Solve {l}{x^2+y^2=16}{y=4x+2}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 16 } \\ { y = 4 x + 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{-2\sqrt{67}-8}{17}\approx -1.433570914\text{, }y=\frac{2-8\sqrt{67}}{17}\approx -3.734283657$$

$$x=\frac{2\sqrt{67}-8}{17}\approx 0.492394444\text{, }y=\frac{8\sqrt{67}+2}{17}\approx 3.969577775$$

Steps Using Substitution and Quadratic Formula

Consider the second equation. Subtract 4x from both sides.

$$y-4x=2$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$y-4x=2,x^{2}+y^{2}=16$$

Solve y-4x=2 for y by isolating y on the left hand side of the equal sign.

$$y-4x=2$$

Subtract -4x from both sides of the equation.

$$y=4x+2$$

Substitute 4x+2 for y in the other equation, x^{2}+y^{2}=16.

$$x^{2}+\left(4x+2\right)^{2}=16$$

Square 4x+2.

$$x^{2}+16x^{2}+16x+4=16$$

Add x^{2} to 16x^{2}.

$$17x^{2}+16x+4=16$$

Subtract 16 from both sides of the equation.

$$17x^{2}+16x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 4^{2} for a, 1\times 2\times 2\times 4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 17\left(-12\right)}}{2\times 17}$$

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

$$x=\frac{-16±\sqrt{256-4\times 17\left(-12\right)}}{2\times 17}$$

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

$$x=\frac{-16±\sqrt{256-68\left(-12\right)}}{2\times 17}$$

Multiply -68 times -12.

$$x=\frac{-16±\sqrt{256+816}}{2\times 17}$$

Add 256 to 816.

$$x=\frac{-16±\sqrt{1072}}{2\times 17}$$

Take the square root of 1072.

$$x=\frac{-16±4\sqrt{67}}{2\times 17}$$

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

$$x=\frac{-16±4\sqrt{67}}{34}$$

Now solve the equation x=\frac{-16±4\sqrt{67}}{34} when ± is plus. Add -16 to 4\sqrt{67}.

$$x=\frac{4\sqrt{67}-16}{34}$$

Divide -16+4\sqrt{67} by 34.

$$x=\frac{2\sqrt{67}-8}{17}$$

Now solve the equation x=\frac{-16±4\sqrt{67}}{34} when ± is minus. Subtract 4\sqrt{67} from -16.

$$x=\frac{-4\sqrt{67}-16}{34}$$

Divide -16-4\sqrt{67} by 34.

$$x=\frac{-2\sqrt{67}-8}{17}$$

Both solutions for x are the same: \frac{-8+2\sqrt{67}}{17}. Substitute \frac{-8+2\sqrt{67}}{17} for x in the equation y=4x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=4\times \frac{2\sqrt{67}-8}{17}+2$$

Now substitute \frac{-8-2\sqrt{67}}{17} for x in the equation y=4x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=4\times \frac{-2\sqrt{67}-8}{17}+2$$

The system is now solved.

$$y=4\times \frac{2\sqrt{67}-8}{17}+2,x=\frac{2\sqrt{67}-8}{17}\text{ or }y=4\times \frac{-2\sqrt{67}-8}{17}+2,x=\frac{-2\sqrt{67}-8}{17}$$