Solve x=(2-2i)-(5+1i)/-3+6i

$$x = \frac { ( 2 - 2 i ) - ( 5 + 1 i ) } { - 3 + 6 i }$$

Solve for x

$$x=-\frac{1}{5}+\frac{3}{5}i=-0.2+0.6i$$

Steps for Solving Linear Equation

Subtract 5+i from 2-2i by subtracting corresponding real and imaginary parts.

$$x=\frac{2-5+\left(-2-1\right)i}{-3+6i}$$

Subtract 5 from 2. Subtract 1 from -2.

$$x=\frac{-3-3i}{-3+6i}$$

Multiply both numerator and denominator of \frac{-3-3i}{-3+6i} by the complex conjugate of the denominator, -3-6i.

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

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

$$x=\frac{\left(-3-3i\right)\left(-3-6i\right)}{\left(-3\right)^{2}-6^{2}i^{2}}$$

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

$$x=\frac{\left(-3-3i\right)\left(-3-6i\right)}{45}$$

Multiply complex numbers -3-3i and -3-6i like you multiply binomials.

$$x=\frac{-3\left(-3\right)-3\times \left(-6i\right)-3i\left(-3\right)-3\left(-6\right)i^{2}}{45}$$

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

$$x=\frac{-3\left(-3\right)-3\times \left(-6i\right)-3i\left(-3\right)-3\left(-6\right)\left(-1\right)}{45}$$

Do the multiplications in -3\left(-3\right)-3\times \left(-6i\right)-3i\left(-3\right)-3\left(-6\right)\left(-1\right).

$$x=\frac{9+18i+9i-18}{45}$$

Combine the real and imaginary parts in 9+18i+9i-18.

$$x=\frac{9-18+\left(18+9\right)i}{45}$$

Do the additions in 9-18+\left(18+9\right)i.

$$x=\frac{-9+27i}{45}$$

Divide -9+27i by 45 to get -\frac{1}{5}+\frac{3}{5}i.

$$x=-\frac{1}{5}+\frac{3}{5}i$$

Assign x

$$x≔-\frac{1}{5}+\frac{3}{5}i$$