Question

the numerator of a fraction is 3 less than the denominator if both the numerator and denominator are increased by 2 the new fraction becomes 6 /7 and find that original number

Answer 1

Let the denominator be x and the numerator be x - 3

A/Q,
$\frac{x - 3 + 2}{x + 2} = \frac{6}{7} \\ = > \frac{x - 1}{x + 2} = \frac{6}{7} \\ = > 7(x - 1) = 6(x + 2) \\ = > 7x - 7 = 6x + 12 \\ = > 7x - 6x = 12 + 7 \\ = > x = 19$
.°. Denominator = 19
Numerator = 19 - 3 = 16

Hence, the number is
$\frac{16}{19}$

HOPE IT'LL HELP..

Answer 2

$\large \dag$ Question :-

The numerator of a fraction is 3 less than the denominator if both the numerator and denominator are increased by 2 the new fraction becomes 6 /7 and find that original number

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{The   \: Original \:  Fraction  \: is \:  \frac{16}{19} }} }$

$\large \dag$ Step by step Explanation :-

Let Numerator and Denominator of Original Fraction be :

• Numerator = x

As Per the question numerator of the fraction is 3 less than the denominator so,

• Denominator should be = x + 3

So ,

$\text{Original Fraction = } \frac{\text x}{\text x + 3}$

❒ When both the numerator and denominator are increased by 2 :

$\\ \rm \text{Fraction Becomes } :  \frac{x +2}{x + 5}$

⏩ According To Question :

$\\ \large \blue \bigstar  \:   \red{ \bf  \frac{x +2}{x + 5} =  \frac{6}{7}  }$

$:\longmapsto \rm 7(x  +  2) = 6(x + 5)$

$:\longmapsto \rm 7x  +  14 = 6x + 30$

$:\longmapsto \rm 7x  -  6x = 30-14$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf x = 16} }}}$

$\blue\dashrightarrow\underline{\underline{\sf  \orange{Numerator  \: of \:  Original  \: Fraction = 16 }} }$

☆ As numerator of the fraction is 3 less than the denominator

$\rm\therefore \:  Denominator  = 16   + 3$

$\blue\dashrightarrow\underline{\underline{\sf  \orange{Denominator  \: of \:  Original  \: Fraction = 19}} }$

Therefore,

$\large\underline{\pink{\underline{\frak{\pmb{ Original  \: Fraction  =  \dfrac{16}{19} }}}}}$