Question

The set of admissible values of x such that $\dfrac{{2x + 3}}{{2x - 9}} < 0$ is:
a) \[\left( { - \infty , - \dfrac{3}{2}} \right) \cup \left( {\dfrac{9}{2},\infty } \right)\]
b) \[\left( { - \infty ,0} \right) \cup \left( {\dfrac{9}{2},\infty } \right)\]
c) \[\left( { - \dfrac{3}{2},0} \right)\]
d) \[\left( {0,\dfrac{9}{2}} \right)\]
e) \[\left( { - \dfrac{3}{2},\dfrac{9}{2}} \right)\]

Answer 1

Hint: We have to solve the inequality given to us in the problem. So, we make use of the algebraic and simplification rules in order to simplify the inequality and find a solution set to the inequality. Multiplying the sides of an inequality by a positive number does not change the sign of inequality.

Formula used: If a fractional expression $\dfrac{(x-a)}{(x-b)}$ is negative, the solution set of the inequality is (a,b).

Complete step by step answer:
In the given question, we are given an inequality $\dfrac{{2x + 3}}{{2x - 9}} < 0$.
Now, for the whole rational expression to be negative, either the numerator will be positive and the denominator will be negative, or the numerator will be negative and the denominator will be positive. 
So, either $2x + 3 < 0$ and $2x - 9 > 0$
$ \Rightarrow 2x < - 3$ and $ \Rightarrow 2x > 9$
Dividing both sides by two, we get,
$ \Rightarrow x < - \dfrac{3}{2}$ and $ \Rightarrow x > \dfrac{9}{2}$
Since two is a positive number, we didn’t change the sign of inequality.
So, the solution set for the above two inequalities are the values of x greater than $\dfrac{9}{2}$ and less than $ - \dfrac{3}{2}$. No such real value of x exists. So, there is no solution for the above inequalities.
Or, $2x + 3 > 0$ and $2x - 9 < 0$
$ \Rightarrow 2x > - 3$ and $ \Rightarrow 2x < 9$
Since two is a positive number, we didn’t change the sign of inequality.
$ \Rightarrow x > - \dfrac{3}{2}$ and $ \Rightarrow x < \dfrac{9}{2}$
So, the solution set of the above inequalities are the values of x greater than $ - \dfrac{3}{2}$ and less than $\dfrac{9}{2}$.
So, $x \in \left( { - \dfrac{3}{2},\dfrac{9}{2}} \right)$.
Hence, taking union of solutions of both the inequalities, we get,
$x \in \left( { - \dfrac{3}{2},\dfrac{9}{2}} \right)$

Hence, the correct answer is option (e) \[\left( { - \dfrac{3}{2},\dfrac{9}{2}} \right)\].

Note: One must know that when multiplying both sides of the inequality by a positive integer, the sign of inequality doesn't change. The inequality sign changes when multiplying or dividing both sides of the inequality by a negative number. We can shift the terms from one side of inequality to another using the transposition method.