Absolute Value Inequalities with Two Branches

Question

please treat my brackets as absolute value bars, or fix my formatting - I'm new here :)

Q: $2 < |2x-3| < 7$

I tried to break it up,

$2 < 2x-3 < 7$

$5 < 2x < 10$

$5/2 < x < 5$

Other branch,

$2 < -2x+3 < 7$

$-1 < -2x < 4$

$1/2 > x > -2$

$-2 < x < 1/2$

How do I solve this? My answers are wrong?

Answer 1

By the definition of absolute function we get:

$|2x-3| = \max(2x-3, -(2x-3))$

What does it mean? Examples:

1). $x=1 \Rightarrow |2\cdot1 -3| = |-1| = \max(-1,-(-1)) =-(-1) = 1$
2). $x=2 \Rightarrow |2\cdot2 -3| = |1| = \max(1,-(1)) = 1$
As you can see, if $2x-3\ge 0$, the absolute function chooses $2x-3$, and if
$2x-3<0$ it chooses $-(2x-3)$ to make it a positive value.

So, how can we know if $2x-3 < 0$ or $2x-3 \ge 0$?

We can't, because the possible values for $x$ are all the numbers in $\mathbb{R}$. that's why we have to branch the solution.

The solution

$2<|2x-3|<7$

1). $2x-3< 0 \Rightarrow 2<\max(2x-3,-(2x-3)) < 7 \Rightarrow 2<-(2x-3) < 7$
$\Rightarrow 2< -2x+3 < 7\Rightarrow -1<-2x<4 \Rightarrow \frac{1}{2}>x>-2$

2). $2x-3\ge 0 \Rightarrow 2< \max(2x-3,-(2x-3))<7 \Rightarrow 2<2x-3<7$
$\Rightarrow 5<2x<10\Rightarrow 2\frac{1}{2}<x<5$

So, $2\frac{1}{2}<x<5$ or $\frac{1}{2}>x>-2$.

Answer 2

The solution is given by $$-2<x<\frac{1}{2}$$ or $$\frac{5}{2}<x<5$$