Solve {l}{y=3x+1}{x^2+y^2=4}

$$\left. \begin{array} { l } { y = 3 x + 1 } \\ { x ^ { 2 } + y ^ { 2 } = 4 } \end{array} \right.$$

Solve for y, x

$$x=\frac{-\sqrt{39}-3}{10}\approx -0.9244998\text{, }y=\frac{1-3\sqrt{39}}{10}\approx -1.7734994$$

$$x=\frac{\sqrt{39}-3}{10}\approx 0.3244998\text{, }y=\frac{3\sqrt{39}+1}{10}\approx 1.9734994$$

Steps Using Substitution and Quadratic Formula

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

$$y-3x=1$$

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.

$$y-3x=1,x^{2}+y^{2}=4$$

Solve y-3x=1 for y by isolating y on the left hand side of the equal sign.

$$y-3x=1$$

Subtract -3x from both sides of the equation.

$$y=3x+1$$

Substitute 3x+1 for y in the other equation, x^{2}+y^{2}=4.

$$x^{2}+\left(3x+1\right)^{2}=4$$

Square 3x+1.

$$x^{2}+9x^{2}+6x+1=4$$

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

$$10x^{2}+6x+1=4$$

Subtract 4 from both sides of the equation.

$$10x^{2}+6x-3=0$$

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

$$x=\frac{-6±\sqrt{6^{2}-4\times 10\left(-3\right)}}{2\times 10}$$

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

$$x=\frac{-6±\sqrt{36-4\times 10\left(-3\right)}}{2\times 10}$$

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

$$x=\frac{-6±\sqrt{36-40\left(-3\right)}}{2\times 10}$$

Multiply -40 times -3.

$$x=\frac{-6±\sqrt{36+120}}{2\times 10}$$

Add 36 to 120.

$$x=\frac{-6±\sqrt{156}}{2\times 10}$$

Take the square root of 156.

$$x=\frac{-6±2\sqrt{39}}{2\times 10}$$

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

$$x=\frac{-6±2\sqrt{39}}{20}$$

Now solve the equation x=\frac{-6±2\sqrt{39}}{20} when ± is plus. Add -6 to 2\sqrt{39}.

$$x=\frac{2\sqrt{39}-6}{20}$$

Divide -6+2\sqrt{39} by 20.

$$x=\frac{\sqrt{39}-3}{10}$$

Now solve the equation x=\frac{-6±2\sqrt{39}}{20} when ± is minus. Subtract 2\sqrt{39} from -6.

$$x=\frac{-2\sqrt{39}-6}{20}$$

Divide -6-2\sqrt{39} by 20.

$$x=\frac{-\sqrt{39}-3}{10}$$

There are two solutions for x: \frac{-3+\sqrt{39}}{10} and \frac{-3-\sqrt{39}}{10}. Substitute \frac{-3+\sqrt{39}}{10} for x in the equation y=3x+1 to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{\sqrt{39}-3}{10}+1$$

Now substitute \frac{-3-\sqrt{39}}{10} for x in the equation y=3x+1 and solve to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{-\sqrt{39}-3}{10}+1$$

The system is now solved.

$$y=3\times \frac{\sqrt{39}-3}{10}+1,x=\frac{\sqrt{39}-3}{10}\text{ or }y=3\times \frac{-\sqrt{39}-3}{10}+1,x=\frac{-\sqrt{39}-3}{10}$$