\sqrt{64-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$x=\frac{-8±\sqrt{64-96}}{2\times 2}$$

Add 64 to -96.

$$x=\frac{-8±\sqrt{-32}}{2\times 2}$$

Take the square root of -32.

$$x=\frac{-8±4\sqrt{2}i}{2\times 2}$$

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

$$x=\frac{-8±4\sqrt{2}i}{4}$$

Now solve the equation x=\frac{-8±4\sqrt{2}i}{4} when ± is plus. Add -8 to 4i\sqrt{2}.

$$x=\frac{-8+2^{\frac{5}{2}}i}{4}$$

Divide -8+i\times 2^{\frac{5}{2}} by 4.

$$x=-2+\sqrt{2}i$$

Now solve the equation x=\frac{-8±4\sqrt{2}i}{4} when ± is minus. Subtract 4i\sqrt{2} from -8.

$$x=\frac{-2^{\frac{5}{2}}i-8}{4}$$

Divide -8-i\times 2^{\frac{5}{2}} by 4.

$$x=-\sqrt{2}i-2$$

There are two solutions for x: -2+i\sqrt{2} and -2-i\sqrt{2}. Substitute -2+i\sqrt{2} for x in the equation y=\frac{1}{2}x+2 to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{2}\left(-2+\sqrt{2}i\right)+2$$

Multiply \frac{1}{2} times -2+i\sqrt{2}.

$$y=\frac{-2+\sqrt{2}i}{2}+2$$

Now substitute -2-i\sqrt{2} for x in the equation y=\frac{1}{2}x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{1}{2}\left(-\sqrt{2}i-2\right)+2$$

Multiply \frac{1}{2} times -2-i\sqrt{2}.

$$y=\frac{-\sqrt{2}i-2}{2}+2$$

The system is now solved.

$$y=\frac{-2+\sqrt{2}i}{2}+2,x=-2+\sqrt{2}i\text{ or }y=\frac{-\sqrt{2}i-2}{2}+2,x=-\sqrt{2}i-2$$