# How do you solve the absolute value inequality abs(2x + 1)< abs(3x - 2)?

Apr 7, 2015

No solution

The inequality would hold good, even if we square both sides, but then absolute value sign would not be material, as both sides would be positive only. Accordingly,
${\left(2 x + 1\right)}^{2}$ < ${\left(3 x - 2\right)}^{2}$
4${x}^{2}$ +4x+1 < 9${x}^{2}$ - 12x +4
0< 5${x}^{2}$-16x +3. On factoring the quadratic expression, it would be (5x-1)(x-3)>0 . Now there two options, either both factors are positive or both are negative.
Case1. Both factors are positive would mean x>3, x>1/5 . If x >3 , it automatically implies it is >1/5. It is ,therefore, concluded that x>3.

case2. Both factors are negative would mean x<1/5 , x<3. If x<1/5, it implies that it is <3. Hence conclusion is x<1/5.

Now x>3 and x<1/5 cannot happen at the same time. Hence there is no solution to the given inequality. If this is represented on the number line the conclusion would become more obvious.