Find the real values of x and y for which:

Solution:

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

$\Rightarrow 3 x-2 i x+3 i y-2 i^{2} y=12+5 i$

$\Rightarrow 3 x+i(-2 x+3 y)-2(-1) y=12+5 i\left[\because i^{2}=-1\right]$

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

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

Comparing the real parts, we get

$3 x+2 y=12 \ldots(i)$

Comparing the imaginary parts, we get

$-2 x+3 y=5 \ldots$ (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

$6 x+4 y=24 \ldots$ (iii)

$-6 x+9 y=15 \ldots$ (iv)

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

$6 x+4 y-6 x+9 y=24+15$

$\Rightarrow 13 y=39$

$\Rightarrow y=3$

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

3x + 2(3) = 12

$\Rightarrow 3 x+6=12$

$\Rightarrow 3 x=12-6$

$\Rightarrow 3 x=6$

$\Rightarrow x=2$

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