Solve {l}{x^2+y^2=16}{x-y=24}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 16 } \\ { x - y = 24 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-2\sqrt{34}i+12\approx 12-11.66190379i\text{, }y=-2\sqrt{34}i-12\approx -12-11.66190379i$$

$$x=12+2\sqrt{34}i\approx 12+11.66190379i\text{, }y=-12+2\sqrt{34}i\approx -12+11.66190379i$$

Steps Using Substitution and Quadratic Formula

Solve x-y=24 for x by isolating x on the left hand side of the equal sign.

$$x-y=24$$

Subtract -y from both sides of the equation.

$$x=y+24$$

Substitute y+24 for x in the other equation, y^{2}+x^{2}=16.

$$y^{2}+\left(y+24\right)^{2}=16$$

Square y+24.

$$y^{2}+y^{2}+48y+576=16$$

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

$$2y^{2}+48y+576=16$$

Subtract 16 from both sides of the equation.

$$2y^{2}+48y+560=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 24\times 1\times 2 for b, and 560 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-48±\sqrt{48^{2}-4\times 2\times 560}}{2\times 2}$$

Square 1\times 24\times 1\times 2.

$$y=\frac{-48±\sqrt{2304-4\times 2\times 560}}{2\times 2}$$

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

$$y=\frac{-48±\sqrt{2304-8\times 560}}{2\times 2}$$

Multiply -8 times 560.

$$y=\frac{-48±\sqrt{2304-4480}}{2\times 2}$$

Add 2304 to -4480.

$$y=\frac{-48±\sqrt{-2176}}{2\times 2}$$

Take the square root of -2176.

$$y=\frac{-48±8\sqrt{34}i}{2\times 2}$$

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

$$y=\frac{-48±8\sqrt{34}i}{4}$$

Now solve the equation y=\frac{-48±8\sqrt{34}i}{4} when ± is plus. Add -48 to 8i\sqrt{34}.

$$y=\frac{-48+8\sqrt{34}i}{4}$$

Divide -48+8i\sqrt{34} by 4.

$$y=-12+2\sqrt{34}i$$

Now solve the equation y=\frac{-48±8\sqrt{34}i}{4} when ± is minus. Subtract 8i\sqrt{34} from -48.

$$y=\frac{-8\sqrt{34}i-48}{4}$$

Divide -48-8i\sqrt{34} by 4.

$$y=-2\sqrt{34}i-12$$

There are two solutions for y: -12+2i\sqrt{34} and -12-2i\sqrt{34}. Substitute -12+2i\sqrt{34} for y in the equation x=y+24 to find the corresponding solution for x that satisfies both equations.

$$x=-12+2\sqrt{34}i+24$$

Add 1\left(-12+2i\sqrt{34}\right) to 24.

$$x=12+2\sqrt{34}i$$

Now substitute -12-2i\sqrt{34} for y in the equation x=y+24 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\sqrt{34}i-12+24$$

Add 1\left(-12-2i\sqrt{34}\right) to 24.

$$x=-2\sqrt{34}i+12$$

The system is now solved.

$$x=12+2\sqrt{34}i,y=-12+2\sqrt{34}i\text{ or }x=-2\sqrt{34}i+12,y=-2\sqrt{34}i-12$$