Solve {l}{x^2+y^2=1}{3x+6y=2}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { 3 x + 6 y = 2 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\sqrt{41}+2}{15}\approx 0.987083232\text{, }y=\frac{4-\sqrt{41}}{15}\approx -0.160208282$$

$$x=\frac{2-2\sqrt{41}}{15}\approx -0.720416565\text{, }y=\frac{\sqrt{41}+4}{15}\approx 0.693541616$$

Steps Using Substitution and Quadratic Formula

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.

$$3x+6y=2,y^{2}+x^{2}=1$$

Solve 3x+6y=2 for x by isolating x on the left hand side of the equal sign.

$$3x+6y=2$$

Subtract 6y from both sides of the equation.

$$3x=-6y+2$$

Divide both sides by 3.

$$x=-2y+\frac{2}{3}$$

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

$$y^{2}+\left(-2y+\frac{2}{3}\right)^{2}=1$$

Square -2y+\frac{2}{3}.

$$y^{2}+4y^{2}-\frac{8}{3}y+\frac{4}{9}=1$$

Add y^{2} to 4y^{2}.

$$5y^{2}-\frac{8}{3}y+\frac{4}{9}=1$$

Subtract 1 from both sides of the equation.

$$5y^{2}-\frac{8}{3}y-\frac{5}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-2\right)^{2} for a, 1\times \frac{2}{3}\left(-2\right)\times 2 for b, and -\frac{5}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\left(-\frac{8}{3}\right)^{2}-4\times 5\left(-\frac{5}{9}\right)}}{2\times 5}$$

Square 1\times \frac{2}{3}\left(-2\right)\times 2.

$$y=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-4\times 5\left(-\frac{5}{9}\right)}}{2\times 5}$$

Multiply -4 times 1+1\left(-2\right)^{2}.

$$y=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64}{9}-20\left(-\frac{5}{9}\right)}}{2\times 5}$$

Multiply -20 times -\frac{5}{9}.

$$y=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{64+100}{9}}}{2\times 5}$$

Add \frac{64}{9} to \frac{100}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{8}{3}\right)±\sqrt{\frac{164}{9}}}{2\times 5}$$

Take the square root of \frac{164}{9}.

$$y=\frac{-\left(-\frac{8}{3}\right)±\frac{2\sqrt{41}}{3}}{2\times 5}$$

The opposite of 1\times \frac{2}{3}\left(-2\right)\times 2 is \frac{8}{3}.

$$y=\frac{\frac{8}{3}±\frac{2\sqrt{41}}{3}}{2\times 5}$$

Multiply 2 times 1+1\left(-2\right)^{2}.

$$y=\frac{\frac{8}{3}±\frac{2\sqrt{41}}{3}}{10}$$

Now solve the equation y=\frac{\frac{8}{3}±\frac{2\sqrt{41}}{3}}{10} when ± is plus. Add \frac{8}{3} to \frac{2\sqrt{41}}{3}.

$$y=\frac{2\sqrt{41}+8}{3\times 10}$$

Divide \frac{8+2\sqrt{41}}{3} by 10.

$$y=\frac{\sqrt{41}+4}{15}$$

Now solve the equation y=\frac{\frac{8}{3}±\frac{2\sqrt{41}}{3}}{10} when ± is minus. Subtract \frac{2\sqrt{41}}{3} from \frac{8}{3}.

$$y=\frac{8-2\sqrt{41}}{3\times 10}$$

Divide \frac{8-2\sqrt{41}}{3} by 10.

$$y=\frac{4-\sqrt{41}}{15}$$

There are two solutions for y: \frac{4+\sqrt{41}}{15} and \frac{4-\sqrt{41}}{15}. Substitute \frac{4+\sqrt{41}}{15} for y in the equation x=-2y+\frac{2}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{\sqrt{41}+4}{15}+\frac{2}{3}$$

Now substitute \frac{4-\sqrt{41}}{15} for y in the equation x=-2y+\frac{2}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{4-\sqrt{41}}{15}+\frac{2}{3}$$

The system is now solved.

$$x=-2\times \frac{\sqrt{41}+4}{15}+\frac{2}{3},y=\frac{\sqrt{41}+4}{15}\text{ or }x=-2\times \frac{4-\sqrt{41}}{15}+\frac{2}{3},y=\frac{4-\sqrt{41}}{15}$$