Solve $|2x−3| < |x+5|$ I've tried solving $$|2x-3|=|x+5|$$ but the same method does not look applicable here.
$$2x-3 = x+5, \text{if}\; x>3/2$$
$$2x-3= -(x+5), \text{if} \; x<3/2$$
This method was not working here.
 A: You have\begin{align}|2x-3|<|x+5|&\iff|2x-3|^2<|x+5|^2\\&\iff3x^2-22x-16<0\\&\iff3\left(x+\frac23\right)(x-8)<0\\&\iff x\in\left(-\frac23,8\right).\end{align}
A: I'll just leave a less conventional approach for discussion. Once you solve $|2x-3| =|x+5|$, leading to $x=-\frac 23 \vee x = 8$, by continuity, you know that on each interval
$$
(-\infty,-\frac23), \quad (-\frac 23, 8), \quad (8,+\infty)
$$
Either the ">" or "<" inequalities are satisfied. To determine which holds on each interval, just probe it with specific points. For instance, using $x = -1, 0, 9$, you conclude that

*

*$|2(-1)-3| > |-1+5| \Rightarrow |2x-3|>|x+5| $ on $(-\infty, -\frac 23)$

*$|2(0)-3| < |0+5| \Rightarrow |2x-3|< |x+5|$ on $(-\frac 23, 8)$

*$|2(9)-3| > |9+5| \Rightarrow |2x-3| > |x+5|$ on $(8,+\infty)$.

A: There are two nodes $x=3/2$ and $x=-5$, so we have to consider the in-equation $|2x-3|< |x+5|$ in three regions three regions
I: $x\le -5 \implies 3-2x<-(x+5) \implies x>8$ (A contradiction).
II:$-5 <x\le 3/2 \implies 3-2x<x+5 \implies x>-2/3 \implies -2/3<x\le 3/2$
III: $x>3/2 \implies 2x-3 <x+5 \implies x<8 \implies 3/2<x \le 8$
Finally, by combining I and II we get the total solution as $x\in (-2/3,8].$
A: You have
$$|a|=a,\text{if } a\ge0$$
and
$$|a|=-a, \text{if } a\le0$$
Similar holds for $b$.
So to solve
$$|a|<|b$$
you have four cases
case 1:
$$a<b, a\ge0, b\ge 0$$
case 2:
$$-a<b, a\le0, b\ge 0$$
case 3:
$$a<-b, a\ge0, b\le0$$
case 4:
$$-a<-b,a\le0,b\le0$$
So can you solve it noe?
