Question

If the numerator and denominator of a fraction are increased by 2 and 1

respectively, it becomes ¾  If the numerator and denominator are decreased by 2

and 1 repectively, it becomes ½ Find the fraction​

Answer 1

❍ Let's Assume the Numerator and Denominator of fraction be x and y respectively.

Then ,

• The Original Fraction = $\dfrac{x}{y}$

Case I ) If the numerator and denominator of a fraction are increased by 2 and 1 respectively, it becomes ¾ .

Therefore,

• $\therefore The \:Fraction = \dfrac{x + 2 }{y + 1} = \dfrac{3}{4}$

Then ,

$\qquad\qquad :\implies \sf{ \dfrac{x + 2 }{y + 1} = \dfrac{3}{4}}$

• By Cross Multiplication :

$\qquad\qquad :\implies \sf{ \dfrac{x + 2 }{y + 1} = \dfrac{3}{4}}$

$\qquad\qquad :\implies \sf{ 4(x + 2) = 3 (y + 1) }$

$\qquad\qquad :\implies \sf{ 4x-3y + 8 = 1 }$

$\qquad\qquad :\implies \sf{ 4x-3y = 1-8 }$

$\qquad\qquad :\implies \bf{\bigg( 4x-3y = -7\bigg) }\longrightarrow \sf{Eq.1}$

Case II ) If the numerator and denominator are decreased by 2 and 1 repectively, it becomes ½

Therefore,

• $\therefore The \:Fraction = \dfrac{x -2 }{y - 1} = \dfrac{1}{2}$

Then ,

$\qquad\qquad :\implies \sf{ \dfrac{x - 2 }{y - 1} = \dfrac{1}{2}}$

• By Cross Multiplication :

$\qquad\qquad :\implies \sf{ \dfrac{x- 2 }{yy- 1} = \dfrac{1}{2}}$

$\qquad\qquad :\implies \sf{ 2(x - 2) = 1 (y - 1) }$

$\qquad\qquad :\implies \sf{ 2x - 4 = y - 1 }$

$\qquad\qquad :\implies \sf{ 2x - 4 - y = - 1 }$

$\qquad\qquad :\implies \sf{ 2x - y = - 1 + 4 }$

$\qquad\qquad :\implies \bf{\bigg( 2x-y = 3\bigg) }\longrightarrow \sf{Eq.2}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: doing\:Elimination \;method\: in \: Eq.1\:\&\:Eq.2 \: \::}}$

• $\qquad\qquad :\implies \bf{\bigg( 4x-3y = -5\bigg) }\longrightarrow \sf{Eq.1}$

• $\qquad\qquad :\implies \bf{\bigg( 2x-y = 3\bigg) }\longrightarrow \sf{Eq.2}$

Now ,

• Multiply Eq.2 by 2 to get it's one variable is equal to the Eq.1 :

$\qquad\qquad :\implies \bf{\bigg( 2x-y = 3\bigg) }\longrightarrow \sf{Eq.2}$

$\qquad\qquad :\implies \sf{ 2x - y = 3 }$

• Multiplying by 2 we get as Eq.2 ,

• $\qquad\qquad :\implies \bf{\bigg( 4x-2y = 6\bigg) }\longrightarrow \sf{Eq.2}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Eliminating \: one \: variable \: \::}}$

$\boxed{\begin {array}{ccc}\\ \sf{\cancel {4x}} & \sf{-3y} & =\sf{-5}\\\\\sf{\cancel {4x}}&\sf{-2y}&=\sf{\:6} \\\\ \sf{-}&\sf{+}&\sf{-}\\\\ \underline {\qquad \qquad}&\underline {\qquad \quad}&\underline {\qquad \quad}\\\\ \sf{\:}&\sf{-y}&=\sf{-11}\end {array}}$

$\qquad\qquad :\implies \sf{ - y = - 11 }$ [ - sign will be eliminated]

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  y = 11\: }}}}\:\bf{\bigstar}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: y=11 \: in\:Eq.2::}}$

$\qquad\qquad :\implies \sf{ 2x - y = 3 }$

$\qquad\qquad :\implies \sf{ 2x - 11 = 3 }$

$\qquad\qquad :\implies \sf{ 2x = 3 +11}$

$\qquad\qquad :\implies \sf{ 2x = 14 }$

$\qquad\qquad :\implies \sf{ x = \dfrac{14}{2} }$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = 7\: }}}}\:\bf{\bigstar}$

Thus ,

• The Numerator is x = 7

• The Denominator is y = 11

Then ,

• The Original Fraction = $\sf{\dfrac{x}{y}= \dfrac{7}{11}}$

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Hence \: The\:Original \:Fraction \:is\:\bf{\dfrac{7}{11}\: }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀