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

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


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