If $(x + i y) (2 - 3 i) = 4 + i$ then (x, y) =

(1, \frac{1}{13})

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(- \frac{5}{13}, \frac{14}{13})

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(\frac{5}{13}, \frac{14}{13})

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(- \frac{5}{13}, - \frac{14}{13})

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Solution

The correct option is D $(\frac{5}{13}, \frac{14}{13})$
$(x + i y) (2 - 3 i) = 4 + i$

$2 x - (3 x) i + (2 y) i - 3 y i^{2} = 4 + i$

$2 x + 3 y      R e a l + (2 y - 3 x)      I m a g i n a r y i = 4 + i$

Comparing the real & imaginary parts,
$2 x + 3 y = 4$ --------------------------(1)
$2 y - 3 x = 1$ ----------------------------(2)
Solving eq(1) & eq(2),
$4 x + 6 y = 8$
$- 9 x + 6 y = 3$
$13 x = 5 \Rightarrow x = \frac{5}{13}$
$y = \frac{14}{13}$
$∴ (x, y) = (\frac{5}{13}, \frac{14}{13})$

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Question 81 (i)

Simplify: $\frac{13}{11} \times \frac{- 14}{5} + \frac{13}{11} \times \frac{- 7}{5} + \frac{- 13}{11} \times \frac{34}{5}$

I = \frac{13}{4} \div \frac{5}{6}

I I = 13 \div [(4 \div 5) \div 6]

I I I = [13 \div (4 \div 5)] \div 6

I V = 13 \div 4 (5 \div 6)

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