Question

How do you solve $\left| \dfrac{5}{2x-1} \right|\ge \left| \dfrac{1}{x-2} \right|$?

Answer 1

We express the whole domain into three parts for the equation $\left| \dfrac{5}{2x-1} \right|\ge \left| \dfrac{1}{x-2} \right|$. We break it according to the denominators of the equations. We then find the solutions for the equation and check if the intervals satisfy or not. Modulus function $f\left( x \right)=\left| x \right|$ works as the distance of the number from 0. The number can be both positive and negative but the distance of that number will always be positive. Distance can never be negative.

Complete step-by-step solution:
We try to break the whole domain into three parts for the equation $\left| \dfrac{5}{2x-1} \right|\ge \left| \dfrac{1}{x-2} \right|$.
The divisions are $x\in \left( -\infty ,-\dfrac{1}{2} \right)\cup \left[ -\dfrac{1}{2},2 \right)\cup \left[ 2,\infty \right)$.
We found these breaking points based on the denominators of the equations $\left| \dfrac{5}{2x-1} \right|\ge \left| \dfrac{1}{x-2} \right|$
Therefore, for $x\in \left( -\infty ,-\dfrac{1}{2} \right)$, the values of both the denominators are negative which gives $\left| \dfrac{5}{2x-1} \right|=\dfrac{-5}{2x-1},\left| \dfrac{1}{x-2} \right|=\dfrac{-1}{x-2}$.
The inequation becomes $\dfrac{-5}{2x-1}\ge \dfrac{-1}{x-2}$. Simplifying we get

& \dfrac{-5}{2x-1}\ge \dfrac{-1}{x-2} \\\ & \Rightarrow 5x-10\ge 2x-1 \\\ & \Rightarrow 3x\ge 9 \\\ & \Rightarrow x\ge 3 \\\ \end{aligned}$$ The conditions $x\in \left( -\infty ,-\dfrac{1}{2} \right)$ and $$x\ge 3$$ can’t happen together. We don’t have any solution in the interval $x\in \left( -\infty ,-\dfrac{1}{2} \right)$. Therefore, for $x\in \left[ -\dfrac{1}{2},2 \right)$, the values of the denominators are positive and negative respectively which gives $$\left| \dfrac{5}{2x-1} \right|=\dfrac{5}{2x-1},\left| \dfrac{1}{x-2} \right|=\dfrac{-1}{x-2}$$. The inequation becomes $$\dfrac{5}{2x-1}\ge \dfrac{-1}{x-2}$$. Simplifying we get $$\begin{aligned} & \dfrac{5}{2x-1}\ge \dfrac{-1}{x-2} \\\ & \Rightarrow 5x-10\le 1-2x \\\ & \Rightarrow 7x\le 11 \\\ & \Rightarrow x\le \dfrac{11}{7} \\\ \end{aligned}$$ The conditions $x\in \left[ -\dfrac{1}{2},2 \right)$ and $$x\le \dfrac{11}{7}$$ can happen together. We have solutions in the interval $x\in \left[ -\dfrac{1}{2},\dfrac{11}{7} \right]$. Therefore, for $x\in \left[ 2,\infty \right)$, the values of both the denominators are negative which gives $$\left| \dfrac{5}{2x-1} \right|=\dfrac{5}{2x-1},\left| \dfrac{1}{x-2} \right|=\dfrac{1}{x-2}$$. The inequation becomes $$\dfrac{5}{2x-1}\ge \dfrac{1}{x-2}$$. Simplifying we get $$\begin{aligned} & \dfrac{5}{2x-1}\ge \dfrac{1}{x-2} \\\ & \Rightarrow 5x-10\ge 2x-1 \\\ & \Rightarrow 3x\ge 9 \\\ & \Rightarrow x\ge 3 \\\ \end{aligned}$$ **The conditions $x\in \left[ 2,\infty \right)$ and $$x\ge 3$$ can happen together. We have solutions in the interval $x\in \left[ 3,\infty \right)$. Therefore, the solution is $x\in \left[ -\dfrac{1}{2},\dfrac{11}{7} \right]\cup \left[ 3,\infty \right)$.** **Note:** In mathematical notation we express it with modulus value. Let a number be x whose sign is not mentioned. The absolute value of that number will be $\left| x \right|$. We can say $\left| x \right|\ge 0$. We can express the function $f\left( x \right)=\left| x \right|$ as $f\left( x \right)=\left\\{ \begin{aligned} & x\left( x\ge 0 \right) \\\ & -x\left( x<0 \right) \\\ \end{aligned} \right.$. We can write $f\left( x \right)=\left| x \right|=\pm x$ depending on the value of the number x.