Question

Find the range of values of x for which $\dfrac{{x - 3}}{4} - x < \dfrac{{x - 1}}{2} - \dfrac{{x - 2}}{3}$ and $2 - x > 2x - 8$.
\[
  {\text{A}}{\text{. }}\left( { - 1,\dfrac{{10}}{3}} \right) \\
  {\text{B}}{\text{. }}\left( {1,\dfrac{{10}}{3}} \right) \\
  {\text{C}}{\text{. }}R \\
 \]
\[{\text{D}}{\text{.}}\] None of these

Answer 1

Hint- Here, we will be proceeding by simplifying the given inequalities (in variable x) in such a way that we get the range of values of x from each of these inequalities and then combining these values of x (i.e., by taking intersection) in order to find the final range of values of x.

Given, first inequality is $\dfrac{{x - 3}}{4} - x < \dfrac{{x - 1}}{2} - \dfrac{{x - 2}}{3}{\text{ }} \to {\text{(1)}}$
This inequality can be simplified as under
$
   \Rightarrow \dfrac{{x - 3 - 4x}}{4} < \dfrac{{3\left( {x - 1} \right) - 2\left( {x - 2} \right)}}{6} \\
   \Rightarrow \dfrac{{ - 3 - 3x}}{4} < \dfrac{{3x - 3 - 2x + 4}}{6} \\
   \Rightarrow \dfrac{{ - 3 - 3x}}{4} < \dfrac{{x + 1}}{6}{\text{ }} \to {\text{(2)}} \\
 $
By applying cross multiplication in inequality (2), we get
\[
   \Rightarrow 6\left( { - 3 - 3x} \right) < 4\left( {x + 1} \right) \\
   \Rightarrow - 18 - 18x < 4x + 4 \\
   \Rightarrow - 18 - 4 < 4x + 18x \\
   \Rightarrow - 22 < 22x \\
   \Rightarrow \dfrac{{ - 22}}{{22}} < x \\
   \Rightarrow - 1 < x \\
   \Rightarrow x > - 1{\text{ }} \to {\text{(3)}} \\
\]
So, after simplification inequality (1) reduces to inequality (3).
Also, given the second inequality as $2 - x > 2x - 8{\text{ }} \to {\text{(4)}}$
This inequality can be simplified as under
$
   \Rightarrow 2 + 8 > 2x + x \\
   \Rightarrow 10 > 3x \\
   \Rightarrow \dfrac{{10}}{3} > x \\
   \Rightarrow x < \dfrac{{10}}{3}{\text{ }} \to {\text{(5)}} \\
 $
So, after simplification inequality (4) reduces to inequality (5).
The final range of values of x which satisfy both the given inequalities is obtained by taking the intersection between the inequalities (3) and (5).
Intersection of \[x > - 1\] and $x < \dfrac{{10}}{3}{\text{ }}$gives $ - 1 < x < \dfrac{{10}}{3}{\text{ }}$ which means required value of x lies between -1 and $\dfrac{{10}}{3}$.
i.e.,$x \in \left( { - 1,\dfrac{{10}}{3}} \right)$
Hence, option A is correct.

Note- In this particular problem, we obtained the final range of values of x by simply taking the intersection of the range of values of x given by inequalities (3) and (5) because for the final range of values of x both the given inequalities should satisfy. So, we will take the range of values of x which are common to both the range of values of x given by inequalities (3) and (5).