Question

The numerator of a fraction is 2 less than its denominator. If 3 is added to each of the numerator and the denominator, the fraction increases by 1/18. Find the fraction.​

Answer 1

Answer:

let the denominator of the required fraction beX.

so the numerator will be X-2

so.

The required fraction will be

10/12 or 11/9

Answer 2

Answer:

The required fraction is

$\displaystyle{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{6}}}\:\sf\:\quad\:OR\:\quad\:\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{9}}}}$

Step-by-step-explanation:

Let the numerator of the fraction be x.

And the denominator of the fraction be y.

$\displaystyle{\sf\:\therefore\:The\:fraction\:=\:\dfrac{x}{y}}$

From the first condition,

x + 2 = y

⇒ y = x + 2 .... ( 1 )

From the second condition,

$\displaystyle{\sf\:\dfrac{x\:+\:3}{y\:+\:3}\:=\:\dfrac{x}{y}\:+\:\dfrac{1}{18}}$

$\displaystyle{\implies\sf\:\dfrac{x\:+\:3}{x\:+\:2\:+\:3}\:=\:\dfrac{x}{x\:+\:2}\:+\:\dfrac{1}{18}\:\quad\:\dots\:[\:From\:(\:1\:)\:]}$

$\displaystyle{\implies\sf\:\dfrac{x\:+\:3}{x\:+\:5}\:=\:\dfrac{18\:x\:+\:x\:+\:2}{18\:\times\:(\:x\:+\:2\:)}}$

$\displaystyle{\implies\sf\:\dfrac{x\:+\:3}{x\:+\:5}\:=\:\dfrac{19\:x\:+\:2}{18\:x\:+\:36}}$

$\displaystyle{\implies\sf\:(\:18x\:+\:36\:)\:(\:x\:+\:3\:)\:=\:(\:19x\:+\:2\:)\:(\:x\:+\:5\:)}$

$\displaystyle{\implies\sf\:18x^2\:+\:54x\:+\:36x\:+\:108\:=\:19x^2\:+\:95x\:+\:2x\:+\:10}$

$\displaystyle{\implies\sf\:18x^2\:+\:90x\:+\:108\:=\:19x^2\:+\:97x\:+\:10}$

$\displaystyle{\implies\sf\:19x^2\:+\:97x\:+\:10\:-\:(\:18x^2\:+\:90x\:+\:108\:)\:=\:0}$

$\displaystyle{\implies\sf\:19x^2\:+\:97x\:+\:10\:-\:18x^2\:-\:90x\:-\:108\:=\:0}$

$\displaystyle{\implies\sf\:19x^2\:-\:18x^2\:+\:97x\:-\:90x\:+\:10\:-\:108\:=\:0}$

$\displaystyle{\implies\sf\:x^2\:+\:7x\:-\:98\:=\:0}$

$\displaystyle{\implies\sf\:x^2\:+\:14x\:-\:7x\:-\:98\:=\:0}$

$\displaystyle{\implies\sf\:x\:(\:x\:+\:14\:)\:-\:7\:(\:x\:+\:14\:)\:=\:0}$

$\displaystyle{\implies\sf\:(\:x\:+\:14\:)\:(\:x\:-\:7\:)\:=\:0}$

$\displaystyle{\implies\sf\:x\:+\:14\:=\:0\:\quad\:OR\:\quad\:x\:-\:7\:=\:0}$

$\displaystyle{\implies\boxed{\pink{\sf\:x\:=\:-\:14}}\sf\:\quad\:OR\:\quad\:\boxed{\pink{\sf\:x\:=\:7}}}$

Now, by substituting x = - 14 in equation ( 1 ), we get,

y = x + 2

$\displaystyle{\implies\sf\:y\:=\:-\:14\:+\:2}$

$\displaystyle{\implies\boxed{\blue{\sf\:y\:=\:-\:12}}}$

Now,

$\displaystyle{\sf\:The\:fraction\:=\:\dfrac{x}{y}}$

$\displaystyle{\implies\sf\:The\:fraction\:=\:\dfrac{-\:14}{-\:12}}$

$\displaystyle{\implies\sf\:The\:fraction\:=\:\cancel{\dfrac{14}{12}}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{6}}}}}$

Now, by substituting x = 7 in equation ( 1 ), we get,

y = x + 2

$\displaystyle{\implies\sf\:y\:=\:7\:+\:2}$

$\displaystyle{\implies\boxed{\blue{\sf\:y\:=\:9}}}$

Now,

$\displaystyle{\sf\:The\:fraction\:=\:\dfrac{x}{y}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:The\:fraction\:=\:\dfrac{7}{9}}}}}$