Solve {l}{x^2+y^2=-36}{y=x-2}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = - 36 } \\ { y = x - 2 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-\sqrt{19}i+1\approx 1-4.358898944i\text{, }y=-\sqrt{19}i-1\approx -1-4.358898944i$$

$$x=1+\sqrt{19}i\approx 1+4.358898944i\text{, }y=-1+\sqrt{19}i\approx -1+4.358898944i$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract x from both sides.

$$y-x=-2$$

Subtract -x from both sides of the equation.

$$y=x-2$$

Substitute x-2 for y in the other equation, x^{2}+y^{2}=-36.

$$x^{2}+\left(x-2\right)^{2}=-36$$

Square x-2.

$$x^{2}+x^{2}-4x+4=-36$$

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

$$2x^{2}-4x+4=-36$$

Add 36 to both sides of the equation.

$$2x^{2}-4x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\left(-2\right)\times 1\times 2 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\times 40}}{2\times 2}$$

Square 1\left(-2\right)\times 1\times 2.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 40}}{2\times 2}$$

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

$$x=\frac{-\left(-4\right)±\sqrt{16-8\times 40}}{2\times 2}$$

Multiply -8 times 40.

$$x=\frac{-\left(-4\right)±\sqrt{16-320}}{2\times 2}$$

Add 16 to -320.

$$x=\frac{-\left(-4\right)±\sqrt{-304}}{2\times 2}$$

Take the square root of -304.

$$x=\frac{-\left(-4\right)±4\sqrt{19}i}{2\times 2}$$

The opposite of 1\left(-2\right)\times 1\times 2 is 4.

$$x=\frac{4±4\sqrt{19}i}{2\times 2}$$

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

$$x=\frac{4±4\sqrt{19}i}{4}$$

Now solve the equation x=\frac{4±4\sqrt{19}i}{4} when ± is plus. Add 4 to 4i\sqrt{19}.

$$x=\frac{4+4\sqrt{19}i}{4}$$

Divide 4+4i\sqrt{19} by 4.

$$x=1+\sqrt{19}i$$

Now solve the equation x=\frac{4±4\sqrt{19}i}{4} when ± is minus. Subtract 4i\sqrt{19} from 4.

$$x=\frac{-4\sqrt{19}i+4}{4}$$

Divide 4-4i\sqrt{19} by 4.

$$x=-\sqrt{19}i+1$$

There are two solutions for x: 1+i\sqrt{19} and 1-i\sqrt{19}. Substitute 1+i\sqrt{19} for x in the equation y=x-2 to find the corresponding solution for y that satisfies both equations.

$$y=1+\sqrt{19}i-2$$

Add 1\left(1+i\sqrt{19}\right) to -2.

$$y=-1+\sqrt{19}i$$

Now substitute 1-i\sqrt{19} for x in the equation y=x-2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=-\sqrt{19}i+1-2$$

Add 1\left(1-i\sqrt{19}\right) to -2.

$$y=-\sqrt{19}i-1$$

The system is now solved.

$$y=-1+\sqrt{19}i,x=1+\sqrt{19}i\text{ or }y=-\sqrt{19}i-1,x=-\sqrt{19}i+1$$