Solve {l}{y=3x}{y^2-x^2=3c^6}

$$\left\{ \begin{array} { l } { y = 3 x } \\ { y ^ { 2 } - x ^ { 2 } = 3 c ^ { 6 } } \end{array} \right.$$

Solve for x, y

$$x=-\frac{\sqrt{6}c^{3}}{4}\text{, }y=-\frac{3\sqrt{6}c^{3}}{4}$$

$$x=\frac{\sqrt{6}c^{3}}{4}\text{, }y=\frac{3\sqrt{6}c^{3}}{4}$$

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.

$$y-3x=0,-x^{2}+y^{2}=3c^{6}$$

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}=3c^{6}.

$$-x^{2}+\left(3x\right)^{2}=3c^{6}$$

Square 3x.

$$-x^{2}+9x^{2}=3c^{6}$$

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

$$8x^{2}=3c^{6}$$

Subtract 3c^{6} from both sides of the equation.

$$8x^{2}-3c^{6}=0$$

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 -3c^{6} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 8\left(-3c^{6}\right)}}{2\times 8}$$

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

$$x=\frac{0±\sqrt{-4\times 8\left(-3c^{6}\right)}}{2\times 8}$$

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

$$x=\frac{0±\sqrt{-32\left(-3c^{6}\right)}}{2\times 8}$$

Multiply -32 times -3c^{6}.

$$x=\frac{0±\sqrt{96c^{6}}}{2\times 8}$$

Take the square root of 96c^{6}.

$$x=\frac{0±4\sqrt{6}\left(|c|\right)^{3}}{2\times 8}$$

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

$$x=\frac{0±4\sqrt{6}\left(|c|\right)^{3}}{16}$$

Now solve the equation x=\frac{0±4\sqrt{6}\left(|c|\right)^{3}}{16} when ± is plus.

$$x=\frac{\sqrt{6}\left(|c|\right)^{3}}{4}$$

Now solve the equation x=\frac{0±4\sqrt{6}\left(|c|\right)^{3}}{16} when ± is minus.

$$x=-\frac{\sqrt{6}\left(|c|\right)^{3}}{4}$$

There are two solutions for x: \frac{\sqrt{6}\left(|c|\right)^{3}}{4} and -\frac{\sqrt{6}\left(|c|\right)^{3}}{4}. Substitute \frac{\sqrt{6}\left(|c|\right)^{3}}{4} for x in the equation y=3x to find the corresponding solution for y that satisfies both equations.

$$y=3\times \frac{\sqrt{6}\left(|c|\right)^{3}}{4}$$

Now substitute -\frac{\sqrt{6}\left(|c|\right)^{3}}{4} for x in the equation y=3x and solve to find the corresponding solution for y that satisfies both equations.

$$y=3\left(-\frac{\sqrt{6}\left(|c|\right)^{3}}{4}\right)$$

The system is now solved.

$$y=3\times \frac{\sqrt{6}\left(|c|\right)^{3}}{4},x=\frac{\sqrt{6}\left(|c|\right)^{3}}{4}\text{ or }y=3\left(-\frac{\sqrt{6}\left(|c|\right)^{3}}{4}\right),x=-\frac{\sqrt{6}\left(|c|\right)^{3}}{4}$$