Re: What am I doing wrong?

I would write:

Square both sides, and expand:

Can you proceed from here?

Re: What am I doing wrong?

You need to consider three cases.

When we have and .

When we have and .

When we have and .

So to solve , when we have

.

Since we don't have any values of x that are both < -3 and > 2, this is impossible.

When we have

so that means is acceptable.

When we have

so that means is also accetable.

So that means our total solution is .

Re: What am I doing wrong?

whenever i see |anything|, i think that this means:

|a| = a, when a ≥ 0

|a| = -a, when a < 0.

now from appearances, it looks like we have 4 cases to investigate:

1) 2x+1, x+3 ≥ 0

2) 2x+1 ≥ 0, x+3 < 0

3) 2x+1 < 0, x+3 > 0

4) 2x+1,x+3 < 0

let's look at each of these in turn.

in case (1), we have |2x+1| < |x+3| becomes 2x+1 < x+3, which then can be manipulated algebraically:

x+1 < 3

x < 2

but this is not quite all. if 2x+1 ≥ 0, this means that 2x ≥ -1, so x ≥ -1/2. since we need x+3 ≥ 0 as well, we need to simultaneously require x ≥ -3.

the more restrictive of the two requirements is x ≥ -1/2. so if case 1 is the case, then x ≥ -1/2 > -3, in which case the inequality |2x+1| < |x+3| tells us x < 2, so we have:

-1/2 ≤ x < 2.

now we turn to case (2). here, 2x+1 ≥ 0 means that x ≥ -1/2, as before. but x+3 < 0, means x < -3. but x cannot be both ≥ -1/2 AND < -3 at the same time. so this case never occurs.

on to case (3). here, 2x+1 < 0, means that x < -1/2. x+3 ≥ 0 means that x ≥ -3. so for case (3) to apply, we need -3 ≤ x < -1/2. but we still have the original inequality to satisfy, as well. for case (3), it is:

-2x-1 < x+3

-3x-1 < 3

-3x < 4

x > -4/3 (we have to "change direction" when we multiply by -1/3).

since -4/3 is in-between -3 and -1/2, requiring that BOTH x > -4/3 AND -3 ≤ x < -1/2 translates into: -4/3 < x < -1/2.

finally, case (4). here 2x+1 < 0 means x < -1/2, and x+3 < 0, means x < -3. the more restrictive condition is x < -3. in this case, our inequality is:

-2x-1 < -x-3

-x-1 < -3

-x < -2

x > 2.

now since x cannot be both > 2 and < -3, in case (4) we have no x that qualify.

now the choice between the 4 cases (only 2 of which are possible) is an "OR" choice. thus we have, as our final solution:

-4/3 < x < -1/2 OR -1/2 ≤ x < 2 since the "endpoints match", we can combine these into the SINGLE condition:

-4/3 < x < 2.

here is how it works "in reverse":

when x < -3, then 2x+1 < -5 < 0, and x+3 < 0, so condition (4) holds. but the only x that satisfy -2x-1 < -x-3 are x > 2, and no x < -3 is > 2.

so there are no x's that work < -3.

when -3 ≤ x < -1/2, then 2x+1 < 0, and x+3 ≥ 0, so condition (3) holds. but for -2x-1 < x+3, we have to have x > -4/3, so the x's that work are in:

[-3,-1/2) ∩ (-4/3,∞) = (-4/3,-1/2).

when -1/2 ≤ x, then 2x+1 ≥ 0 and x+3 ≥ 5/2 > 0, so condition (1) holds, in which case 2x+1 < x+3 means that x < 2. so that x's that work are in:

[-1/2,∞) ∩ (-∞,2) = [-1/2,2).

so our final answer is (-4/3,-1/2) U [-1/2,2) = (-4/3,2).

**********

let's compare this with MarkFL2's approach:

suppose (3x + 4)(x - 2) < 0. when can this happen? in just two ways:

A) 3x + 4 < 0 and x - 2 > 0

B) 3x + 4 > 0 and x - 2 < 0

(A) first: 3x + 4 < 0 means 3x < -4, and so x < -4/3. x - 2 > 0 means x > 2. it is impossible for both of these to be true.

(B) next: 3x + 4 > 0 means x > -4/3, and x - 2 < 0 means x < 2. this, we can do, so we have: -4/3 < x < 2.

much less work.