Solve {l}{x^2+y^2=1}{x-2y=0}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { x - 2 y = 0 } \end{array} \right.$$

Solve for x, y

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

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

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.

$$x-2y=0,y^{2}+x^{2}=1$$

Solve x-2y=0 for x by isolating x on the left hand side of the equal sign.

$$x-2y=0$$

Subtract -2y from both sides of the equation.

$$x=2y$$

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

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

Square 2y.

$$y^{2}+4y^{2}=1$$

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

$$5y^{2}=1$$

Subtract 1 from both sides of the equation.

$$5y^{2}-1=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square 1\times 0\times 2\times 2.

$$y=\frac{0±\sqrt{-4\times 5\left(-1\right)}}{2\times 5}$$

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

$$y=\frac{0±\sqrt{-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$y=\frac{0±\sqrt{20}}{2\times 5}$$

Take the square root of 20.

$$y=\frac{0±2\sqrt{5}}{2\times 5}$$

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

$$y=\frac{0±2\sqrt{5}}{10}$$

Now solve the equation y=\frac{0±2\sqrt{5}}{10} when ± is plus.

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

Now solve the equation y=\frac{0±2\sqrt{5}}{10} when ± is minus.

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

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

$$x=2\times \frac{\sqrt{5}}{5}$$

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

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

The system is now solved.

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