Question

15. The numerator of a rational number is 5 less than its
denominator. If 2 is subtracted from the numerator and 2
is added to the denominator, the number becomes 2/5.
Find the rational number.​

Answer 1

Answer:

Step-by-step explanation:

$\huge{\sf{\pink{Answer:-}}}$

Let the denominator be x.

So, accordingly the numerator will be x-5.

$\huge{\sf{\green{A.T.Q:-}}}$

2 is added to denominator and subsequently subtracted from Numerator:-

$\sf{\dfrac{(x-5)-2}{x+2}= \dfrac{2}{5}}$

$\sf{\dfrac{x-5-2}{x+2}= \dfrac{2}{5}}$

$\sf{\dfrac{x-7}{x+2}= \dfrac{2}{5}}$

Cross multiplication:-

$\sf{5(x-7)=2(x+2)}$

$\sf{5x-35=2x+4}$

$\sf{5x-2x = 4+35}$

$\sf{3x = 39}$

$\sf{x = \dfrac{39}{3}}$

$\sf{x = 13}$

The required Fraction is:-

$\sf{\dfrac{13-5}{13}}$

$\sf{\purple{=\dfrac{8}{13}}}$

Please Note that:-

The attemption of my answer is considerate or applicable for linear equation in one variable. You may like to refer the second answer for 2 variable answer. ⬇️

Answer 2

Answer :-

$\: \: ✰ \: \: \underline{\boxed{\sf{\blue{Understanding \: the \: concept}}}}$

Here the concept of a fraction and Linear Equations in Two Variables has been used. According to this, the fraction has two parts, that is numerator and denominator. Also, using Linear Equations In Two Variables, we can find the value of two variables by making each other depend on each other. Standard form is given as,

• ax + by + c = 0

• px + qy + d = 0

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★ Question :-

The numerator of a rational number is 5 less than its denominator. If 2 is subtracted from the numerator and 2 is added to the denominator, the number becomes 2/5. Find the rational number.

____________________________________

★ Solution :-

Given,

» The denominator of rational number = 5 + The numerator of that rational number

» If 2 is subtracted from the numerator and 2 is added to the denominator, the number becomes 2/5.

• Let the numerator be 'x' and the denominator be 'y' of the rational number.

Then using these, let us make our equations and solve further.

~ Case I :-

➥ y = x + 5 ... (i)

~ Case II :-

$\: \: : \: \Longrightarrow \rm{(\dfrac{x \: - \: 2}{y \: + \: 2}) \: = \: (\dfrac{2}{5})}$

By cross - multiplication, we get,

➥ 5(x - 2) = 2(y + 2)

➥ 5x - 10 = 2y + 4

➥ 5x - 2y = 4 + 10

➥ 5x - 2y = 14 ....(ii)

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

⟹ 5x - 2(x + 5) = 14

⟹ 5x - 2x - 10 = 14

⟹ 3x = 14 + 10

⟹ 3x = 24

$\: \: \huge{\Longrightarrow \: \: \rm{x \: = \: \dfrac{24}{3}}}$

⟹ x = 8

• Hence, we get, the numerator of the original fraction is = x = 8

Now using the value of x and equation (i), we get,

⟹ y = x + 5

⟹ y = 8 + 5

⟹ y = 13

• Hence, we get, the denominator of the original fraction is = y = 13

$\: \: ✰ \: \: \underline{\boxed{\sf{\green{Thus,\: the \: original \: fraction \: is \: \dfrac{8}{13}}}}}$

______________________________

$\: \: ✰ \: \: \boxed{\rm{\red{Confused?, \: Don't \: worry \: let's \: verify \: it }}}$

For verification, we need to simply apply the values we got, into the equations we formed.

~ Case I :-

=> y = x + 5

=> 13 = 8 + 5

=> 13 = 13

Clearly, LHS = RHS

~ Case II :-

=> 5x - 2y = 14

=> 5(8) - 2(13) = 14

=> 40 - 26 = 14

=> 14 = 14

Clearly, LHS = RHS

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

Hence, verified.

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$\: \: ✰ \: \: \boxed{\sf{\orange{A \: piece \: of \: Supplementary \: Counsel}}}$

• Linear Equations are the equations formed using constant and variable terms. These are a type of Linear Polynomials with highest degree of 1.

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

• Linear Equations in Two Variables have two variables terms which are solved simultaneously.

*Note here I have used two variables for easier procedure. You can also use single variable term, using the first case. For single variable answer, see first ↑ answer. This also gives quick answer with better response. Finally the answer should be correct, whichever method you use.