# 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

Verified

362.7k+ views

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).

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).

Last updated date: 25th Sep 2023

•

Total views: 362.7k

•

Views today: 4.62k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is the IUPAC name of CH3CH CH COOH A 2Butenoic class 11 chemistry CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The dimensions of potential gradient are A MLT 3A 1 class 11 physics CBSE

Write the equation involved in the following reactions class 11 chemistry CBSE

What is the nature of the Gaussian surface involved class 11 physics CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the need for measurement of a physics quan class 11 physics CBSE

State Gay Lusaaccs law of gaseous volume class 11 chemistry CBSE