Double absolute inequality

Question

I wan to determine for which $x$ the following inequality holds $$ |x-1| < |2x+3| $$ Can somebody tell me how to approach this? Shall I analyze the following inequality $$ -(2x+3) < x-1 < +(2x+3) $$ jointly with this other one $$ +(x-1) < 2x+3 < -(x-1) $$ If so, how I should proceed from there? Thanks in advance

Answer 1

For $$x\geq 1$$ we get $$x-1<2x+3$$ and for $$-\frac{3}{2}\le x<1$$ we get $$-x+1<2x +3$$ and so on! Finally we get $$-\frac{2}{3}<x+\infty, -\infty<x<-4$$

Answer 2

For these kind of inequalities we need to consider some cases, notably since the LHS and RHS are zero at $x=-\frac32$ and $x=1$ we can proceed as follows

• $x<-\frac32\implies -x+1 < -2x-3$
• $-\frac32\le x<1\implies -x+1 < 2x+3$
• $x\ge 1 \implies 1-x < 2x+3$

then we need to solve each inequality in the corresponding range and finally take tha union of the single solutions as final solution for the given inequality.

Answer 3

---

==========================================

Answer 4

If you square it (you can since both sides are nonegative) $$ |x-1|^2 < |2x+3|^2 $$ we get $$ 4x^2+12x+9>x^2-2x+1\implies 3x^2+14x+8>0$$

So $$(3x+2)(x+4)>0\implies x \in (-\infty,-4)\cup (-{2\over 3},\infty)$$