ong>Answer:

3/4

Step-by-step explanation:

Let the denominator be x.

A.T.Q., the NUMERATOR will be (x - 1).

Original fraction = {\SF{{\dfrac{x - 1}{x}}}}

x

x−1

Now, as given in the question, 3 is added to both numerator and denominator.

So,

Numerator = x - 1 + 3 = x + 2

Denominator = x + 3

A.T.Q.

{\sf{{\dfrac{x + 2}{x + 3}} = {\dfrac{x - 1}{x}} + {\dfrac{3}{28}} }}

x+3

x+2

=

x

x−1

+

28

3

On further solving, we get

\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28(x - 1) + 3(x)}{(28)(x)}} }}⇒

x+3

x+2

=

(28)(x)

28(x−1)+3(x)

\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{28x - 28 + 3x}{28x}} }}⇒

x+3

x+2

=

28x

28x−28+3x

\Rightarrow{\sf{ {\dfrac{x + 2}{x + 3}} = {\dfrac{31x - 28}{28x}} }}⇒

x+3

x+2

=

28x

31x−28

Cross - multiplication.

\Rightarrow⇒ 28x(x + 2) = (x + 3)(31x - 28)

\Rightarrow⇒ 28x(x + 2) = x(31x - 28) + 3(31x - 28)

\Rightarrow⇒ 28x² + 56x = 31x² - 28x + 93X - 84

\Rightarrow⇒ 28x² + 56x = 31x² + 65x - 84

Transposing the terms.

\Rightarrow⇒ 28x² - 31x² = 65x - 56x - 84

\Rightarrow⇒ - 3x² = 9x - 84

\Rightarrow⇒ 3x² + 9x - 84 = 0

\Rightarrow⇒ 3(x² + 3x - 28) = 0

\Rightarrow⇒ x² + 3x - 28 = 0

Using Middle Term Factorisation, we get

\Rightarrow⇒ x² - 4X + 7x - 28 = 0

\Rightarrow⇒ x(x - 4) + 7(x - 4) = 0

\Rightarrow⇒ (x + 7)(x - 4) = 0

Using zero product rule, we get

\Rightarrow⇒ x + 7 = 0 and x - 4 = 0

\Rightarrow⇒ x = - 7 and x = 4

Here, x is a natural number, where x > 0.

Hence, x = 4

The required fraction is :

\Rightarrow{\sf{ {\dfrac{x - 1}{x}} }}⇒

x

x−1

\Rightarrow{\sf{ {\dfrac{4 - 1}{4}} }}⇒

4

4−1

\Rightarrow{\boxed{\sf{\red{ {\dfrac{3}{4}} }}}}⇒

4

3