Solve {l}{2x+-y=0}{x^2+y^2=9}

$$\left\{ \begin{array} { l } { 2 x + - y = 0 } \\ { x ^ { 2 } + y ^ { 2 } = 9 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{3\sqrt{5}}{5}\approx -1.341640786\text{, }y=-\frac{6\sqrt{5}}{5}\approx -2.683281573$$

$$x=\frac{3\sqrt{5}}{5}\approx 1.341640786\text{, }y=\frac{6\sqrt{5}}{5}\approx 2.683281573$$

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.

$$2x-y=0,y^{2}+x^{2}=9$$

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

$$2x-y=0$$

Subtract -y from both sides of the equation.

$$2x=y$$

Divide both sides by 2.

$$x=\frac{1}{2}y$$

Substitute \frac{1}{2}y for x in the other equation, y^{2}+x^{2}=9.

$$y^{2}+\left(\frac{1}{2}y\right)^{2}=9$$

Square \frac{1}{2}y.

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

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}=9$$

Subtract 9 from both sides of the equation.

$$\frac{5}{4}y^{2}-9=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times \frac{5}{4}\left(-9\right)}}{2\times \frac{5}{4}}$$

Square 1\times 0\times \frac{1}{2}\times 2.

$$y=\frac{0±\sqrt{-4\times \frac{5}{4}\left(-9\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{0±\sqrt{-5\left(-9\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -9.

$$y=\frac{0±\sqrt{45}}{2\times \frac{5}{4}}$$

Take the square root of 45.

$$y=\frac{0±3\sqrt{5}}{2\times \frac{5}{4}}$$

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

$$y=\frac{0±3\sqrt{5}}{\frac{5}{2}}$$

Now solve the equation y=\frac{0±3\sqrt{5}}{\frac{5}{2}} when ± is plus.

$$y=\frac{6\sqrt{5}}{5}$$

Now solve the equation y=\frac{0±3\sqrt{5}}{\frac{5}{2}} when ± is minus.

$$y=-\frac{6\sqrt{5}}{5}$$

There are two solutions for y: \frac{6\sqrt{5}}{5} and -\frac{6\sqrt{5}}{5}. Substitute \frac{6\sqrt{5}}{5} for y in the equation x=\frac{1}{2}y to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\times \frac{6\sqrt{5}}{5}$$

Multiply \frac{1}{2} times \frac{6\sqrt{5}}{5}.

$$x=\frac{6\sqrt{5}}{2\times 5}$$

Now substitute -\frac{6\sqrt{5}}{5} for y in the equation x=\frac{1}{2}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{6\sqrt{5}}{5}\right)$$

Multiply \frac{1}{2} times -\frac{6\sqrt{5}}{5}.

$$x=-\frac{6\sqrt{5}}{2\times 5}$$

The system is now solved.

$$x=\frac{6\sqrt{5}}{2\times 5},y=\frac{6\sqrt{5}}{5}\text{ or }x=-\frac{6\sqrt{5}}{2\times 5},y=-\frac{6\sqrt{5}}{5}$$