Solve (4x+3)-sqrt{3i}=frac{sqrt{3}-3i}{sqrt{3}}

$$( 4 x + 3 ) - \sqrt { 3 i } = \frac { \sqrt { 3 } - 3 i } { \sqrt { 3 } }$$

Solve for x

$$x=-\frac{\sqrt{3}i}{4}+\sqrt{6}\left(\frac{1}{8}+\frac{1}{8}i\right)-\frac{1}{2}\approx -0.193813782-0.126826484i$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{\sqrt{3}-3i}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4x+3-\sqrt{3i}=\frac{\left(\sqrt{3}-3i\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$4x+3-\sqrt{3i}=\frac{\left(\sqrt{3}-3i\right)\sqrt{3}}{3}$$

Use the distributive property to multiply \sqrt{3}-3i by \sqrt{3}.

$$4x+3-\sqrt{3i}=\frac{\left(\sqrt{3}\right)^{2}-3i\sqrt{3}}{3}$$

The square of \sqrt{3} is 3.

$$4x+3-\sqrt{3i}=\frac{3-3i\sqrt{3}}{3}$$

Divide each term of 3-3i\sqrt{3} by 3 to get 1-i\sqrt{3}.

$$4x+3-\sqrt{3i}=1-i\sqrt{3}$$

Subtract 3 from both sides.

$$4x-\sqrt{3i}=1-i\sqrt{3}-3$$

Subtract 3 from 1 to get -2.

$$4x-\sqrt{3i}=-2-i\sqrt{3}$$

Add \sqrt{3i} to both sides.

$$4x=-2-i\sqrt{3}+\sqrt{3i}$$

Reorder the terms.

$$4x=\sqrt{3i}-2-\sqrt{3}i$$

Multiply -1 and i to get -i.

$$4x=\sqrt{3i}-2-i\sqrt{3}$$

Reorder the terms.

$$4x=-\sqrt{3}i+\sqrt{3i}-2$$

Divide both sides by 4.

$$\frac{4x}{4}=\frac{\sqrt{6}\left(\frac{1}{2}+\frac{1}{2}i\right)-\sqrt{3}i-2}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x=\frac{\sqrt{6}\left(\frac{1}{2}+\frac{1}{2}i\right)-\sqrt{3}i-2}{4}$$

Divide -2-i\sqrt{3}+\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt{6} by 4.

$$x=-\frac{\sqrt{3}i}{4}+\sqrt{6}\left(\frac{1}{8}+\frac{1}{8}i\right)-\frac{1}{2}$$