Absolute Value Inequality!

Question

Solve the inequality $$\left| \frac{2x - 1}{x - 1} \right| > 2.$$

I'm really confused on how to solve problems such as this. Logically I think you could take $\frac{2x-1}{x-1}>2$ and $\frac{2x-1}{x-1}<-2$. The first one gives $x \in (-\infty,\infty)$ and the second gives $x \in (-\infty,\frac34)$. However, does this mean that the answer is $x \in (-\infty,\frac34)$, because it is the stricter bound? Thanks in advance.

Answer 1

Since both sides are nonnegative, we can square them in order to obtain an equivalent inequation:

\begin{align*} \left|\frac{2x - 1}{x - 1}\right| > 2 & \Longleftrightarrow \left(\frac{2x-1}{x-1}\right)^{2} > 4\\\\ & \Longleftrightarrow \frac{4x^{2} - 4x + 1}{x^{2} - 2x + 1} - 4 > 0\\\\ & \Longleftrightarrow \frac{4x - 3}{x^{2} - 2x + 1} > 0\\\\ & \Longleftrightarrow \begin{cases} 4x - 3 > 0\\\\ x\neq 1 \end{cases} \end{align*}

Can you take it from here?

Answer 2

Your approach does not work. It is false that $\require{cancel}\left|\frac{2x-1}{x-1}\right|$ is both $\frac{2x-1}{x-1}$ and $-\frac{2x-1}{x-1}$; it is $\frac{2x-1}{x-1}$ when $\frac{2x-1}{x-1}\geqslant0$ and $-\frac{2x-1}{x-1}$ otherwise. Now, note that$$\frac{2x-1}{x-1}\geqslant 0\iff x\in\left(-\infty,\frac12\right]\cup(1,\infty).$$You can also solve the the problem using the fact that\begin{align}\left|\frac{2x-1}{x-1}\right|>2&\iff\left(\frac{2x-1}{x-1}\right)^2>4\\&\iff\cancel{4x^2}-4x+1\geqslant\cancel{4x^2}-8x+4\end{align}

Answer 3

$$\left| \frac{2x - 1}{x - 1} \right| > 2.$$

$$\implies \frac{2x - 1}{x - 1} > 2 \text{ } \text{or} \text{ }\frac{2x - 1}{x - 1} < -2$$

$$\implies \frac{2x - 1}{x - 1} -2>0 \text{ } \text{or} \text{ }\frac{2x - 1}{x - 1} +2 < 0$$

$$\implies x\in (1,\infty) \text{ } \text{or} \text{ }x\in(0.75,1)$$

$$\implies x\in (1,\infty) \text{ } \cup \text{ }x\in(0.75,1)$$

$$\implies x\in(0.75,1)\cup(1,\infty)$$

Answer 4

Logically I think you could take $\frac{2x-1}{x-1}>2$ and $\frac{2x-1}{x-1}<-2$.

That is a mistake: This would be correct if you had said "or" rather than "and".

If $|a|>b$ that means $\Bigg(a>b \text{ }\underline{\text{or}}\text{ (not “and”) } a<-b\Bigg).$

The first one does not give $x\in(-\infty,+\infty).$ Here is the next mistake: $$ \require{cancel} \xcancel{\text{If } \frac{2x-1}{x-1}>2 \text{ then } 2x-1>2(x-1).} $$

You can multiply both sides of an inequality by the same thing if it is positive, and also if it is negative if you interchange the roles of "greater than" and "less than". But $x-1$ is positive if $x>1$ and negative if $x<1.$

One way to solve this is this: \begin{align} & \frac{2x-1}{x-1} > 2 \\[8pt] & \frac{2x-1}{x-1} - 2 > 0 \\[8pt] & \frac{2x-1}{x-1} - \frac{2x-2}{x-1} > 0 \\[8pt] & \frac 1 {x-1}>0 \\[8pt] & x-1>0 \end{align}