Question

Find the values of x and y for which the following values are satisfied :
$\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$
​

Answer 1

Answer:

x = 3,

y = -1

Step-by-step explanation:

Given :

To solve for x & y if,

$\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i$

Solution :

⇒ $\frac{(1+i)x - 2i}{3+i} + \frac{(2-3i)y+i}{3-i} = i$

⇒ $\frac{[(1+i)x - 2i](3-i)+[(2-3i)y+i](3+i)}{(3-i)(3+i)} = i$

⇒ $\frac{[3(1+i)x - i(1+i)x - 2i(-i) -2i(3)]+ [3(2 - 3i)y - i(2-3i)y +i(3-i)}{(3-i)(3+i)} = i$

⇒ $\frac{[3x + 3ix - ix - i^2x + 2i^2 -6i]+ [6y - 9iy + 2iy + 3i^2y +3i+i^2]}{9-i^2} = i$

⇒ $\frac{(4x + 9y - 3) +i(2x - 7y- 3)}{9-i^2} = i$

⇒ $(4x + 9y - 3) +i(2x - 7y- 3) = i(9+1)$

⇒ $(4x + 9y - 3) +i(2x - 7y- 3) = 10i$

⇒  $(4x + 9y - 3) = 0$ and $i(2x - 7y- 3) = 10i$ ⇒ $2x - 7y- 3) = 10$

⇒ $4x + 9y = 3$  ...(i) and $2x - 7y = 13$ ...(ii)

By adding both the equation (i) × 7 & (ii) × 9

⇒ $7 ( 4x + 9y ) + 9 (2x - 7y) = 7 × 3 + 9 × 13$

⇒ $28x + 63y + 18x - 63y = 21 + 117$

⇒ $46x = 138$

⇒ $x = 3$

By substituting the value of x in (i),

We get,

⇒ $4x + 9y = 3$

⇒ $4(3) + 9y = 3$

⇒  $12+ 9y = 3$

⇒  $9y = 3 - 12 = -9$

⇒ $y = -1$