**Question:**

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

**Solution:**

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

We know that,

Conjugate of $-6-24 i=-6+24 i$

∴ According to the given condition,

$(x-i y)(3+5 i)=-6+24 i$

$\Rightarrow x(3+5 i)-i y(3+5 i)=-6+24 i$

$\Rightarrow 3 x+5 i x-3 i y-5 i^{2} y=-6+24 i$

$\Rightarrow 3 x+i(5 x-3 y)-5(-1) y=-6+24 i\left[\because i^{2}=-1\right]$

$\Rightarrow 3 x+i(5 x-3 y)+5 y=-6+24 i$

$\Rightarrow(3 x+5 y)+i(5 x-3 y)=-6+24 i$

Comparing the real parts, we get

$3 x+5 y=-6 \ldots$ (i)

Comparing the imaginary parts, we get

$5 x-3 y=24 \ldots$ (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

$15 x+25 y=-30 \ldots$ (iii)

$15 x-9 y=72 \ldots$ (iv)

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

$15 x-9 y-15 x-25 y=72-(-30)$

$\Rightarrow-34 y=72+30$

$\Rightarrow-34 y=102$

$\Rightarrow y=-3$

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

$3 x+5(-3)=-6$

$\Rightarrow 3 x-15=-6$

$\Rightarrow 3 x=-6+15$

$\Rightarrow 3 x=9$

$\Rightarrow x=3$

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