What is the answer to ((2z)}{(1+i)}-\frac{(2z))/i = 5/((2+i)) ?

The answer to ((2z)}{(1+i)}-\frac{(2z))/i = 5/((2+i)) is z= 1/2-i 3/2

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((2z)}{(1+i)}-\frac{(2z))/i = 5/((2+i))

\frac{( 2 z )}{( 1 + i )} - \frac{( 2 z )}{i} = \frac{5}{( 2 + i )}

z = \frac{1}{2} - i \frac{3}{2}

Solution steps

\frac{( 2 z )}{( 1 + i )} - \frac{( 2 z )}{i} = \frac{5}{( 2 + i )}

\frac{2 z}{1 + i} - \frac{2 z}{i} = \frac{5}{2 + i}

Multiply by LCM

2 i (2 + i) z - 2 (1 + i) (2 + i) z = 5 i (1 + i)

Expand

2 i (2 + i) z - 2 (1 + i) (2 + i) z : - 2 i z - 4 z

Expand

5 i (1 + i) : - 5 + 5 i

- 2 i z - 4 z = - 5 + 5 i

Factor

- 2 i z - 4 z : - 2 (i + 2) z

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

Divide both sides by

- 2 (i + 2)

z = \frac{1}{2} - i \frac{3}{2}

x, 23 x + 29 y = 25

3 y \frac{1}{2} - y - 2 = 0

4 (18 - 5 x) - 12 (13 x - 7) = 15 (2 x - 16) - 6 (x + 14)

\frac{5}{4} = \frac{- 4 c + 1}{4}

\frac{8}{3} = 6, x = 1

What is the answer to ((2z)}{(1+i)}-\frac{(2z))/i = 5/((2+i)) ?
The answer to ((2z)}{(1+i)}-\frac{(2z))/i = 5/((2+i)) is z= 1/2-i 3/2