How do you solve #6 + 2(3j - 2) = 4(1 + j) #?

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Answer 1 · 2017-05-10T16:08:50

#1 = j#

#6 + 2(3j - 2 ) = 4(1 + j)#

# 6 + 6j - 4 = 4 + 4j#

#6j + 6 = 4 + 4 + 4j#

#6j + 6 = 8 + 4j#

#6j = 8 - 6 + 4j#

#6j = 2 + 4j#

Since

#(x + 1)j = j + xj#

#6j = 2j + 4j#

#therefore 2 = 2j#

#j = 2/2 = 1#