Question

in certain fraction the dinominator is greatot than the number by 2 if 5 is added to both numerator and the dinominator the fraction increase by ⅕ find the fraction​

Answer 1

Answer :-

$\: \: \large{\underline{\underline{\rm{\mapsto \: \: \: Firstly \: let's \: understand \: the \: concept \: used}}}}$

Here the concept of Linear Equations in Two Variables has been used. According to this, if the two unknown quantities and their dependencies are given, we can take them as variables, form equations and solve it. We will take both the numerator, denominator and the fraction as variables.

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★ Question ::

In certain fraction the denominator is greater than the number by 2 if 5 is added to both numerator and the denominator the fraction increase by ⅕. Find the fraction.

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★ Solution :-

Given,

» Denominator of the fraction = 2 + Numerator

» (5 + Numerator) and (5 + Denominator) = Initial Fraction + ⅕

• Let the numerator of the fraction be 'x'

• Let the denominator of the fraction be 'y'

$\: \: \\ \sf{\mapsto \: \: Then, \; the \; fraction \; is \; = \; \bf{\dfrac{x}{y}}}$

Then, according to the question :-

~ Case I :-

➣ y = 2 + x ... (i)

~ Case II :-

$\: \\\: \large{\rm{\Longrightarrow \: \: \: \dfrac{x \: + \: 5}{y \: + \: 5} \: = \: \bf{\dfrac{x}{y} \: + \: \dfrac{1}{5}}}}$

$\: \\ \: \large{\rm{\Longrightarrow \: \: \: \dfrac{x \: + \: 5}{y \: + \: 5} \: = \: \bf{\dfrac{5x \: + \: y}{5y}}}}$

By cross multiplication, we get,

➣ 5y(x + 5) = (5x + y)(y + 5)

➣ 5xy + 25y = 5xy + 25x + y² + 5y

Cancelling 5xy on both sides, we get,

➣ y² + 5y + 25x - 25y = 0

➣ y² + 25x - 20y = 0 .... (ii)

From equation (i) and (ii), we get,

➣ (2 + x)² + 25x - 20(2 + x) = 0

➣ 4x + x² + 4 + 25x - 40 - 20x = 0

➣ x² + 9x - 36 = 0

➣ x² + 12x - 3x - 36 = 0

➣ x(x + 12) - 3(x + 12) = 0

➣ (x - 3)(x + 12) = 0

Here, either (x - 3) = 0 or (x + 12) = 0.

Then,

➣ (x - 3) = 0 or (x + 12) = 0

➣ x = 3 or x = -12

$\: \\ \large{\boxed{\boxed{\tt{x \: = \: 3 \: \: or \: \: -12}}}}$

Now applying the values of x in equation (i), we get,

➣ y = 2 + x

➣ y = (2 + 3) or (2 - 12)

➣ y = 5 or y = -10

$\: \\ \large{\boxed{\boxed{\tt{y \: = \: 5 \: \: or \: \: -10}}}}$

From the above results we see that,

• when, x = 3 , then y = 5

• when, x = -12 , then y = -10 (cancelling negative sign from numerator and denominator)

Then,

$\\ \: \large{\boxed{\boxed{\sf{\dfrac{x}{y} \: = \: \dfrac{3}{5} \: \: \: or \: \: \: \dfrac{\not{-}12}{\not{-}10}}}}}$

We see that, when we cancel negative sign from numerator and denominator from second value of fraction, we see

=> y = 2 + x

=> 10 = 2 + 12

=> 10 ≠ 14

Here the case doesn't satisfies. So we have neglect this fraction as correct option. Hence, 12/10 is not the correct fraction.

$\: \\ \large{\underline{\underline{\rm{\mapsto \: \: \: Thus, \: the \: required \: correct \: fraction \: is \: \boxed{\bf{\dfrac{3}{5}}}}}}}$

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$\: \\ \: \large{\underline{\underline{\underline{\sf{Confused? \: Don't \: worry \: let's \: verify \: it \: :-}}}}}$

Here x = 3 and y = 5

~ Case I :-

=> y = 2 + x

=> 5 = 2 + 3

=> 5 = 5

Clearly, LHS = RHS

~ Case II :-

$\: \\\: \large{\rm{\Longrightarrow \: \: \: \dfrac{x \: + \: 5}{y \: + \: 5} \: = \: \bf{\dfrac{x}{y} \: + \: \dfrac{1}{5}}}}$

$\: \\ \: \large{\rm{\Longrightarrow \: \: \: \dfrac{x \: + \: 5}{y \: + \: 5} \: = \: \bf{\dfrac{5x \: + \: y}{5y}}}}$

=> 5xy + 25y = 5xy + 25x + y² + 5y

=> 5(3)(5) + 25(5) = 5(3)(5) + 25(3) + 5² + 5(5)

=> 75 + 125 = 75 + 75 + 25 + 25

=> 200 = 200

Clearly, LHS = RHS

Here both the conditions satisfy, so our answer is correct.

Hence, Verified.

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$\: \qquad \qquad \large{\underline{\leadsto \: \: More \: to \: Know \: :-}}$

• Polynomials are the equations formed using variable and constant terms but variable term can be of many degrees.

• Linear Equations are the equations formed using variable and constant terms but variable term is of single degree only.

Answer 2

Given :

• Denominator is greater than numerator by 2
• If 5 is added to both numerator and denominator , fraction increases by 1/5.

To find :

• The fraction

Solution :

Let numerator be m and denominator be (m + 2)

∴ Fraction = m/(m + 2)

Now atq,

⇒  (m + 5)/(m + 2 + 5) = m/(m + 2) + 1/5

⇒  (m + 5)/(m + 7) = (5m + m + 2)/5(m + 2)

⇒  (m + 5)/(m + 7) = (6m + 2)/(5m + 10)

⇒  (m + 5) (5m + 10) = (m + 7)(6m + 2)

⇒  5m² + 10m + 25m + 50 = 6m² + 2m + 42m + 14

⇒  5m² + 35m + 50 = 6m² + 44m + 14

⇒  6m² - 5m² + 44m - 35m + 14 - 50 = 0

⇒  m² + 9m - 36 = 0

⇒  m² + 12m - 3m - 36 = 0

⇒  m(m + 12) - 3(m + 12) = 0

⇒  (m - 3)(m + 12) = 0

⇒  m = 3  or,  m = - 12

When m = 3 :

Numerator = m = 3

Denominator = m + 2 = 3 + 2 = 5

∴ Fraction = 3/5

When m = - 12 :

Numerator = m = - 12

Denominator = m + 2 = - 12 + 2 = - 10

∴ Fraction = -12/-10 = 6/5

∵ 2nd fraction i.e 6/5 doesn't fulfill our first condition, so the required fraction = 3/5

Therefore,

The fraction = 3/5