Find the real values of $x$ and $y$ for which the following equation is satisfied
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$

x = 3, y = - 1

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x = - 3, y = - 1

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x = 3, y = 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 3, y = 1

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Solution

The correct option is A $x = 3, y = - 1$
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i \Rightarrow (1 + i) (3 - i) x - 2 i (3 - i) + (3 + i) (2 - 2 i) y + i (3 + i) = 10 i \Rightarrow 4 x + 2 i x - 6 i - 2 + 9 y - 7 i y + 3 i - 1 = 10 i$
$\Rightarrow 4 x + 9 y - 3 = 0$ and $2 x - 7 y - 3 = 10$
$\Rightarrow x = 3$ and $y = - 1$

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Similar questions

x

y

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

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

If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then the real values of x and y are given by :

x

y

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

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

x, y

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