Solve {l}{x^2-y^2=400}{y=2x-10}

$$\left. \begin{array} { l } { x ^ { 2 } - y ^ { 2 } = 400 } \\ { y = 2 x - 10 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{-10\sqrt{11}i+20}{3}\approx 6.666666667-11.055415968i\text{, }y=\frac{-20\sqrt{11}i+10}{3}\approx 3.333333333-22.110831936i$$

$$x=\frac{20+10\sqrt{11}i}{3}\approx 6.666666667+11.055415968i\text{, }y=\frac{10+20\sqrt{11}i}{3}\approx 3.333333333+22.110831936i$$

Steps Using Substitution and Quadratic Formula

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

$$y-2x=-10$$

Subtract -2x from both sides of the equation.

$$y=2x-10$$

Substitute 2x-10 for y in the other equation, x^{2}-y^{2}=400.

$$x^{2}-\left(2x-10\right)^{2}=400$$

Square 2x-10.

$$x^{2}-\left(4x^{2}-40x+100\right)=400$$

Multiply -1 times 4x^{2}-40x+100.

$$x^{2}-4x^{2}+40x-100=400$$

Add x^{2} to -4x^{2}.

$$-3x^{2}+40x-100=400$$

Subtract 400 from both sides of the equation.

$$-3x^{2}+40x-500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-2^{2} for a, -\left(-10\right)\times 2\times 2 for b, and -500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-3\right)\left(-500\right)}}{2\left(-3\right)}$$

Square -\left(-10\right)\times 2\times 2.

$$x=\frac{-40±\sqrt{1600-4\left(-3\right)\left(-500\right)}}{2\left(-3\right)}$$

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

$$x=\frac{-40±\sqrt{1600+12\left(-500\right)}}{2\left(-3\right)}$$

Multiply 12 times -500.

$$x=\frac{-40±\sqrt{1600-6000}}{2\left(-3\right)}$$

Add 1600 to -6000.

$$x=\frac{-40±\sqrt{-4400}}{2\left(-3\right)}$$

Take the square root of -4400.

$$x=\frac{-40±20\sqrt{11}i}{2\left(-3\right)}$$

Multiply 2 times 1-2^{2}.

$$x=\frac{-40±20\sqrt{11}i}{-6}$$

Now solve the equation x=\frac{-40±20\sqrt{11}i}{-6} when ± is plus. Add -40 to 20i\sqrt{11}.

$$x=\frac{-40+20\sqrt{11}i}{-6}$$

Divide -40+20i\sqrt{11} by -6.

$$x=\frac{-10\sqrt{11}i+20}{3}$$

Now solve the equation x=\frac{-40±20\sqrt{11}i}{-6} when ± is minus. Subtract 20i\sqrt{11} from -40.

$$x=\frac{-20\sqrt{11}i-40}{-6}$$

Divide -40-20i\sqrt{11} by -6.

$$x=\frac{20+10\sqrt{11}i}{3}$$

There are two solutions for x: \frac{20-10i\sqrt{11}}{3} and \frac{20+10i\sqrt{11}}{3}. Substitute \frac{20-10i\sqrt{11}}{3} for x in the equation y=2x-10 to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-10\sqrt{11}i+20}{3}-10$$

Now substitute \frac{20+10i\sqrt{11}}{3} for x in the equation y=2x-10 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{20+10\sqrt{11}i}{3}-10$$

The system is now solved.

$$y=2\times \frac{-10\sqrt{11}i+20}{3}-10,x=\frac{-10\sqrt{11}i+20}{3}\text{ or }y=2\times \frac{20+10\sqrt{11}i}{3}-10,x=\frac{20+10\sqrt{11}i}{3}$$