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