Solve xcosalpha+sqrt{3}y+2=0

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

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{3}y+2}{\cos(\alpha )}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}+\frac{\pi }{2}\\x\in \mathrm{R}\text{, }&y=-\frac{2\sqrt{3}}{3}\text{ and }\exists n_{1}\in \mathrm{Z}\text{ : }\alpha =\pi n_{1}+\frac{\pi }{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract \sqrt{3}y from both sides. Anything subtracted from zero gives its negation.

$$x\cos(\alpha )+2=-\sqrt{3}y$$

Subtract 2 from both sides.

$$x\cos(\alpha )=-\sqrt{3}y-2$$

The equation is in standard form.

$$\cos(\alpha )x=-\sqrt{3}y-2$$

Divide both sides by \cos(\alpha ).

$$\frac{\cos(\alpha )x}{\cos(\alpha )}=\frac{-\sqrt{3}y-2}{\cos(\alpha )}$$

Dividing by \cos(\alpha ) undoes the multiplication by \cos(\alpha ).

$$x=\frac{-\sqrt{3}y-2}{\cos(\alpha )}$$

Divide -\sqrt{3}y-2 by \cos(\alpha ).

$$x=-\frac{\sqrt{3}y+2}{\cos(\alpha )}$$

Solve for y

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

Steps for Solving Linear Equation

Subtract x\cos(\alpha ) from both sides. Anything subtracted from zero gives its negation.

$$\sqrt{3}y+2=-x\cos(\alpha )$$

Subtract 2 from both sides.

$$\sqrt{3}y=-x\cos(\alpha )-2$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}y}{\sqrt{3}}=\frac{-x\cos(\alpha )-2}{\sqrt{3}}$$

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

$$y=\frac{-x\cos(\alpha )-2}{\sqrt{3}}$$

Divide -x\cos(\alpha )-2 by \sqrt{3}.

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