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If $\;4x+i(3x-y)=3+i (-6)\;$ where x and y are real numbers , then find the values of x and y

$(a)\;\large\frac{23}{4} , \large\frac{3}{4}\qquad(b)\;\large\frac{3}{4} , \large\frac{33}{4}\qquad(c)\;\large\frac{1}{4} , \large\frac{33}{4}\qquad(d)\;\large\frac{3}{4} , \large\frac{13}{4}$

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Answer : $\;\large\frac{3}{4} , \large\frac{33}{4}$
Explanation :
$\;4x+i(3x-y)=3+i (-6)\;$----(1)
Equating the real and imaginary parts of (1) , we obtain
$4x=3 \; , 3x-y = -6$
Which , on solving simultaneously give $\;x= \large\frac{3}{4}\; , y= \large\frac{33}{4}$
answered Apr 16, 2014 by yamini.v

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