Find the real values of x and y for which:
Question:

Find the real values of x and y for which:

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

 

Solution:

Consider,

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

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

Taking LCM

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

$\Rightarrow \frac{3 x+3 x i-6 i-x i-x i^{2}+2 i^{2}+6 y-9 i y+3 i+2 i y-3 i^{2} y+i^{2}}{(3)^{2}-(i)^{2}}=i$

Putting $i^{2}=-1$

$\Rightarrow \frac{3 x+2 x i-6 i-x(-1)+2(-1)+6 y-7 i y+3 i-3(-1) y+(-1)}{9-(-1)}=i$

$\Rightarrow \frac{3 x+2 x i-6 i+x-2+6 y-7 i y+3 i+3 y-1}{9+1}=i$

$\Rightarrow \frac{4 x+2 x i-3 i-3+9 y-7 i y}{10}=i$

$\Rightarrow 4 x+2 x i-3 i-3+9 y-7 i y=10 i$

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

Comparing the real parts, we get

$4 x-3+9 y=0$

$\Rightarrow 4 x+9 y=3 \ldots$ (i)

Comparing the imaginary parts, we get

$2 x-3-7 y=10$

$\Rightarrow 2 x-7 y=10+3$

$\Rightarrow 2 x-7 y=13 \ldots$ (ii)

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

$4 x-14 y=26 \ldots$ (iii)

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

$4 x-14 y-(4 x+9 y)=26-3$

$\Rightarrow 4 x-14 y-4 x-9 y=23$

$\Rightarrow-23 y=23$

$\Rightarrow y=-1$

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

$4 x+9(-1)=3$

$\Rightarrow 4 x-9=3$

$\Rightarrow 4 x=12$

$\Rightarrow x=3$

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

 

 

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