Solve sqrt{6}x+3y+sqrt{2}=0

$$\sqrt { 6 } x + 3 y + \sqrt { 2 } = 0$$

Solve for x

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

Steps for Solving Linear Equation

Subtract 3y from both sides. Anything subtracted from zero gives its negation.

$$\sqrt{6}x+\sqrt{2}=-3y$$

Subtract \sqrt{2} from both sides.

$$\sqrt{6}x=-3y-\sqrt{2}$$

Divide both sides by \sqrt{6}.

$$\frac{\sqrt{6}x}{\sqrt{6}}=\frac{-3y-\sqrt{2}}{\sqrt{6}}$$

Dividing by \sqrt{6} undoes the multiplication by \sqrt{6}.

$$x=\frac{-3y-\sqrt{2}}{\sqrt{6}}$$

Divide -3y-\sqrt{2} by \sqrt{6}.

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

Solve for y

$$y=\frac{-\sqrt{6}x-\sqrt{2}}{3}$$

Steps for Solving Linear Equation

Subtract \sqrt{6}x from both sides. Anything subtracted from zero gives its negation.

$$3y+\sqrt{2}=-\sqrt{6}x$$

Subtract \sqrt{2} from both sides.

$$3y=-\sqrt{6}x-\sqrt{2}$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{-\sqrt{6}x-\sqrt{2}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{-\sqrt{6}x-\sqrt{2}}{3}$$