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

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

Solve for y, x

$$x=-\frac{2\sqrt{5}}{5}\approx -0.894427191\text{, }y=-\frac{4\sqrt{5}}{5}\approx -1.788854382$$

$$x=\frac{2\sqrt{5}}{5}\approx 0.894427191\text{, }y=\frac{4\sqrt{5}}{5}\approx 1.788854382$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 2x from both sides.

$$y-2x=0$$

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.

$$y-2x=0,x^{2}+y^{2}=4$$

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

$$y-2x=0$$

Subtract -2x from both sides of the equation.

$$y=2x$$

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

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

Square 2x.

$$x^{2}+4x^{2}=4$$

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

$$5x^{2}=4$$

Subtract 4 from both sides of the equation.

$$5x^{2}-4=0$$

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 -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

$$x=\frac{0±\sqrt{-20\left(-4\right)}}{2\times 5}$$

Multiply -20 times -4.

$$x=\frac{0±\sqrt{80}}{2\times 5}$$

Take the square root of 80.

$$x=\frac{0±4\sqrt{5}}{2\times 5}$$

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

$$x=\frac{0±4\sqrt{5}}{10}$$

Now solve the equation x=\frac{0±4\sqrt{5}}{10} when ± is plus.

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

Now solve the equation x=\frac{0±4\sqrt{5}}{10} when ± is minus.

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

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

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

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

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

The system is now solved.

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