Question

the denominator of a rational number is less than its ne numerator than 5 if the five is added to the numerator the new number become 11 by 6 find the original rational number​

Answer 1

Answer:

I assume you mean your question to read this way:

The denominator of a rational number is less than the numerator by 5. If 5 is added to the numerator, the new number becomes 11/6. Find the rational number. Note: your English grammar needs as much help as your math!

D = N - 5

(N + 5)/D = 11/6

(N + 5)/(N - 5) = 11/6

6N + 30 = 11N - 55

5N = 85

N = 17

D = 12

The fraction of rational number is 17/12

Answer 2

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

$\large\underline\red{GIVEN,}$

$\sf\dashrightarrow \blue{THE\:GIVEN\:FRACTION\:IS\:A\:RATIONAL\:NUMBER.}$

$\sf\dashrightarrow {\blue{\mathbb{\text{ denominator is less than its numerator by 5.}}}}$

$\sf\therefore \blue{let\:the\:numerator\:be\:x}$

$\sf\dashrightarrow \blue{denominator= x-5}$

$\sf\dashrightarrow \blue{\dfrac{x}{x-5}}$

$\sf\dashrightarrow \bold\pink{if \:5 \:is \:added\: to \:the \:numerator, \:numerator \:becomes\: ; \dfrac{11}{6}}$

$\sf\dashrightarrow \red{numerator= x+5}$

THE EQUATION FORM IS,

$\rm{\boxed{\sf{\green{ \circ\:\: \dfrac{x+5}{x-5} = \dfrac{11}{6}\:\: \circ}}}}$

$\large\underline\purple{TO\:FIND,}$

$\sf\dashrightarrow \red{\:THE\:ORIGINAL\:RATIONAL\:NUMBER}$

$\sf\implies \green{\dfrac{x + 5}{x - 5} = \dfrac{11}{6}}$

$\sf\implies \green{6 \times (x+5)= 11 \times (x-5)}$

$\sf\implies \green{6x+30= 11x-55}$

$\sf\implies \green{30+55=11x-6x}$

$\sf\implies \green{85= 5x}$

$\sf\implies \green{x= \dfrac{85}{5}}$

$\sf\implies \green{x= \cancel \dfrac{85}{5}}$

$\sf\implies \orange{x = 17}$

$\rm{\boxed{\sf{ \circ\:\: x= 17 \:\: \circ}}}$

THE ORIGINAL NUMBERS ARE,

$\sf\implies \red{numerator=x =17}$

$\sf\implies \red{d enominator= x-5}$

$\sf\implies \red{denominator=17-5}$

$\sf\implies \pink{denominator=12}$

$\large\underline\orange{FRACTION,}$

$\sf\dashrightarrow \purple{\dfrac{x}{x-5}= \dfrac{17}{12}}$

$\sf\dashrightarrow \purple{\dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12}}$

$\rm\underline\blue{NUMERATOR\:IS\:17\:AND\: DENOMINATOR\:IS\:12}$

$\rm{\boxed{\sf{ \circ\:\: \dfrac{NUMERATOR}{DENOMINATOR}= \dfrac{17}{12} \:\: \circ}}}$