Solve (10-2i)z+(5+i)=2i

$$( 10 - 2 i ) z + ( 5 + i ) = 2 i$$

Solve for z

$$z=-\frac{1}{2}=-0.5$$

Steps for Solving Linear Equation

Subtract 5+i from both sides.

$$\left(10-2i\right)z=2i-\left(5+i\right)$$

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

$$\left(10-2i\right)z=5+\left(2-1\right)i$$

Subtract 1 from 2.

$$\left(10-2i\right)z=-5+i$$

Divide both sides by 10-2i.

$$z=\frac{-5+i}{10-2i}$$

Multiply both numerator and denominator of \frac{-5+i}{10-2i} by the complex conjugate of the denominator, 10+2i.

$$z=\frac{\left(-5+i\right)\left(10+2i\right)}{\left(10-2i\right)\left(10+2i\right)}$$

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

$$z=\frac{\left(-5+i\right)\left(10+2i\right)}{10^{2}-2^{2}i^{2}}$$

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

$$z=\frac{\left(-5+i\right)\left(10+2i\right)}{104}$$

Multiply complex numbers -5+i and 10+2i like you multiply binomials.

$$z=\frac{-5\times 10-5\times \left(2i\right)+10i+2i^{2}}{104}$$

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

$$z=\frac{-5\times 10-5\times \left(2i\right)+10i+2\left(-1\right)}{104}$$

Do the multiplications in -5\times 10-5\times \left(2i\right)+10i+2\left(-1\right).

$$z=\frac{-50-10i+10i-2}{104}$$

Combine the real and imaginary parts in -50-10i+10i-2.

$$z=\frac{-50-2+\left(-10+10\right)i}{104}$$

Do the additions in -50-2+\left(-10+10\right)i.

$$z=\frac{-52}{104}$$

Divide -52 by 104 to get -\frac{1}{2}.

$$z=-\frac{1}{2}$$