Absolute Values and its Inequalities

Question

Find $x \in \mathbb{R}$ that satisfies both $|2x - 3| < 5$ and $|x + 1| > 2$ simultaneously.

Answer 1

First we solve $|2x-3|<5$. We have $2x-3<5$ and $-(2x-3)<5$

So, \begin{align}2x-3&<5\\ 2x&<8\\ x&<4\end{align}

and \begin{align}-(2x-3)&<5\\ -2x+3&<5\\ -2x&<2\\ x&>-1\end{align}

Therefore, $-1<x<4$

Then we solve $|x+1|>2$. We have $x+1>2$ and $-(x+1)>2$

So \begin{align}x+1&>2\\ x&>1\end{align}

and \begin{align}-(x+1)&>2\\ -x-1&>2\\ -x&>3\\ x&<-3\end{align}

Therefore $x>1$ or $x<-3$

We can then combine these two, to give us the final answer of $$1<x<4$$

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This (rather messy, sorry!) number line shows how we have combined the two results - the red line is the result of $|2x-3|<5$, the green line is the result of $|x+1|>2$, and the blue line is the answer to the question (anywhere there is a red and green line):

Answer 2

Hint: if you can solve both inequalities, it shouldn't be too hard to combine the results.

$$|2x - 3| < 5 \iff -5 < 2x - 3 < 5 \iff \ldots \iff x \in S_1$$

$$|x + 1| > 2 \iff x + 1 < -2 \;\vee\; x + 1 > 2 \iff \ldots \iff x \in S_2$$

Once you have the solutions sets to both inequalities, take the intersection: $$\left\{ \begin{array}{l} |2x - 3| < 5 \\ |x + 1| > 2\end{array}\right. \iff x \in S_1 \cap S_2$$

Answer 3

$|2x-3|<5\Leftrightarrow -5<2x-3<5\Leftrightarrow -1<x<4$

$|x+1|>2 \Leftrightarrow x+1<-2 \;\text{or}\; x+1>2 \Leftrightarrow x<-3 \;\text{or}\; x>1$

Hence solutions of inequalities are $$ 1<x<4 $$