Solve {l}{sqrt{2}x+sqrt{3}y=0}{y-3x-sqrt{8}y=0}

$$\left. \begin{array} { l } { \sqrt { 2 } x + \sqrt { 3 } y = 0 } \\ { y - 3 x - \sqrt { 8 } y = 0 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$y-3x-2\sqrt{2}y=0$$

Combine all terms containing x,y.

$$\left(1-2\sqrt{2}\right)y-3x=0$$

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.

$$\sqrt{2}x+\sqrt{3}y=0,-3x+\left(1-2\sqrt{2}\right)y=0$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$\sqrt{2}x+\sqrt{3}y=0$$

Subtract \sqrt{3}y from both sides of the equation.

$$\sqrt{2}x=\left(-\sqrt{3}\right)y$$

Divide both sides by \sqrt{2}.

$$x=\frac{\sqrt{2}}{2}\left(-\sqrt{3}\right)y$$

Multiply \frac{\sqrt{2}}{2} times -\sqrt{3}y.

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

Substitute -\frac{\sqrt{6}y}{2} for x in the other equation, -3x+\left(1-2\sqrt{2}\right)y=0.

$$-3\left(-\frac{\sqrt{6}}{2}\right)y+\left(1-2\sqrt{2}\right)y=0$$

Multiply -3 times -\frac{\sqrt{6}y}{2}.

$$\frac{3\sqrt{6}}{2}y+\left(1-2\sqrt{2}\right)y=0$$

Add \frac{3\sqrt{6}y}{2} to y-2y\sqrt{2}.

$$\left(\frac{3\sqrt{6}}{2}+1-2\sqrt{2}\right)y=0$$

Divide both sides by \frac{3\sqrt{6}}{2}+1-2\sqrt{2}.

$$y=0$$

Substitute 0 for y in x=\left(-\frac{\sqrt{6}}{2}\right)y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,y=0$$

Steps Using Elimination

Consider the second equation. Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$y-3x-2\sqrt{2}y=0$$

Combine all terms containing x,y.

$$\left(1-2\sqrt{2}\right)y-3x=0$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$\sqrt{2}x+\sqrt{3}y=0,-3x+\left(1-2\sqrt{2}\right)y=0$$

To make \sqrt{2}x and -3x equal, multiply all terms on each side of the first equation by -3 and all terms on each side of the second by \sqrt{2}.

$$-3\sqrt{2}x-3\sqrt{3}y=0,\sqrt{2}\left(-3\right)x+\sqrt{2}\left(1-2\sqrt{2}\right)y=0$$

Simplify.

$$\left(-3\sqrt{2}\right)x+\left(-3\sqrt{3}\right)y=0,\left(-3\sqrt{2}\right)x+\left(\sqrt{2}-4\right)y=0$$

Subtract \left(-3\sqrt{2}\right)x+\left(\sqrt{2}-4\right)y=0 from \left(-3\sqrt{2}\right)x+\left(-3\sqrt{3}\right)y=0 by subtracting like terms on each side of the equal sign.

$$\left(-3\sqrt{2}\right)x+3\sqrt{2}x+\left(-3\sqrt{3}\right)y+\left(4-\sqrt{2}\right)y=0$$

Add -3\sqrt{2}x to 3\sqrt{2}x. Terms -3\sqrt{2}x and 3\sqrt{2}x cancel out, leaving an equation with only one variable that can be solved.

$$\left(-3\sqrt{3}\right)y+\left(4-\sqrt{2}\right)y=0$$

Add -3\sqrt{3}y to \left(-\sqrt{2}+4\right)y.

$$\left(4-\sqrt{2}-3\sqrt{3}\right)y=0$$

Divide both sides by -3\sqrt{3}-\sqrt{2}+4.

$$y=0$$

Substitute 0 for y in -3x+\left(1-2\sqrt{2}\right)y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-3x=0$$

Divide both sides by -3.

$$x=0$$

The system is now solved.

$$x=0,y=0$$