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Modulus Inequalities

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by maihuna » Sun Apr 19, 2009 12:28 am
Find the solution to the inequality :
2x+5
------------ <3
|x| +1

1.x < -2/5 or x > 1
2.x < -1 or x > 2/5
3.x < -2/5 or x > 2
4.x > -2/5 or x < 2
5.x > 1 or x < -1
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by DanaJ » Sun Apr 19, 2009 1:53 am
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.
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by maihuna » Sun Apr 19, 2009 3:21 am
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
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by aj5105 » Tue Apr 28, 2009 9:45 pm
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."
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by DanaJ » Wed Apr 29, 2009 12:19 am
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.
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by kanha81 » Thu Apr 30, 2009 8:28 am
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]
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