Question

How do you solve the expression $\dfrac{3}{\left| 2x-1 \right|}\ge 4$ ?

Answer 1

We are given an absolute value inequality in this problem. We shall first rearrange the given equation to a form where the absolute value function is not in the denominator and has some constant term in front of its inequality. Then, we will open the modulus function with both its positive as well as negative values one by one to get our required interval value of x-variable.

Complete step by step solution:
This is a linear inequality in one variable. In order to solve this, we must have prior knowledge about solving the absolute value inequalities. The term $2x-1$ has been put under the modulus sign. This implies that this term must be checked for both its negative and positive values.
We shall first rearrange the given equation, $\dfrac{3}{\left| 2x-1 \right|}\ge 4$. Dividing both the sides by 4 and then taking $\left| 2x-1 \right|$ to the right-hand side, we get
$\begin{aligned} & \Rightarrow \dfrac{3}{\left| 2x-1 \right|}\times \dfrac{1}{4}\ge \dfrac{4}{4} \\\ & \Rightarrow \dfrac{1}{\left| 2x-1 \right|}\times \dfrac{3}{4}\ge 1 \\\ \end{aligned}$
$\Rightarrow \dfrac{3}{4}\ge \left| 2x-1 \right|$
Now, we will first open the modulus function with a positive value, that is, we will substitute $\left| 2x-1 \right|=2x-1$ in the equation.
$\Rightarrow \dfrac{3}{4}\ge 2x-1$
Multiplying both sides by 4, we get
$\begin{aligned} & \Rightarrow \dfrac{3}{4}\times 4\ge \left( 2x-1 \right)\times 4 \\\ & \Rightarrow 3\ge 8x-4 \\\ \end{aligned}$
We shall transpose -4 to the left-hand side of inequality.
$\begin{aligned} & \Rightarrow 3+4\ge 8x \\\ & \Rightarrow 7\ge 8x \\\ \end{aligned}$
Finally dividing both sides by 8, we get
$\Rightarrow \dfrac{7}{8}\ge \dfrac{8x}{8}$
$\Rightarrow \dfrac{7}{8}\ge x$ …………… equation (1)
Similarly, we will first open the modulus function with a negative sign, that is, we will substitute $\left| 2x-1 \right|=-\left( 2x-1 \right)$ in the equation. Also, we will reverse the sign of inequality.
$\begin{aligned} & \Rightarrow \dfrac{3}{4}\le -\left( 2x-1 \right) \\\ & \Rightarrow \dfrac{3}{4}\le -2x+1 \\\ \end{aligned}$
Multiplying both sides by 4, we get
$\begin{aligned} & \Rightarrow \dfrac{3}{4}\times 4\le \left( -2x+1 \right)\times 4 \\\ & \Rightarrow 3\le -8x+4 \\\ \end{aligned}$
We shall transpose 4 to the left-hand side of inequality.
$\begin{aligned} & \Rightarrow 3-4\le -8x \\\ & \Rightarrow -1\le -8x \\\ \end{aligned}$
Finally dividing both sides by -8 and changing the inequality, we get
$\Rightarrow -\dfrac{1}{-8}\ge -\dfrac{8x}{-8}$
$\Rightarrow \dfrac{1}{8}\ge x$ …………… equation (2)
From equation (1) and (2), we get
$x\le \dfrac{1}{8}$
Therefore, for the function, $\dfrac{3}{\left| 2x-1 \right|}\ge 4$, x lies in the interval $\left( -\infty ,\dfrac{1}{8} \right]$.

Note: Earlier in the question, we reversed the sign of the inequality because a negative sign was being multiplied to the right-hand side. The value of this term completely changed. This is because the function $2x-1$ which was some units on the right of zero on the number line went the same number of units to the left of zero on the number line after opening it with a negative sign.