If (x + iy)(2 - 3i) = 4 - i, find the real values of x and y.

Correct option is D

Solution:

1. Expand the expression:
$$(x + iy)(2 - 3i) = 2x - 3xi + 2iy - 3iy^2$$​
Since $$i^2$$​ = -1, this simplifies to:
2x - 3xi + 2iy + 3y
Rearrange into real and imaginary parts:
(2x + 3y) + (-3x + 2y)i

2. Equate to the given value 4 – i :
(2x + 3y) + (-3x + 2y)i = 4 - i
Equating real and imaginary parts:
$$2x + 3y = 4 \tag{1}$$​
$$-3x + 2y = -1 \tag{2}$$​

3. Solve the system of equations:
From equation (1):
$$x = \frac{4 - 3y}{2} \tag{3}$$​
Substitute equation (3) into equation (2):
$$-3\left(\frac{4 - 3y}{2}\right) + 2y = -1$$​
Simplify:
$$-\frac{12 - 9y}{2} + 2y = -1$$​
Multiply through by 2 to eliminate the fraction:
-12 + 9y + 4y = -2
Combine terms:
$$13y = 10 \quad \Rightarrow \quad y = \frac{10}{13}$$​

4. Substitute y = $$\frac{10}{13}$$​ into equation (3):
$$x = \frac{4 - 3\left(\frac{10}{13}\right)}{2}$$​
Simplify:
$$x = \frac{4 - \frac{30}{13}}{2} = \frac{\frac{52}{13} - \frac{30}{13}}{2} = \frac{\frac{22}{13}}{2} = \frac{11}{13}$$​

Final Answer:

The values are:
$$x = \frac{11}{13}, \quad y = \frac{10}{13}$$​
Correct Option: **(D)**