Question 459891: 9+3b+(3a-b)i=6+6i
I'm thinking of this question for 1 hour, and i can't find the answer, can u please help me
and thanks
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
9+3b+(3a-b)i=6+6i (separate real and imaginary values also separate variable on left side and constants on right side.)
=> 3b + (3a-b)i = 6 + 6i -9
=> 3b + (3a-b)i = -3 + 6i
now compare real and imaginary values..
i.e 3b = -3
and 3a-b = 6
from first equation b = -1 now, put this value on second
3a+1 = 6
=> a = 5/3
so, a= 5/3, b = -1
if any doubt, you are welcome to contact me..
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