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

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

Solve for x

$$x=-\frac{8}{13}-\frac{1}{13}i\approx -0.615384615-0.076923077i$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3-2i by 1+x.

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

Multiply both numerator and denominator of \frac{1-i}{i} by imaginary unit i.

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

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

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

Multiply 1-i times i.

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

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

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

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

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

Divide 1+i by -1 to get -1-i.

$$3-2i+\left(3-2i\right)x=-1-i+2$$

Combine the real and imaginary parts in numbers -1-i and 2.

$$3-2i+\left(3-2i\right)x=-1+2-i$$

Add -1 to 2.

$$3-2i+\left(3-2i\right)x=1-i$$

Subtract 3-2i from both sides.

$$\left(3-2i\right)x=1-i-\left(3-2i\right)$$

Subtract 3-2i from 1-i by subtracting corresponding real and imaginary parts.

$$\left(3-2i\right)x=1-3+\left(-1-\left(-2\right)\right)i$$

Subtract 3 from 1. Subtract -2 from -1.

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

Divide both sides by 3-2i.

$$x=\frac{-2+i}{3-2i}$$

Multiply both numerator and denominator of \frac{-2+i}{3-2i} by the complex conjugate of the denominator, 3+2i.

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

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

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

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

$$x=\frac{\left(-2+i\right)\left(3+2i\right)}{13}$$

Multiply complex numbers -2+i and 3+2i like you multiply binomials.

$$x=\frac{-2\times 3-2\times \left(2i\right)+3i+2i^{2}}{13}$$

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

$$x=\frac{-2\times 3-2\times \left(2i\right)+3i+2\left(-1\right)}{13}$$

Do the multiplications in -2\times 3-2\times \left(2i\right)+3i+2\left(-1\right).

$$x=\frac{-6-4i+3i-2}{13}$$

Combine the real and imaginary parts in -6-4i+3i-2.

$$x=\frac{-6-2+\left(-4+3\right)i}{13}$$

Do the additions in -6-2+\left(-4+3\right)i.

$$x=\frac{-8-i}{13}$$

Divide -8-i by 13 to get -\frac{8}{13}-\frac{1}{13}i.

$$x=-\frac{8}{13}-\frac{1}{13}i$$