Question

2.
+3=0
$\frac{2}{x} + \frac{2}{3y} = \frac{1}{6} \\ \frac{3}{x} + \frac{2}{y } = 0$
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Answer 1

Answer

${\sf \dashrightarrow \dfrac{2}{x} + \dfrac{2}{3y} = \dfrac{1}{6} \: \: --- (i)}$

${\sf \dashrightarrow \dfrac{3}{x} + \dfrac{2}{y} = 0 \: \: --- (ii)}$

Let,

${\sf \dashrightarrow \dfrac{1}{x} = u}$

${\sf \dashrightarrow \dfrac{1}{y} = v}$

Now, the equations become

${\sf \dashrightarrow 2u + \dfrac{2y}{3} = \dfrac{1}{6}}$

${\sf \dashrightarrow 3u + 2v = 0}$

Now, from equation (ii),

${\sf \dashrightarrow 3u + 2v = 0}$

${\sf \dashrightarrow 3u = -2v}$

${\sf \dashrightarrow u = \dfrac{-2v}{3}}$

Now, let's find the value of v by first equation.

${\sf \dashrightarrow 2 \bigg( \dfrac{-2v}{3} \bigg) + \dfrac{2v}{3} = \dfrac{1}{6}}$

${\sf \dashrightarrow \dfrac{-4v}{3} + \dfrac{2v}{3} = \dfrac{1}{6}}$

${\sf \dashrightarrow \dfrac{-2v}{3} = \dfrac{1}{6}}$

${\sf \dashrightarrow -2v = \dfrac{3}{6}}$

${\sf \dashrightarrow -2v = \dfrac{1}{2}}$

${\sf \dashrightarrow -v = \dfrac{1}{4}}$

${\sf \dashrightarrow v = \dfrac{-1}{4}}$

Now, let's find the value of u.

${\sf \dashrightarrow u = \dfrac{-2v}{3}}$

${\sf \dashrightarrow u = \dfrac{-2 \bigg( \dfrac{-1}{4} \bigg)}{3}}$

${\sf \dashrightarrow u = \dfrac{\dfrac{1}{2}}{3}}$

${\sf \dashrightarrow u = \dfrac{1}{6}}$

Now, let's find the value of x.

${\sf \dashrightarrow \dfrac{1}{x} = u}$

${\sf \dashrightarrow \dfrac{1}{x} = \dfrac{1}{6}}$

${\sf \dashrightarrow x = 6}$

Now, let's find the value of y.

${\sf \dashrightarrow \dfrac{1}{y} = v}$

${\sf \dashrightarrow \dfrac{1}{y} = \dfrac{-1}{4}}$

${\sf \dashrightarrow y = (-4)}$

Hence, the values of x and y are 6 and (-4) respectively.