Solve {l}{x^2+y^2=3}{x-y=4}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 3 } \\ { x - y = 4 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-\frac{\sqrt{10}i}{2}+2\approx 2-1.58113883i\text{, }y=-\frac{\sqrt{10}i}{2}-2\approx -2-1.58113883i$$

$$x=\frac{\sqrt{10}i}{2}+2\approx 2+1.58113883i\text{, }y=\frac{\sqrt{10}i}{2}-2\approx -2+1.58113883i$$

Steps Using Substitution and Quadratic Formula

Solve x-y=4 for x by isolating x on the left hand side of the equal sign.

$$x-y=4$$

Subtract -y from both sides of the equation.

$$x=y+4$$

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

$$y^{2}+\left(y+4\right)^{2}=3$$

Square y+4.

$$y^{2}+y^{2}+8y+16=3$$

Add y^{2} to y^{2}.

$$2y^{2}+8y+16=3$$

Subtract 3 from both sides of the equation.

$$2y^{2}+8y+13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 4\times 1\times 2 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-8±\sqrt{8^{2}-4\times 2\times 13}}{2\times 2}$$

Square 1\times 4\times 1\times 2.

$$y=\frac{-8±\sqrt{64-4\times 2\times 13}}{2\times 2}$$

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

$$y=\frac{-8±\sqrt{64-8\times 13}}{2\times 2}$$

Multiply -8 times 13.

$$y=\frac{-8±\sqrt{64-104}}{2\times 2}$$

Add 64 to -104.

$$y=\frac{-8±\sqrt{-40}}{2\times 2}$$

Take the square root of -40.

$$y=\frac{-8±2\sqrt{10}i}{2\times 2}$$

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

$$y=\frac{-8±2\sqrt{10}i}{4}$$

Now solve the equation y=\frac{-8±2\sqrt{10}i}{4} when ± is plus. Add -8 to 2i\sqrt{10}.

$$y=\frac{-8+2\sqrt{10}i}{4}$$

Divide -8+2i\sqrt{10} by 4.

$$y=\frac{\sqrt{10}i}{2}-2$$

Now solve the equation y=\frac{-8±2\sqrt{10}i}{4} when ± is minus. Subtract 2i\sqrt{10} from -8.

$$y=\frac{-2\sqrt{10}i-8}{4}$$

Divide -8-2i\sqrt{10} by 4.

$$y=-\frac{\sqrt{10}i}{2}-2$$

There are two solutions for y: -2+\frac{i\sqrt{10}}{2} and -2-\frac{i\sqrt{10}}{2}. Substitute -2+\frac{i\sqrt{10}}{2} for y in the equation x=y+4 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{10}i}{2}-2+4$$

Add 1\left(-2+\frac{i\sqrt{10}}{2}\right) to 4.

$$x=\frac{\sqrt{10}i}{2}+2$$

Now substitute -2-\frac{i\sqrt{10}}{2} for y in the equation x=y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{10}i}{2}-2+4$$

Add 1\left(-2-\frac{i\sqrt{10}}{2}\right) to 4.

$$x=-\frac{\sqrt{10}i}{2}+2$$

The system is now solved.

$$x=\frac{\sqrt{10}i}{2}+2,y=\frac{\sqrt{10}i}{2}-2\text{ or }x=-\frac{\sqrt{10}i}{2}+2,y=-\frac{\sqrt{10}i}{2}-2$$