Question

How do you solve the inequality $\dfrac{{\left( {2x + 1} \right)}}{{\left( {x - 5} \right)}} \leqslant 0$ ?

Answer 1

Hint: In order to solve the given inequality, we first put the denominator as less than zero. We can never put the denominator as equal to zero as doing so will result in the fraction being undefined. Then, we solve our inequality to get the first value of $x$, after which we multiply both sides of our given inequality with the denominator and change the signs of the inequality as the condition is inherently negative. We solve further to get our required answer.

Complete step-by-step solution:
The given inequality is $\dfrac{{\left( {2x + 1} \right)}}{{\left( {x - 5} \right)}} \leqslant 0$.
Now according to the given condition, the given inequality is less than equal to zero. This means that the entire fraction has to be negative.
According to the given fraction the denominator needs to be less than zero, but we should keep in mind that the denominator can never be equal to zero, as then our fraction will become unidentified, since any number divided by zero is undefined.
Therefore, $x - 5 \ne 0$
Let try to solve the sum by taking $x - 5 < 0$
On adding $ + 5$ to both the sides, we get:
$ \Rightarrow x < 5$
Now let’s multiply both sides of our original inequality with $\left( {x - 5} \right)$
Thus we get, $\left( {x - 5} \right)\dfrac{{\left( {2x + 1} \right)}}{{\left( {x - 5} \right)}} \leqslant 0 \times \left( {x - 5} \right)$
Since we have used the condition that our denominator is less than zero i.e $x - 5 < 0$, therefore it is inherently negative.
Since we are multiplying our inequality with a negative number, therefore we change the inequality sign to opposite of what it originally was.
Thus, our new inequality is: $\left( {x - 5} \right)\dfrac{{\left( {2x + 1} \right)}}{{\left( {x - 5} \right)}} \geqslant 0 \times \left( {x - 5} \right)$
$ \Rightarrow \not{{\left( {x - 5} \right)}}\dfrac{{\left( {2x + 1} \right)}}{{\not{{\left( {x - 5} \right)}}}} \geqslant 0 \times \left( {x - 5} \right)$
On rewriting we get
$ \Rightarrow \left( {2x + 1} \right) \geqslant 0$
On adding $ - 1$ to both sides, we get:
$ \Rightarrow 2x \geqslant - 1$
On dividing both sides with $2$, we get:
$ \Rightarrow x \geqslant \dfrac{{ - 1}}{2}$
Thus we have two conditions now: $x < 5$ and $x \geqslant \dfrac{{ - 1}}{2}$

Both conditions satisfy each other if we only choose $x < 5$ even.

Note: In mathematics, inequality is simply a relation which makes a comparison between two unequal numbers. A rational inequality is simply an inequality containing rational expressions and having zeroed on one side. After solving inequalities, we can simply express them on number lines. Inequalities are expressed using the greater than’ <’ or less than ’ > ‘signs.