Answer :

x(3 – 2i) + iy(3 – 2i) = 12 + 5i

3x – 2ix + 3iy – 2i2y = 12 + 5i


3x + i(-2x + 3y) – 2(-1)y = 12 + 5i [ i2 = -1]


3x + i(-2x + 3y) + 2y = 12 + 5i


(3x + 2y) + i(-2x + 3y) = 12 + 5i


Comparing the real parts, we get


3x + 2y = 12 …(i)


Comparing the imaginary parts, we get


–2x + 3y = 5 …(ii)


Solving eq. (i) and (ii) to find the value of x and y


Multiply eq. (i) by 2 and eq. (ii) by 3, we get


6x + 4y = 24 …(iii)


–6x + 9y = 15 …(iv)


Adding eq. (iii) and (iv), we get


6x + 4y – 6x + 9y = 24 + 15


13y = 39


y = 3


Putting the value of y = 3 in eq. (i), we get


3x + 2(3) = 12


3x + 6 = 12


3x = 12 – 6


3x = 6


x = 2


Hence, the value of x = 2 and y = 3


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