Question

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

Answer 1

Hint: Assume that the number that should be added is x. Now, first find the value of twice the rational number $\dfrac{-7}{3}$ by multiplying the numerator with 2. Now, take the sum of the obtained rational number with x and equate it with $\dfrac{3}{7}$ to form a linear equation in x. Solve this equation for the value of x to get the answer.

Complete step by step solution:
Here we have been asked to find the number that must be added to twice the rational number $\dfrac{-7}{3}$ to get $\dfrac{3}{7}$ as the resultant fraction.
Now, let us assume that the number that should be added is x. The meaning of twice of $\dfrac{-7}{3}$ means we need to multiply the numerator of this rational number with 2, so we get,
$\Rightarrow 2\times \left( \dfrac{-7}{3} \right)=\dfrac{-14}{3}$
So we need to add $\dfrac{-14}{3}$ with x to get $\dfrac{3}{7}$, therefore the expression will be given as:
$\Rightarrow x+\left( \dfrac{-14}{3} \right)=\dfrac{3}{7}$
Clearly this is a linear equation in x so we need to solve for the value of x. Leaving x in the L.H.S and taking all other terms to the R.H.S we get,
$\Rightarrow x=\dfrac{3}{7}-\left( \dfrac{-14}{3} \right)$
We know that $\left( -1 \right)\times \left( -1 \right)=1$ so we get,
$\Rightarrow x=\dfrac{3}{7}+\dfrac{14}{3}$
Taking the L.C.M of 7 and 3 which is 21 we get,
$\begin{align}
  & \Rightarrow x=\dfrac{\left( 3\times 3 \right)+\left( 14\times 7 \right)}{21} \\
 & \Rightarrow x=\dfrac{9+98}{21} \\
 & \therefore x=\dfrac{107}{21} \\
\end{align}$
Hence the required number that must be added is $\dfrac{107}{21}$.

Note: You must remember how to add and subtract two rational numbers. To find the value of twice the rational number always multiply the numerator with 2 and not the denominator. Note that to add fractions we take the L.C.M of their denominators and not numerators. Here both 3 and 7 were prime numbers and we know that the L.C.M of two prime numbers is simply their product.