Solving absolute value inequalities

[Nitrate]

Homework Statement

[abs(x)]/[abs(x+2)]<2

Homework Equations

The Attempt at a Solution

case 1: [abs(x)]/[abs(x+2)]<2
case 2: [abs(x)]<2[abs(x+2)]
is this right so far?
if so, why is there two cases
and what do i do next?

 

[HallsofIvy]

What you have written is NOT two cases. It is two versions of the same inequality.

Because an absolute value is always positive, multiplying both sides of "case 1" by |x+2| does not change the direction of the inequality sign and leads to "case 2".

To solve this inequality, you should consider three cases:
a) $x\le -2$ so that x and x+2 are both less than 0. |x+ 2|= -(x+2) and |x|= -x.
b) $-2< x\le 0$ so that x+ 2 is positive but x is still less than 0. |x+ 2|= x+ 2 and |x|= -x.
c) $0< x$ so that both x and x+ 2 are both positive. |x+ 2|= x+ 2 and |x|= x.

 

[SammyS]

Homework Statement

[abs(x)]/[abs(x+2)]<2

Homework Equations

The Attempt at a Solution

case 1: [abs(x)]/[abs(x+2)]<2
case 2: [abs(x)]<2[abs(x+2)]
is this right so far?
if so, why is there two cases
and what do i do next?

Those are not two different cases.

The inequality $\displaystyle\frac{|x|}{|x+2|}<2$ is equivalent to $\displaystyle|x|<2|x+2|\ .$