Solve {l}{y=3x}{x^2+y^2=4}

$$\left\{ \begin{array} { l } { y = 3 x } \\ { x ^ { 2 } + y ^ { 2 } = 4 } \end{array} \right.$$

Solve for y, x

$$x=-\frac{\sqrt{10}}{5}\approx -0.632455532\text{, }y=-\frac{3\sqrt{10}}{5}\approx -1.897366596$$

$$x=\frac{\sqrt{10}}{5}\approx 0.632455532\text{, }y=\frac{3\sqrt{10}}{5}\approx 1.897366596$$

Steps Using Substitution and Quadratic Formula

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

$$y-3x=0$$

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.

$$y-3x=0,x^{2}+y^{2}=4$$

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

$$y-3x=0$$

Subtract -3x from both sides of the equation.

$$y=3x$$

Substitute 3x for y in the other equation, x^{2}+y^{2}=4.

$$x^{2}+\left(3x\right)^{2}=4$$

Square 3x.

$$x^{2}+9x^{2}=4$$

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

$$10x^{2}=4$$

Subtract 4 from both sides of the equation.

$$10x^{2}-4=0$$

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

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

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

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

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

$$x=\frac{0±\sqrt{-40\left(-4\right)}}{2\times 10}$$

Multiply -40 times -4.

$$x=\frac{0±\sqrt{160}}{2\times 10}$$

Take the square root of 160.

$$x=\frac{0±4\sqrt{10}}{2\times 10}$$

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

$$x=\frac{0±4\sqrt{10}}{20}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{20} when ± is plus.

$$x=\frac{\sqrt{10}}{5}$$

Now solve the equation x=\frac{0±4\sqrt{10}}{20} when ± is minus.

$$x=-\frac{\sqrt{10}}{5}$$

There are two solutions for x: \frac{\sqrt{10}}{5} and -\frac{\sqrt{10}}{5}. Substitute \frac{\sqrt{10}}{5} for x in the equation y=3x to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{\sqrt{10}}{5}$$

Now substitute -\frac{\sqrt{10}}{5} for x in the equation y=3x and solve to find the corresponding solution for y that satisfies both equations.

$$y=3\left(-\frac{\sqrt{10}}{5}\right)$$

The system is now solved.

$$y=3\times \frac{\sqrt{10}}{5},x=\frac{\sqrt{10}}{5}\text{ or }y=3\left(-\frac{\sqrt{10}}{5}\right),x=-\frac{\sqrt{10}}{5}$$