Question

In a certain positive fraction , the denominator is greater than the numerator by 3 . If 1 is subtracted from both numerator and denominator the fraction is decreased by 1/14 . Find the fraction

Answer 1

Hi there !!
Here's your answer


Let the numerator be x
and the denominator be y
the fraction is x/y

also given that the denominator is greater than the numerator by 3
So denominator = y = x + 3

The fraction formed will be

$\frac{x}{x + 3}$
Given,

If 1 is subtracted from both numerator and denominator the fraction is decreased by 1/14

So,
the new fraction will be

$\frac{x - 1}{x + 3 - 1} = \frac{x - 1}{x + 2}$

So,
the following balanced equation will be formed


$\frac{x - 1}{x + 2} = \frac{x}{x + 3} - \frac{1}{14}$

Transposing x/x+3 to LHS ,
we have,


$\frac{x - 1}{x + 2} - \frac{x}{x + 3} = \frac{ - 1}{14}$

Taking LCM as (x-2)(x+3),
we have,


$\frac{(x - 1)(x + 3) - x(x + 2)}{(x + 2)(x + 3)} = \frac{ - 1}{14}$

Multiplying the binomials in the numerator and denominator, we have,


$\frac{x {}^{2} + 2x - 3 - {x}^{2} - 2x }{ {x}^{2} + 5x + 6 } = \frac{ - 1}{14}$


$\frac{ - 3}{ {x}^{2} + 5x + 6} = \frac{ - 1}{14}$

$- 3 \times 14 = - 1 \times {x}^{2} + 5x + 6$

$- 42 = - {x}^{2} + 5x + 6$
cancelling the negative sign in the RHS and LHS,
we have,

42 = x² + 5x + 6

0 = x² + 5x + 6 - 42

0 = x² + 5x - 36

splitting the middle term
we know that ab = -36 and a + b = 5
so,
a = -4
b = 9
, we have,

0 = x² - 4x + 9x - 36

0 = x(x - 4) + 9(x - 4)

0 = (x + 9)(x - 4)

x + 9 = 0
x = -9

x - 4 = 0
x = 4

So,



x = -9
y = x + 3 = -9 + 3 = -6

The fraction will be

$\frac{ - 9}{ - 6} = \frac{9}{6} \\ = \frac{2}{3}$
But this is not possible as the conditions are not fulfilled
So,

we will take
x = 4
y = x + 3 = 4 + 3 = 7

Hence,

the fraction is

$\frac{4}{7}$

Answer 2

Hi there

Here is your answer

Step-by-step explanation: