Question

if we add one to the numerator and subtract I from denominator fraction reduces to one it becomes
$\frac{1}{2}$
if we add l to the denominator what is fraction

Answer 1

Here is ur answer hope it helps....

Answer 2

Let the Numerator of Fraction =x
Denominator of fraction. =y

$required \: fraction \: is = \frac{x}{y}$
According to lst condition

$\: \: \: \: \: \: = \frac{x + 1}{y - 1} = 1 \\ \: \: \: \: \: \: = x + 1 = y - y \\ \: \: \: \: \: \: = x - y + 2 = 0 \: \: \: \: \: (1)$
According to the second condition
$\: \: \: \: \: \: \: \: \: \: \frac{x}{y + 1} = \frac{1}{2} \\ \: \: \: \: \: \: \: \: \: 2x \: \: \: \: \: \: \: = y + 1\\ 2x - y - 1 =0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$


Now (2)-(1) gives
$2x - y - 1 = 0 \\ x - y + 2 \: \: =0 \\ - \: \: \: + \: \: \: \: \: \: - \\ \: \: \: \: \: \: \: \: \: \: \: x - 3 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 3$
substiutue the value of x in (2) we get
$2 \times 3 - y - 1 = 0 \\ \: \: \: \: \: \: \: \: 6 - y - 1 = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 5 - y = 0 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = 5 \\ hence \: required \: fraction \: is \: \frac{3}{5}$