Solve {l}{y=3x+8}{x^2+y^2=4}

$$\left\{ \begin{array} { l } { y = 3 x + 8 } \\ { x ^ { 2 } + y ^ { 2 } = 4 } \end{array} \right.$$

Solve for y, x (complex solution)

$$x=\frac{-\sqrt{6}i-12}{5}\approx -2.4-0.489897949i\text{, }y=\frac{-3\sqrt{6}i+4}{5}\approx 0.8-1.469693846i$$

$$x=\frac{-12+\sqrt{6}i}{5}\approx -2.4+0.489897949i\text{, }y=\frac{4+3\sqrt{6}i}{5}\approx 0.8+1.469693846i$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3x from both sides.

$$y-3x=8$$

Subtract -3x from both sides of the equation.

$$y=3x+8$$

Substitute 3x+8 for y in the other equation, x^{2}+y^{2}=4.

$$x^{2}+\left(3x+8\right)^{2}=4$$

Square 3x+8.

$$x^{2}+9x^{2}+48x+64=4$$

Add x^{2} to 9x^{2}.

$$10x^{2}+48x+64=4$$

Subtract 4 from both sides of the equation.

$$10x^{2}+48x+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 3^{2} for a, 1\times 8\times 2\times 3 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-48±\sqrt{48^{2}-4\times 10\times 60}}{2\times 10}$$

Square 1\times 8\times 2\times 3.

$$x=\frac{-48±\sqrt{2304-4\times 10\times 60}}{2\times 10}$$

Multiply -4 times 1+1\times 3^{2}.

$$x=\frac{-48±\sqrt{2304-40\times 60}}{2\times 10}$$

Multiply -40 times 60.

$$x=\frac{-48±\sqrt{2304-2400}}{2\times 10}$$

Add 2304 to -2400.

$$x=\frac{-48±\sqrt{-96}}{2\times 10}$$

Take the square root of -96.

$$x=\frac{-48±4\sqrt{6}i}{2\times 10}$$

Multiply 2 times 1+1\times 3^{2}.

$$x=\frac{-48±4\sqrt{6}i}{20}$$

Now solve the equation x=\frac{-48±4\sqrt{6}i}{20} when ± is plus. Add -48 to 4i\sqrt{6}.

$$x=\frac{-48+4\sqrt{6}i}{20}$$

Divide -48+4i\sqrt{6} by 20.

$$x=\frac{-12+\sqrt{6}i}{5}$$

Now solve the equation x=\frac{-48±4\sqrt{6}i}{20} when ± is minus. Subtract 4i\sqrt{6} from -48.

$$x=\frac{-4\sqrt{6}i-48}{20}$$

Divide -48-4i\sqrt{6} by 20.

$$x=\frac{-\sqrt{6}i-12}{5}$$

There are two solutions for x: \frac{-12+i\sqrt{6}}{5} and \frac{-12-i\sqrt{6}}{5}. Substitute \frac{-12+i\sqrt{6}}{5} for x in the equation y=3x+8 to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{-12+\sqrt{6}i}{5}+8$$

Now substitute \frac{-12-i\sqrt{6}}{5} for x in the equation y=3x+8 and solve to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{-\sqrt{6}i-12}{5}+8$$

The system is now solved.

$$y=3\times \frac{-12+\sqrt{6}i}{5}+8,x=\frac{-12+\sqrt{6}i}{5}\text{ or }y=3\times \frac{-\sqrt{6}i-12}{5}+8,x=\frac{-\sqrt{6}i-12}{5}$$