Question

the dominator of a rational number is greater than numerator by 8 if then numerator in increased by 17 and the denominator is decreased by 1 the number obtained is 3/2 find the rational numbers =solutions​

Answer 1

Let the numerator of the rational number be x.

So as per the given condition, the denominator will be x + 8.

The rational number will be $\sf{\(\frac{x}{x+8}\)}$

According to the given condition,

$\implies\sf{\(\frac{x+17}{x+8-1} = \frac{3}{2}\)}$

$\implies\sf{\(\frac{x+17}{x+7} = \frac{3}{2}\)}$

$\implies$ 3(x + 7) = 2(x + 17)

$\implies$ 3x + 21 = 2x + 34

$\implies$ 3x – 2x + 21 – 34 = 0

$\implies$ x – 13 = 0

$\implies$ x = 13

The rational number will be

$\sf{= \(\frac{x}{x+8}\)}$

$\sf{= \(\frac{13}{13+8}\)}$

Rational number = $\bf{\frac{13}{21}}$

Answer 2

$\frak{Given}\begin{cases}\sf{\quad Denominator = \bf{Numerator + 8}}\\\\\sf{\quad \dfrac{Numerator + 17}{Denominator - 1} = \bf{\dfrac{3}{2}}}\end{cases}$

$\frak{To\; find:}$ The rational number?

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❍ Let's say, that the Numerator of the fraction be x. Then, the Denominator of the fraction would be (x + 8).

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$\underline{\bigstar\:\boldsymbol{According\;to\;the\; Question\; :}}$⠀

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• If the numerator is increased by 17 and the denominator of the fraction is decreased by 1 then the number obtained is 3⁄2.

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$:\implies\sf \Bigg\{\dfrac{Numerator + 17}{Denominator - 1}\Bigg\} = \Bigg\{\dfrac{3}{2}\Bigg\} \\\\\\:\implies\sf \Bigg\{\dfrac{x + 17}{x + 8 - 1}\Bigg\} = \Bigg\{\dfrac{3}{2}\Bigg\} \\\\\\:\implies\sf \Bigg\{\dfrac{x + 17}{x + 7}\Bigg\} = \Bigg\{\dfrac{3}{2}\Bigg\} \\\\\\:\implies\sf 2\Big\{x + 17\Big\} = 3\Big\{x + 7\Big\} \\\\\\:\implies\sf 2x + 34 = 3x + 21 \\\\\\:\implies\sf 2x - 3x = 21 - 34 \\\\\\:\implies\sf \cancel{-}x = \cancel{-} 13 \\\\\\:\implies\underline{\boxed{\pmb{\frak{x = 13}}}}\:\bigstar$

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Therefore,

• Numerator of the fraction, x = 13
• Denominator of the fraction, (x + 8) = (13 + 8) = 21.

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$\therefore{\underline{ \sf{Hence,\; the \;rational\; number \;is \;{\sf{\pmb{\dfrac{13}{21}}}}.}}}$