Solve {l}{x^2-y^2=4}{y=2(x-1)}

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

Solve for x, y (complex solution)

$$x=\frac{-2\sqrt{2}i+4}{3}\approx 1.333333333-0.942809042i\text{, }y=\frac{-4\sqrt{2}i+2}{3}\approx 0.666666667-1.885618083i$$

$$x=\frac{4+2\sqrt{2}i}{3}\approx 1.333333333+0.942809042i\text{, }y=\frac{2+4\sqrt{2}i}{3}\approx 0.666666667+1.885618083i$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Use the distributive property to multiply 2 by x-1.

$$y=2x-2$$

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

$$x^{2}-\left(2x-2\right)^{2}=4$$

Square 2x-2.

$$x^{2}-\left(4x^{2}-8x+4\right)=4$$

Multiply -1 times 4x^{2}-8x+4.

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

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

$$-3x^{2}+8x-4=4$$

Subtract 4 from both sides of the equation.

$$-3x^{2}+8x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-2^{2} for a, -\left(-2\right)\times 2\times 2 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

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

$$x=\frac{-8±\sqrt{64-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

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

$$x=\frac{-8±\sqrt{64+12\left(-8\right)}}{2\left(-3\right)}$$

Multiply 12 times -8.

$$x=\frac{-8±\sqrt{64-96}}{2\left(-3\right)}$$

Add 64 to -96.

$$x=\frac{-8±\sqrt{-32}}{2\left(-3\right)}$$

Take the square root of -32.

$$x=\frac{-8±4\sqrt{2}i}{2\left(-3\right)}$$

Multiply 2 times 1-2^{2}.

$$x=\frac{-8±4\sqrt{2}i}{-6}$$

Now solve the equation x=\frac{-8±4\sqrt{2}i}{-6} when ± is plus. Add -8 to 4i\sqrt{2}.

$$x=\frac{-8+2^{\frac{5}{2}}i}{-6}$$

Divide -8+i\times 2^{\frac{5}{2}} by -6.

$$x=\frac{-2\sqrt{2}i+4}{3}$$

Now solve the equation x=\frac{-8±4\sqrt{2}i}{-6} when ± is minus. Subtract 4i\sqrt{2} from -8.

$$x=\frac{-2^{\frac{5}{2}}i-8}{-6}$$

Divide -8-i\times 2^{\frac{5}{2}} by -6.

$$x=\frac{4+2\sqrt{2}i}{3}$$

There are two solutions for x: \frac{4-2i\sqrt{2}}{3} and \frac{4+2i\sqrt{2}}{3}. Substitute \frac{4-2i\sqrt{2}}{3} for x in the equation y=2x-2 to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{-2\sqrt{2}i+4}{3}-2$$

Now substitute \frac{4+2i\sqrt{2}}{3} for x in the equation y=2x-2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{4+2\sqrt{2}i}{3}-2$$

The system is now solved.

$$y=2\times \frac{-2\sqrt{2}i+4}{3}-2,x=\frac{-2\sqrt{2}i+4}{3}\text{ or }y=2\times \frac{4+2\sqrt{2}i}{3}-2,x=\frac{4+2\sqrt{2}i}{3}$$