Modulus Inequalities

What is the OA and the source for this question? I'm getting a slightly different answer than the choices.

Edit: Sorry, I was wrong.
I uploaded my solution in a Word Document, since it would have been extremely hard to write it without the Equation Editor.

I saw your answer it is correct.

I just found another way to solve it:

2x + 5 < 3(|x| + 1)

2x + 5 < 3|x| + 3

2x + 2 < 3|x|

3x < -(2x + 2) or 3x > 2x + 2

5x < -2 or x > 2

The solution is x < -2/5 or x > 2

Could you please explain this part?

"Again, it all comes down to the sign of the fraction. Since –x + 1 is always positive, then 5x + 2 must be negative for the fraction to be negative as well."

When you have a fraction x/y, its sign is determined by the signs of x and y. If x and y have the same sign, then the fraction will be positive. When x and y have different signs, then the fraction will be negative.

Now, since -x is always positive (since we've established that x is negative for that particular case), -x + 1 will also be positive. This means that 5x + 2 has to be negative for the whole fraction to be positive.

maihuna wrote:I saw your answer it is correct.

I just found another way to solve it:

2x + 5 < 3(|x| + 1)
2x + 5 < 3|x| + 3
2x + 2 < 3|x|
3x < -(2x + 2) or 3x > 2x + 2
5x < -2 or x > 2

The solution is x < -2/5 or x > 2

Nice! However, my question is why can't we solve it as:

x>0:
2x + 5 < 3(x + 1)
2x + 5 < 3x + 3
2 < x
x > 2 .........(1)


x<0:
2x + 5 < 3(-x + 1)
2x + 5 < -3x + 3
5x < -2
x < -2/5

Hence [spoiler](C)[/spoiler]