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

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

Solve for x, y

$$x=-\frac{3\sqrt{10}}{10}\approx -0.948683298\text{, }y=-\frac{\sqrt{10}}{10}\approx -0.316227766$$

$$x=\frac{3\sqrt{10}}{10}\approx 0.948683298\text{, }y=\frac{\sqrt{10}}{10}\approx 0.316227766$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 3y from both sides.

$$x-3y=0$$

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.

$$x-3y=0,y^{2}+x^{2}=1$$

Solve x-3y=0 for x by isolating x on the left hand side of the equal sign.

$$x-3y=0$$

Subtract -3y from both sides of the equation.

$$x=3y$$

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

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

Square 3y.

$$y^{2}+9y^{2}=1$$

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

$$10y^{2}=1$$

Subtract 1 from both sides of the equation.

$$10y^{2}-1=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\times 10\left(-1\right)}}{2\times 10}$$

Square 1\times 0\times 2\times 3.

$$y=\frac{0±\sqrt{-4\times 10\left(-1\right)}}{2\times 10}$$

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

$$y=\frac{0±\sqrt{-40\left(-1\right)}}{2\times 10}$$

Multiply -40 times -1.

$$y=\frac{0±\sqrt{40}}{2\times 10}$$

Take the square root of 40.

$$y=\frac{0±2\sqrt{10}}{2\times 10}$$

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

$$y=\frac{0±2\sqrt{10}}{20}$$

Now solve the equation y=\frac{0±2\sqrt{10}}{20} when ± is plus.

$$y=\frac{\sqrt{10}}{10}$$

Now solve the equation y=\frac{0±2\sqrt{10}}{20} when ± is minus.

$$y=-\frac{\sqrt{10}}{10}$$

There are two solutions for y: \frac{\sqrt{10}}{10} and -\frac{\sqrt{10}}{10}. Substitute \frac{\sqrt{10}}{10} for y in the equation x=3y to find the corresponding solution for x that satisfies both equations.

$$x=3\times \frac{\sqrt{10}}{10}$$

Now substitute -\frac{\sqrt{10}}{10} for y in the equation x=3y and solve to find the corresponding solution for x that satisfies both equations.

$$x=3\left(-\frac{\sqrt{10}}{10}\right)$$

The system is now solved.

$$x=3\times \frac{\sqrt{10}}{10},y=\frac{\sqrt{10}}{10}\text{ or }x=3\left(-\frac{\sqrt{10}}{10}\right),y=-\frac{\sqrt{10}}{10}$$