What is the answer to (3z}{1-i}+\frac{2z)/i = 6/(4+i) ?

The answer to (3z}{1-i}+\frac{2z)/i = 6/(4+i) is z= 78/85+i 6/85

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(3z}{1-i}+\frac{2z)/i = 6/(4+i)

\frac{3 z}{1 - i} + \frac{2 z}{i} = \frac{6}{4 + i}

z = \frac{78}{85} + i \frac{6}{85}

Solution steps

\frac{3 z}{1 - i} + \frac{2 z}{i} = \frac{6}{4 + i}

Multiply by LCM

3 i (4 + i) z + 2 (1 - i) (4 + i) z = 6 i (1 - i)

Expand

3 i (4 + i) z + 2 (1 - i) (4 + i) z : 6 i z + 7 z

Expand

6 i (1 - i) : 6 + 6 i

6 i z + 7 z = 6 + 6 i

Factor

6 i z + 7 z : (6 i + 7) z

(6 i + 7) z = 6 + 6 i

Divide both sides by

6 i + 7

z = \frac{78}{85} + i \frac{6}{85}

\frac{6 n + 6}{n + 9} \cdot \frac{n ^{2} + 6 n - 27}{6 n + 6}

2 - 3 (x + 5) = 5 x - 13

\frac{2 x - 7}{4} \leq \frac{1}{5}

(\frac{1}{4 ab})^{- 2}

What is the answer to (3z}{1-i}+\frac{2z)/i = 6/(4+i) ?
The answer to (3z}{1-i}+\frac{2z)/i = 6/(4+i) is z= 78/85+i 6/85