Solve {l}{x-3y=sqrt{5}}{x^2+y^2=1}

$$\left. \begin{array} { l } { x - 3 y = \sqrt { 5 } } \\ { x ^ { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{\sqrt{5}}{5}\approx -0.447213595\text{, }y=-\frac{2\sqrt{5}}{5}\approx -0.894427191$$

$$x=\frac{2\sqrt{5}}{5}\approx 0.894427191\text{, }y=-\frac{\sqrt{5}}{5}\approx -0.447213595$$

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.

$$x-3y=\sqrt{5},y^{2}+x^{2}=1$$

Solve x-3y=\sqrt{5} for x by isolating x on the left hand side of the equal sign.

$$x-3y=\sqrt{5}$$

Subtract -3y from both sides of the equation.

$$x=3y+\sqrt{5}$$

Substitute 3y+\sqrt{5} for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(3y+\sqrt{5}\right)^{2}=1$$

Square 3y+\sqrt{5}.

$$y^{2}+9y^{2}+6\sqrt{5}y+\left(\sqrt{5}\right)^{2}=1$$

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

$$10y^{2}+6\sqrt{5}y+\left(\sqrt{5}\right)^{2}=1$$

Subtract 1 from both sides of the equation.

$$10y^{2}+6\sqrt{5}y+\left(\sqrt{5}\right)^{2}-1=0$$

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

$$y=\frac{-6\sqrt{5}±\sqrt{\left(6\sqrt{5}\right)^{2}-4\times 10\times 4}}{2\times 10}$$

Square 1\times 2\times 3\sqrt{5}.

$$y=\frac{-6\sqrt{5}±\sqrt{180-4\times 10\times 4}}{2\times 10}$$

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

$$y=\frac{-6\sqrt{5}±\sqrt{180-40\times 4}}{2\times 10}$$

Multiply -40 times 4.

$$y=\frac{-6\sqrt{5}±\sqrt{180-160}}{2\times 10}$$

Add 180 to -160.

$$y=\frac{-6\sqrt{5}±\sqrt{20}}{2\times 10}$$

Take the square root of 20.

$$y=\frac{-6\sqrt{5}±2\sqrt{5}}{2\times 10}$$

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

$$y=\frac{-6\sqrt{5}±2\sqrt{5}}{20}$$

Now solve the equation y=\frac{-6\sqrt{5}±2\sqrt{5}}{20} when ± is plus. Add -6\sqrt{5} to 2\sqrt{5}.

$$y=-\frac{4\sqrt{5}}{20}$$

Divide -4\sqrt{5} by 20.

$$y=-\frac{\sqrt{5}}{5}$$

Now solve the equation y=\frac{-6\sqrt{5}±2\sqrt{5}}{20} when ± is minus. Subtract 2\sqrt{5} from -6\sqrt{5}.

$$y=-\frac{8\sqrt{5}}{20}$$

Divide -8\sqrt{5} by 20.

$$y=-\frac{2\sqrt{5}}{5}$$

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

$$x=3\left(-\frac{\sqrt{5}}{5}\right)+\sqrt{5}$$

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

$$x=3\left(-\frac{2\sqrt{5}}{5}\right)+\sqrt{5}$$

The system is now solved.

$$x=3\left(-\frac{\sqrt{5}}{5}\right)+\sqrt{5},y=-\frac{\sqrt{5}}{5}\text{ or }x=3\left(-\frac{2\sqrt{5}}{5}\right)+\sqrt{5},y=-\frac{2\sqrt{5}}{5}$$