Respuesta :
❍ Let's say, that the numerator of the fraction be x and denominator of the fraction be y.
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[tex]\underline{\sf{Case\;}\it{1}:}[/tex]
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- If 2 is added to the numerator of the fraction, it reduces to 1/2.
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Therefore,
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[tex]\begin{gathered}:\implies\sf{\dfrac{x+2}{y}=\dfrac{1}{2}}\\\\\\:\implies\sf{2x+4=y}\\\\\\:\implies\sf{2x=y-4}\qquad\qquad\bigg\lgroup\sf{eq^n\;1}\bigg\rgroup\end{gathered}[/tex]
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[tex]\underline{\sf{Case\;}\it{2}:}[/tex]
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- If 1 is subtracted from the denominator of the fraction, it reduces to 1/3.
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Therefore,
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[tex]\begin{gathered}:\implies\sf{\dfrac{x}{y-1}=\dfrac{1}{3}}\\\\\\:\implies\sf{3x=y-1}\qquad\qquad\bigg\lgroup\sf{eq^n\;2}\bigg\rgroup\end{gathered}[/tex]
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[tex]\qquad\qquad\footnotesize{\underline{\bf{\dag\;}\frak{Subtracting\;eq^n\;2\;from\;eq^n\;1:}}}[/tex]
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[tex]\begin{gathered}:\implies\sf{3x-2x=y-1-y-4}\\\\\\:\implies\underline{\boxed{\pink{\frak{\pmb{x=3}}}}}\;\bigstar\end{gathered}[/tex]
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[tex]\qquad\qquad\footnotesize{\underline{\bf{\dag\;}\frak{Substituting\;value\;of\;x\;in\;eq^n\;1:}}}[/tex]
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[tex]\begin{gathered}:\implies\sf{2x=y-4}\\\\\\:\implies\sf{2(3)=y-4}\\\\\\:\implies\sf{6=y-4}\\\\\\:\implies\sf{6=y-4}\\\\\\:\implies\sf{y=-4-6}\\\\\\:\implies\underline{\boxed{\purple{\frak{\pmb{y=10}}}}}\;\bigstar\end{gathered}[/tex]
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Hence,
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- Numerator = x = 3
- Denominator = y = 10
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[tex]\therefore\;{\underline{\sf{Hence,\;the\;fraction\;is\;\frak{\dfrac{3}{10}}}.}}[/tex]⠀⠀