Solve 2x+sqrt{3}y=-2

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

Solve for x

$$x=-\frac{\sqrt{3}y}{2}-1$$

Steps for Solving Linear Equation

Subtract \sqrt{3}y from both sides.

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

Reorder the terms.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

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

$$x=-\frac{\sqrt{3}y}{2}-1$$

Solve for y

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

Steps for Solving Linear Equation

Subtract 2x from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

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

Divide -2-2x by \sqrt{3}.

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