What is the answer to (8-12i)((8+4i)/8 x)-8x=-40i ?

The answer to (8-12i)((8+4i)/8 x)-8x=-40i is x= 16/5-12/5 i

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(8-12i)((8+4i)/8 x)-8x=-40i

(8 - 12 i) (\frac{8 + 4 i}{8} x) - 8 x = - 40 i

x = \frac{16}{5} - \frac{12}{5} i

Solution steps

(8 - 12 i) (\frac{8 + 4 i}{8} x) - 8 x = - 40 i

Substitute

x = a + bi

(8 - 12 i) \frac{8 + 4 i}{8} (a + bi) - 8 (a + bi) = - 40 i

Expand

(8 - 12 i) \frac{8 + 4 i}{8} (a + bi) - 8 (a + bi) : (6 a + 8 b) + i (6 b - 8 a)

(6 a + 8 b) + i (6 b - 8 a) = - 40 i

Rewrite

- 40 i

in standard complex form:

0 - 40 i

(6 a + 8 b) + i (6 b - 8 a) = 0 - 40 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[6 a + 8 b = 0 6 b - 8 a = - 40]

[6 a + 8 b = 0 6 b - 8 a = - 40] : a = \frac{16}{5}, b = - \frac{12}{5}

Substitute back

x = a + bi

x = \frac{16}{5} - \frac{12}{5} i

3 x^{2} - 6 x

m - 1 + 20 \geq 77

\frac{5}{3} \cdot \frac{4}{3}

\frac{- \frac{1}{2}}{- \frac{3}{2}}

\frac{1}{2} (- 1)

What is the answer to (8-12i)((8+4i)/8 x)-8x=-40i ?
The answer to (8-12i)((8+4i)/8 x)-8x=-40i is x= 16/5-12/5 i