​ Find the real values of x and y for which: (1+i)x - 2i/(3+i)+(2 - 3i)y+i/(3 - i) = i ​

Question

Find the real values of x and y for which:

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

(1+i)x - 2i/(3+i)+(2 - 3i)y+i/(3 - i) = i

Answer 1

Consider,

Taking LCM

⇒ 4x + 2xi – 3i – 3 + 9y – 7iy = 10i

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

Comparing the real parts, we get

4x – 3 + 9y = 0

⇒ 4x + 9y = 3 …(i)

Comparing the imaginary parts, we get

2x – 3 – 7y = 10

⇒ 2x – 7y = 10 + 3

⇒ 2x – 7y = 13 …(ii)

Multiply the eq. (ii) by 2, we get

4x – 14y = 26 …(iii)

Subtracting eq. (i) from (iii), we get

4x – 14y – (4x + 9y) = 26 – 3

⇒ 4x – 14y – 4x – 9y = 23

⇒ - 23y = 23

⇒ y = - 1

Putting the value of y = -1 in eq. (i), we get

4x + 9(-1) = 3

⇒ 4x – 9 = 3

⇒ 4x = 12

⇒ x = 3

Hence, the value of x = 3 and y = -1