Solve 1+3i/25z+2-3i/1+2i=-7i/5

$$\frac { 1 + 3 i } { 25 } z + \frac { 2 - 3 i } { 1 + 2 i } = - \frac { 7 i } { 5 }$$

Solve for z

$$z=2-6i$$

Steps for Solving Linear Equation

Divide 1+3i by 25 to get \frac{1}{25}+\frac{3}{25}i.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z+\frac{2-3i}{1+2i}=-\frac{7i}{5}$$

Multiply both numerator and denominator of \frac{2-3i}{1+2i} by the complex conjugate of the denominator, 1-2i.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z+\frac{\left(2-3i\right)\left(1-2i\right)}{\left(1+2i\right)\left(1-2i\right)}=-\frac{7i}{5}$$

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

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z+\frac{-4-7i}{5}=-\frac{7i}{5}$$

Divide -4-7i by 5 to get -\frac{4}{5}-\frac{7}{5}i.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z+\left(-\frac{4}{5}-\frac{7}{5}i\right)=-\frac{7i}{5}$$

Divide 7i by 5 to get \frac{7}{5}i.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z+\left(-\frac{4}{5}-\frac{7}{5}i\right)=-\frac{7}{5}i$$

Add \frac{4}{5}+\frac{7}{5}i to both sides.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z=-\frac{7}{5}i+\left(\frac{4}{5}+\frac{7}{5}i\right)$$

Add -\frac{7}{5}i and \frac{4}{5}+\frac{7}{5}i to get \frac{4}{5}.

$$\left(\frac{1}{25}+\frac{3}{25}i\right)z=\frac{4}{5}$$

Divide both sides by \frac{1}{25}+\frac{3}{25}i.

$$z=\frac{\frac{4}{5}}{\frac{1}{25}+\frac{3}{25}i}$$

Multiply both numerator and denominator of \frac{\frac{4}{5}}{\frac{1}{25}+\frac{3}{25}i} by the complex conjugate of the denominator, \frac{1}{25}-\frac{3}{25}i.

$$z=\frac{\frac{4}{5}\left(\frac{1}{25}-\frac{3}{25}i\right)}{\left(\frac{1}{25}+\frac{3}{25}i\right)\left(\frac{1}{25}-\frac{3}{25}i\right)}$$

Do the multiplications in \frac{\frac{4}{5}\left(\frac{1}{25}-\frac{3}{25}i\right)}{\left(\frac{1}{25}+\frac{3}{25}i\right)\left(\frac{1}{25}-\frac{3}{25}i\right)}.

$$z=\frac{\frac{4}{125}-\frac{12}{125}i}{\frac{2}{125}}$$

Divide \frac{4}{125}-\frac{12}{125}i by \frac{2}{125} to get 2-6i.

$$z=2-6i$$