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

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Answer 1 · 2016-02-17T18:14:07

#j=1#

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

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

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

#2 + 6j = 4 + 4j#

#6j - 4j = 4-2 #

#2j = 2 #

#j=2/2#

#j=1#