Solve {c}{x^2+y^2=16}{y=3x+2}

$$\left. \begin{array} { c } { x ^ { 2 } + y ^ { 2 } = 16 } \\ { y = 3 x + 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{-\sqrt{39}-3}{5}\approx -1.8489996\text{, }y=\frac{1-3\sqrt{39}}{5}\approx -3.546998799$$

$$x=\frac{\sqrt{39}-3}{5}\approx 0.6489996\text{, }y=\frac{3\sqrt{39}+1}{5}\approx 3.946998799$$

Steps Using Substitution and Quadratic Formula

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

$$y-3x=2$$

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.

$$y-3x=2,x^{2}+y^{2}=16$$

Solve y-3x=2 for y by isolating y on the left hand side of the equal sign.

$$y-3x=2$$

Subtract -3x from both sides of the equation.

$$y=3x+2$$

Substitute 3x+2 for y in the other equation, x^{2}+y^{2}=16.

$$x^{2}+\left(3x+2\right)^{2}=16$$

Square 3x+2.

$$x^{2}+9x^{2}+12x+4=16$$

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

$$10x^{2}+12x+4=16$$

Subtract 16 from both sides of the equation.

$$10x^{2}+12x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 3^{2} for a, 1\times 2\times 2\times 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 10\left(-12\right)}}{2\times 10}$$

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

$$x=\frac{-12±\sqrt{144-4\times 10\left(-12\right)}}{2\times 10}$$

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

$$x=\frac{-12±\sqrt{144-40\left(-12\right)}}{2\times 10}$$

Multiply -40 times -12.

$$x=\frac{-12±\sqrt{144+480}}{2\times 10}$$

Add 144 to 480.

$$x=\frac{-12±\sqrt{624}}{2\times 10}$$

Take the square root of 624.

$$x=\frac{-12±4\sqrt{39}}{2\times 10}$$

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

$$x=\frac{-12±4\sqrt{39}}{20}$$

Now solve the equation x=\frac{-12±4\sqrt{39}}{20} when ± is plus. Add -12 to 4\sqrt{39}.

$$x=\frac{4\sqrt{39}-12}{20}$$

Divide -12+4\sqrt{39} by 20.

$$x=\frac{\sqrt{39}-3}{5}$$

Now solve the equation x=\frac{-12±4\sqrt{39}}{20} when ± is minus. Subtract 4\sqrt{39} from -12.

$$x=\frac{-4\sqrt{39}-12}{20}$$

Divide -12-4\sqrt{39} by 20.

$$x=\frac{-\sqrt{39}-3}{5}$$

There are two solutions for x: \frac{-3+\sqrt{39}}{5} and \frac{-3-\sqrt{39}}{5}. Substitute \frac{-3+\sqrt{39}}{5} for x in the equation y=3x+2 to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{\sqrt{39}-3}{5}+2$$

Now substitute \frac{-3-\sqrt{39}}{5} for x in the equation y=3x+2 and solve to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{-\sqrt{39}-3}{5}+2$$

The system is now solved.

$$y=3\times \frac{\sqrt{39}-3}{5}+2,x=\frac{\sqrt{39}-3}{5}\text{ or }y=3\times \frac{-\sqrt{39}-3}{5}+2,x=\frac{-\sqrt{39}-3}{5}$$