Question

what must be added in 7/3 to get 6/8 ​

Answer 1

$\underline{\underline{\bf{\bigstar\: Solution : }}}$

$\:\:$

${\footnotesize{ Let, \:the\:no.\: added\:be\:x}}$

$\:\:$

$\red{\footnotesize{ Sum\: of\:no. = \dfrac{6}{8}}}$

${\footnotesize{\implies \dfrac{7}{3} + x = \dfrac{6}{8}}}$

${\footnotesize{\implies x = \dfrac{6}{8} - \dfrac{7}{3} }}$

${\footnotesize{\implies x = \dfrac{18 - 56}{24}}}$

${\footnotesize{\implies x = \dfrac{-38}{24}}}$

$\green{\footnotesize{\implies x = \dfrac{-19}{12}}}$

$\:\:$

$\underline{\underline{\bf{\bigstar\: Answer : \dfrac{-19}{12} }}}$

Answer 2

$\huge\mathfrak{Answer:}$

Given:

• We have been given two rational numbers, 7/3 and 6/8.

To Find:

• We need to find a number that must be added to 7/3 to get 6/8.

Solution:

Let the number that must be added to 7/3 to get 6/8 be x.

By Question

$\sf{ \dfrac{7}{3} + x = \dfrac{6}{8} }$

$\implies\sf{x = \dfrac{6}{8} - \dfrac{7}{3} }$

$\implies\sf{ x = \dfrac{6 \times 3 - 7 \times 8}{24} }$

$\implies\sf{x = \dfrac{18 - 56}{24} }$

$\implies\sf{x = \dfrac{ - 38}{24}}$

$\implies\sf{x = \dfrac{ - 19}{12} }$

Hence, the required number is -19/12.

Verification:

$\implies\sf{ \dfrac{7}{3} + ( \dfrac{ - 19}{12} )}$

$\implies\sf{ \dfrac{7}{3} - \dfrac{19}{12} }$

$\implies\sf{ \dfrac{(7 \times12) - (3 \times 19)}{3 \times 12} }$

$\implies\sf{ \dfrac{84 - 57}{36} }$

$\implies\sf{ \dfrac{27}{36} }$

$\implies\sf{ \dfrac{3}{4} }$

$\implies\sf{\dfrac{3}{4} = \dfrac{6}{8}}$

$\implies\sf{\dfrac{3}{4} = \dfrac{3}{4}}$

Hence verified!!

Hence, -19/12 is the required number.