Question

Solve the following inequality: $\dfrac{{\left| {x - 1} \right|}}{{x + 2}} < 1$

Answer 1

In the question, we are given an inequation. A statement involving variable(s) and the sign of inequalities is called an inequation or inequality. To solve the given question, we will first shift $1$ to the left-hand side of an inequality, and take LCM on the left-hand side of the equation given in the question. As we can see $x - 1$ is inside the modulus. Therefore, here two cases arise: $\left( {x - 1} \right)$ and $- \left( {x - 1} \right)$. We will solve these two cases separately and find the solution set of them. The interval which is common in both will be our solution set for the given inequality.

Complete step-by-step solution:
We have, $\dfrac{{\left| {x - 1} \right|}}{{x + 2}} < 1$
Shift $1$ to left-hand side
$\Rightarrow \dfrac{{\left| {x - 1} \right|}}{{x + 2}} - 1 < 0$
We will take LCM on the left-hand side of the above inequation. Thus, we will get the following inequation.
$\Rightarrow \dfrac{{\left| {x - 1} \right| - \left( {x + 2} \right)}}{{x + 2}} < 0$
Now the following cases arise.
Case 1 When $x - 1 \geqslant 0$ i.e. $x \geqslant 1$
In this case we have $\left| {x - 1} \right| = x - 1$
$\therefore \dfrac{{\left| {x - 1} \right| - \left( {x + 2} \right)}}{{x + 2}} < 0$
Substitute $\left| {x - 1} \right| = x - 1$
$\Rightarrow \dfrac{{\left( {x - 1} \right) - \left( {x + 2} \right)}}{{x + 2}} < 0$
$\Rightarrow \dfrac{{x - 1 - x - 2}}{{x + 2}} < 0$
On addition and subtraction of like terms, we get
$\Rightarrow \dfrac{{ - 3}}{{x + 2}} < 0$
$\Rightarrow x + 2 > 0$ $\left[ {\because \dfrac{a}{b}\text{ and } a < 0 \Rightarrow b > 0} \right]$
$\Rightarrow x > - 2$
But, $x \geqslant 1$. Therefore, $x \geqslant 1$.
Thus, in this case the solution set of the given inequation is $\left[ {1,\infty } \right)......\left( i \right)$
Case 2 When $x - 1 < 0$ i.e. $x < 1$
In this case we have $\left| {x - 1} \right| = - \left( {x - 1} \right)$
$\therefore \dfrac{{\left| {x - 1} \right| - \left( {x + 2} \right)}}{{x + 2}} < 0$
Substitute $\left| {x - 1} \right| = - \left( {x - 1} \right)$
$\Rightarrow \dfrac{{ - \left( {x - 1} \right) - \left( {x + 2} \right)}}{{x + 2}} < 0$
$\Rightarrow \dfrac{{ - x + 1 - x - 2}}{{x + 2}} < 0$
On addition and subtraction of like terms, we get
$\Rightarrow \dfrac{{ - 2x - 1}}{{x + 2}} < 0$
It can also be written as
$\Rightarrow - \dfrac{{2x + 1}}{{x + 2}} < 0$
On multiplying $- 1$ on both sides, we get
$\Rightarrow \dfrac{{2x + 1}}{{x + 2}} > 0$
But, $x < 1$. Therefore,
$x \in \left( { - \infty , - 2} \right) \cup \left( \dfrac{-1}{2}, 1 \right)$
Thus, in this case the solution set of the given inequation is $\left( { - \infty , - 2} \right) \cup \left( \dfrac{-1}{2}, 1 \right) - - - - \left( {ii} \right)$
Combining $\left( i \right)$ and $\left( {ii} \right)$, we obtain that the solution set of the given inequation is $\left( { - \infty , - 2} \right) \cup \left( \dfrac{-1}{2}, 1 \right)$
The solution set can be represented on number line as follows:

Note: Addition or subtraction of the same number to both sides of an inequation doesn’t affect the sign of inequality. Both sides of an inequation can be multiplied or divided by the same positive real number without changing the sign of inequality but the sign of inequality is reversed when both sides of an inequation are multiplied or divided by a negative number. Any term of an inequation can be taken to the other side with its sign changed without affecting the sign of inequality.