Question

How do you simplify the given division $\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}$?

Answer 1

Hint: We start solving the problem by equating the given division to a variable. We then multiply the numerator and denominator of the division with x to proceed through the problem. We then multiply the numerator and denominator of the obtained result with 3 to proceed further through the problem. We then make the necessary calculations to get the required simplified form of the given division.

Complete step-by-step solution:
According to the problem, we are asked to simplify the given term $\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}$.
Let us assume $d=\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}$ ---(1).
Let us multiply the numerator and denominator of equation (1) with x.
$\Rightarrow d=\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}\times \dfrac{x}{x}$.
$\Rightarrow d=\dfrac{\left( \dfrac{{{x}^{2}}}{3} \right)-6x}{10x+4}$ ---(2).
Let us multiply the numerator and denominator of equation (2) with 3.
$\Rightarrow d=\dfrac{\left( \dfrac{{{x}^{2}}}{3} \right)-6x}{10x+4}\times \dfrac{3}{3}$.
$\Rightarrow d=\dfrac{{{x}^{2}}-18x}{30x+12}$.
So, we have found the result of the division $\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}$ as $\dfrac{{{x}^{2}}-18x}{30x+12}$.
$\therefore $ The simplified form of the term $\dfrac{\left( \dfrac{x}{3} \right)-6}{10+\left( \dfrac{4}{x} \right)}$ is $\dfrac{{{x}^{2}}-18x}{30x+12}$.

Note: Here we have assumed that the value of x given in the division is not equal to ‘0’ and $\dfrac{-2}{5}$ as the given function is not valid at those values of x. We can also verify the obtained result by checking the result at the value of x other than ‘0’ and $\dfrac{-2}{5}$. We should not make mistakes while performing the multiplication operations in this problem. Similarly, we can expect problems to find the limit of the function when the value of x approaches or tends to 0 and $\dfrac{-2}{5}$.