Solve -2ix-left(3-3iright)x+3+i=0

$$-2ix- \left( 3-3i \right) x+3+i = 0$$

Solve for x

$$x=\frac{4}{5}+\frac{3}{5}i=0.8+0.6i$$

Steps for Solving Linear Equation

Combine -2ix and \left(-3+3i\right)x to get \left(-3+i\right)x.

$$\left(-3+i\right)x+3+i=0$$

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

$$\left(-3+i\right)x+i=-3$$

Subtract i from both sides.

$$\left(-3+i\right)x=-3-i$$

Divide both sides by -3+i.

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

Multiply both numerator and denominator of \frac{-3-i}{-3+i} by the complex conjugate of the denominator, -3-i.

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

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

By definition, i^{2} is -1. Calculate the denominator.

$$x=\frac{\left(-3-i\right)\left(-3-i\right)}{10}$$

Multiply complex numbers -3-i and -3-i like you multiply binomials.

$$x=\frac{-3\left(-3\right)-3\left(-i\right)-i\left(-3\right)-\left(-i^{2}\right)}{10}$$

By definition, i^{2} is -1.

$$x=\frac{-3\left(-3\right)-3\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right)}{10}$$

Do the multiplications in -3\left(-3\right)-3\left(-i\right)-i\left(-3\right)-\left(-\left(-1\right)\right).

$$x=\frac{9+3i+3i-1}{10}$$

Combine the real and imaginary parts in 9+3i+3i-1.

$$x=\frac{9-1+\left(3+3\right)i}{10}$$

Do the additions in 9-1+\left(3+3\right)i.

$$x=\frac{8+6i}{10}$$

Divide 8+6i by 10 to get \frac{4}{5}+\frac{3}{5}i.

$$x=\frac{4}{5}+\frac{3}{5}i$$