Question

What should be added to twice the rational number $\frac{-7}{3}$ to get $\frac{3}{7}$?

Answer 1

Answer:

Step-by-step explanation:

$\frac{-7}{3} +2x =\frac{3}{7} \\2x=\frac{+49+9}{21} \\x=\frac{58}{21*2} \\x=\frac{29}{21}$

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Answer 2

QuestioN :

What should be added to twice the rational number $\sf \frac{-7}{3}$ to get $\sf \frac{3}{7}$?

GiveN :

• $\sf Rational \:number= \dfrac{-7}{3} \: to\:get\dfrac{3}{7}$

To FiNd :

• What should be added?

ANswer :

$\sf \dfrac{107}{21}\: should\:be\:added.$

SolutioN :

$\sf Twice\:the\: rational\: number\:\dfrac{-7}{3} \:is \:2 \times \left[\begin{array}{ccc}\dfrac{-7}{3}\end{array}\right] = \dfrac{-14}{3}.$

$\sf Suppose \:x\:added\:to\:this\:number\:gives\: \dfrac37;\:i.e.,$

$\sf \implies x+\left[\begin{array}{ccc}\dfrac{-14}{3}\end{array}\right]= \dfrac{3}{7}$

or,

$\sf \implies x- \dfrac{14}{3}= \dfrac{3}{7}$

or,

$\sf \implies x= \dfrac{3}{7}+\dfrac{14}{3}$

             $\sf (transposting \: \dfrac{14}{3}\:to \:RHS)$

$\sf \implies \dfrac{(3\times3)+(14\times7)}{21}=\dfrac{9+98}{21}=\dfrac{107}{21}.$

Hence, $\dfrac{107}{21}$ Should be added.

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