Solve (x+2texttt{i})(2-texttt{i})=4+3texttt{i}

$$(x+2 \texttt{i} )(2- \texttt{i} )=4+3 \texttt{i}$$

Solve for x

$$x=1$$

Steps for Solving Linear Equation

Divide both sides by 2-i.

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

Multiply both numerator and denominator of \frac{4+3i}{2-i} by the complex conjugate of the denominator, 2+i.

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

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

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

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

$$x+2i=\frac{\left(4+3i\right)\left(2+i\right)}{5}$$

Multiply complex numbers 4+3i and 2+i like you multiply binomials.

$$x+2i=\frac{4\times 2+4i+3i\times 2+3i^{2}}{5}$$

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

$$x+2i=\frac{4\times 2+4i+3i\times 2+3\left(-1\right)}{5}$$

Do the multiplications in 4\times 2+4i+3i\times 2+3\left(-1\right).

$$x+2i=\frac{8+4i+6i-3}{5}$$

Combine the real and imaginary parts in 8+4i+6i-3.

$$x+2i=\frac{8-3+\left(4+6\right)i}{5}$$

Do the additions in 8-3+\left(4+6\right)i.

$$x+2i=\frac{5+10i}{5}$$

Divide 5+10i by 5 to get 1+2i.

$$x+2i=1+2i$$

Subtract 2i from both sides.

$$x=1+2i-2i$$

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

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

Add 2 to -2.

$$x=1$$