Question

First we increase the denominator of a positive fraction by 3 and next
time we decrease it by 5. The sum of the resulting fractions proves to be equal to 2/3 . Find the denominator of the fraction if its numerator is 2 .
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Answer 1

Question:-

• First we increase the denominator of a positive fraction by 3 and next time we decrease it by 5. The sum of the resulting fractions proves to be equal to 2/3 . Find the denominator of the fraction if its numerator is 2 .

To Find:-

• Find the denominator .

Given:-

• First we increase the denominator of a positive fraction by 3 and next time we decrease it by 5. The sum of the resulting fractions proves to be equal to 2/3 .

Solution:-

Let the denominator be " x "

According to the Question:-

$\tt\implies \: \dfrac { 2 } { x + 3 } + \dfrac { 2 } { x - 5 } = \dfrac { 2 } { 3 }$

$\tt\implies \: \dfrac { 2( x - 5 ) + 2( x + 3 ) } { ( x + 3 ) ( x - 5 ) } = \dfrac { 2 } { 3 }$

$\tt\implies \: \dfrac { 2x - 10 + 2x + 6 } { { x }^{ 2 } - 5x +3x - 15 } = \dfrac { 2 } { 3 }$

$\tt\implies \: \dfrac { 4x - 4 } { { x }^{ 2 } - 2x - 15 } = \dfrac { 2 } { 3 }$

$\tt\implies \: 3( 4x - 4 ) = 2( { x }^{ 2 } - 2x - 15 )$

$\tt\implies \: 12x - 12 = { 2x }^{ 2 } - 4x - 30$

$\tt\implies \: { 2x }^{ 2 } - 16x - 18 = 0$

$\tt\implies \: { 2x }^{ 2 } - 18x + 2x - 18 = 0$

$\tt\implies \: 2x( x - 9 ) + 2( x - 9 ) = 0$

$\tt\implies \: ( x - 9 ) ( 2x + 2 )$

$\tt\implies \: x = 9 ; - 1$

Answer 2

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

• First we increase the denominator of a positive fraction by 3 and next time we decrease it by 5. The sum of the resulting fractions proves to be equal to 2/3 . Find the denominator of the fraction if its numerator is 2 .

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

• $\sf{Let \:the \:given \:fraction \:be \:\dfrac{x}{y}.}$

• $\textsf{According to the question, x = 2.}$

• $\textsf{First, the denominator increases by 3.}$

• $\sf{So, \: the \: fraction \: is \: \: \dfrac{2}{y \: + \: 3}}$

• $\textsf{Second, the denominator is decreased by 5.}$

• $\sf{So, \: the \: fraction \: is \: \: \dfrac{2}{y \: - \: 5}}$

• $\sf{The\: sum\: of \:the \:fractions \:is\: \: \dfrac{2}{3}}$

• $\textsf{Solving,}$

• $\sf{\dfrac{2}{y \: + \: 3} \: + \: \dfrac{2}{y \: - \: 5} \: = \: \dfrac{2}{3}}$

• $\textsf{Taking LCM,}$

• $\sf{\dfrac{2(y \: - \: 5) \: + \: 2(y \: + \: 3)}{(y \: - \: 5) \: (y \: + \: 3)} \: = \: \dfrac{2}{3}}$

• $\sf{\dfrac{2y \: - \: 10 \: + \: 2y \: + \: 6}{y(y \: + \: 3) \: - \: 5(y \: + \: 3)} \: = \: \dfrac{2}{3}}$

• $\sf{\dfrac{4y \: - \: 4}{y^{2} \: + \: 3y \: - \: 5y \: - \: 15} \: = \: \dfrac{2}{3}}$

• $\sf{3(4y \: - \: 4) \: = \: 2(y^{2} \: - \: 2y \: - \: 15)}$

• $\sf{12y \: - \: 12 \: = \: 2y^{2} \: - \: 4y \: - \: 30}$

• $\sf{2y^{2} \: - \: 16y \: - \: 18 \: = \: 0}$

• $\textsf{Finding factors by PSF method,}$

• $\sf{P \: = \: - \: 36y^{2}}$

• $\sf{S \: = \: - \: 16y}$

• $\sf{F \: = \: - \: 18y \: , \: + \: 2y}$

• $\sf{2y^{2} \: - \: 18y \: + \: 2y \: - \: 18 \: = \: 0}$

• $\sf{2y(y \: - \: 9) \: + \: 2(y \: - \: 9) \: = \: 0}$

• $\sf{(2y \: + \: 2) \: (y \: - \: 9) \: = \: 0}$

• $\sf{2y \: + \: 2 \: = \: 0 \: , \: y \: - \: 9 \: = \: 0}$

• $\sf{2y \: = \: - \: 2 \: , \: y \: = \: 9}$

• $\boxed{\sf{y \: = \: - \: 1 \: , \: y \: = \: 9}}$

$\huge{\bf{\red{\mathfrak{Conclusion:-}}}}$

• $\boxed{\therefore{\sf{The \: denominators \: of \: the \: fraction \: are \: \: 9 \: , \: - \: 1.}}}$