Solve (3+i2)*X=(2+i4)*(2-i3)

$$( 3 + i 2 ) \times X = ( 2 + i 4 ) \times ( 2 - i 3 )$$

Solve for X

$$X=4-2i$$

Steps for Solving Linear Equation

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

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

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

$$\left(3+2i\right)X=2\times 2+2\times \left(-3i\right)+4i\times 2+4\left(-3\right)\left(-1\right)$$

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

$$\left(3+2i\right)X=4-6i+8i+12$$

Combine the real and imaginary parts in 4-6i+8i+12.

$$\left(3+2i\right)X=4+12+\left(-6+8\right)i$$

Do the additions in 4+12+\left(-6+8\right)i.

$$\left(3+2i\right)X=16+2i$$

Divide both sides by 3+2i.

$$X=\frac{16+2i}{3+2i}$$

Multiply both numerator and denominator of \frac{16+2i}{3+2i} by the complex conjugate of the denominator, 3-2i.

$$X=\frac{\left(16+2i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}$$

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

$$X=\frac{\left(16+2i\right)\left(3-2i\right)}{3^{2}-2^{2}i^{2}}$$

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

$$X=\frac{\left(16+2i\right)\left(3-2i\right)}{13}$$

Multiply complex numbers 16+2i and 3-2i like you multiply binomials.

$$X=\frac{16\times 3+16\times \left(-2i\right)+2i\times 3+2\left(-2\right)i^{2}}{13}$$

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

$$X=\frac{16\times 3+16\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right)}{13}$$

Do the multiplications in 16\times 3+16\times \left(-2i\right)+2i\times 3+2\left(-2\right)\left(-1\right).

$$X=\frac{48-32i+6i+4}{13}$$

Combine the real and imaginary parts in 48-32i+6i+4.

$$X=\frac{48+4+\left(-32+6\right)i}{13}$$

Do the additions in 48+4+\left(-32+6\right)i.

$$X=\frac{52-26i}{13}$$

Divide 52-26i by 13 to get 4-2i.

$$X=4-2i$$