The real numbers x and y are such that x-2-y-8-3 and y-2-x-3 - The value of xy- is

The correct option is B 2 x+2/y=8/3 #160; #160; #160; #160; #160; ...(i) y+2/x=3⇒ y=3x 2/x #160; #160; ...(ii)From (i) ...

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Quantitative Aptitude

Divisibility Rule for 3 & 9

The real numb...

Question

The real numbers $x$ and $y$ are such that $x + \frac{2}{y} = \frac{8}{3}$ and $y + \frac{2}{x} = 3$ .
The value of $x y$ , is

\frac{4}{3}

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\frac{16}{9}

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Solution

The correct option is B $2$
$x + \frac{2}{y} = \frac{8}{3}$ ...(i)

$y + \frac{2}{x} = 3 \Rightarrow y = \frac{3 x - 2}{x}$ ...(ii)
From (i)
$x + \frac{2 x}{3 x - 2} = \frac{8}{3}$
$\Rightarrow 3 (3 x^{2} - 2 x + 2 x) = 8 (3 x - 2)$
$\Rightarrow 9 x^{2} - 24 x + 16 = 0$
$\Rightarrow (3 x - 4)^{2} = 0$
$\Rightarrow x = \frac{4}{3}$
From (ii), we get
$y = \frac{3 \times \frac{4}{3} - 2}{\frac{4}{3}} = \frac{3}{2}$
Now, $x y = \frac{4}{3} \times \frac{3}{2} = 2$
Hence, option C is correct.

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The real numbers x and y are such that x+2y=83 and y+2x=3 . The value of xy, is

The correct option is B 2x+2y=83 ...(i)y+2x=3⇒y=3x−2x ...(ii)From (i)x+2x3x−2=83 ⇒3(3x2−2x+2x)=8(3x−2) ⇒9x2−24x+16=0⇒(3x−4)2=0⇒x=43From (ii), we gety=3×43−243=32Now, xy=43×32=2Hence, option C is correct.

The real numbers $x$ and $y$ are such that $x + \frac{2}{y} = \frac{8}{3}$ and $y + \frac{2}{x} = 3$ .
The value of $x y$ , is