Solve 2+(3-i)z=2i-1

$$2 + ( 3 - i ) z = 2 i - 1$$

Solve for z

$$z=-\frac{11}{10}+\frac{3}{10}i=-1.1+0.3i$$

Steps for Solving Linear Equation

Subtract 2 from both sides.

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

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

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

Add -1 to -2.

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

Divide both sides by 3-i.

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

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

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

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

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

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

$$z=\frac{\left(-3+2i\right)\left(3+i\right)}{10}$$

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

$$z=\frac{-3\times 3-3i+2i\times 3+2i^{2}}{10}$$

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

$$z=\frac{-3\times 3-3i+2i\times 3+2\left(-1\right)}{10}$$

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

$$z=\frac{-9-3i+6i-2}{10}$$

Combine the real and imaginary parts in -9-3i+6i-2.

$$z=\frac{-9-2+\left(-3+6\right)i}{10}$$

Do the additions in -9-2+\left(-3+6\right)i.

$$z=\frac{-11+3i}{10}$$

Divide -11+3i by 10 to get -\frac{11}{10}+\frac{3}{10}i.

$$z=-\frac{11}{10}+\frac{3}{10}i$$