# Find the real values of x and y for which (x – iy) (3 + 5i) is the conjugate of (-6 – 24i).

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Find the real values of x and y for which (x – iy) (3 + 5i) is the conjugate of (-6 – 24i).

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Given: (x – iy) (3 + 5i) is the conjugate of (-6 – 24i)

We know that,

Conjugate of – 6 – 24i = - 6 + 24i

∴ According to the given condition,

(x – iy) (3 + 5i) = -6 + 24i

⇒ x(3 + 5i) – iy(3 + 5i) = -6 + 24i

⇒ 3x + 5ix – 3iy – 5i2y = -6 + 24i

⇒ 3x + i(5x – 3y) – 5(-1)y = -6 + 24i [∵ i2 = -1]

⇒ 3x + i(5x – 3y) + 5y = - 6 + 24i

⇒ (3x + 5y) + i(5x – 3y) = - 6 + 24i

Comparing the real parts, we get

3x + 5y = - 6 …(i)

Comparing the imaginary parts, we get

5x – 3y = 24 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Multiply eq. (i) by 5 and eq. (ii) by 3, we get

15x + 25y = -30 …(iii)

15x – 9y = 72 …(iv)

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

15x – 9y – 15x – 25y = 72 – (-30)

⇒ - 34y = 72 + 30

⇒ - 34y = 102

⇒ y = -3

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

3x + 5(-3) = - 6

⇒ 3x – 15 = - 6

⇒ 3x = - 6 + 15

⇒ 3x = 9

⇒ x = 3

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