Question

What should be added to twice the rational number-7 apon 3 and 3 apon 7​

Answer 1

$\huge{\underline{\sf{\color{red}{Answer:-}}}}$

Let the number added be x,

★ According to question :—

= $\sf{\dfrac{-7}{3} + \dfrac{3}{7} + x = 2(\dfrac{-7}{3} + \dfrac{3}{7} ) }$

= $\sf{\dfrac{-49+9}{21}+ x = 2(\dfrac{-49+9}{21} ) }$

= $\sf{\dfrac{-40}{21}+ x = 2(\dfrac{-40}{21} ) }$

= $\sf{\dfrac{-40}{21}+ x =(\dfrac{-80}{21} ) }$

= $\sf{x = \dfrac{-80}{21} -(\dfrac{-40}{21} ) }$

= $\sf{x = \dfrac{-80+40}{21} }$

= $\sf{x = \dfrac{-40}{21}}$

Thus , 0 should be added.

_____________________________

Answer 2

ANSWER✔

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

$\sf\therefore let\:the\:number\:be\:x$

$\sf\dashrightarrow the \:rational\:number\:are= \dfrac{-7}{3} \:and\: \dfrac{3}{7}$

$\sf\dashrightarrow twice\:the\:rational\:number\:added.\:then,$

$\sf\therefore 2 \times \bigg( \dfrac{-7}{3} + \dfrac{3}{7} \bigg)$

$\large\underline\bold{SOLUTION,}$

✯ACCORDING TO THE QUESTION

THE EQUATION FORMED IS,

$\sf\therefore \dfrac{-7}{3} + \dfrac{3}{7} +x = 2 \times \bigg( \dfrac{-7}{3} + \dfrac{3}{7} \bigg)$

$\sf\implies \dfrac{(-49)+ (9)}{21} +x = 2 \times \bigg( \dfrac{(-49)+ (9)}{21} \bigg)$

$\sf\implies \dfrac{-40}{21}+x= 2 \times \bigg( \dfrac{-40}{21} \bigg)$

$\sf\implies \dfrac{-40}{21}+x= \dfrac{-80}{21}$

$\sf\implies x= \dfrac{-80}{21} -\dfrac{-40}{21}$

$\sf\implies x= \dfrac{(-80)- (-40)}{21}$

$\sf\implies x= \dfrac{(-80)+40)}{21}$

$\sf\implies x= \dfrac{(-40)}{21}$

$\large{\boxed{\bf{ \star\:\: x= \dfrac{(-40)}{21}\:\: \star}}}$

$\large\underline\bold{THE\:NUMBER\:OBTAINED\:IS\:\: \dfrac{-40}{21}}$

______________