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.
Modulus Inequalities
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Source: Beat The GMAT — Problem Solving |
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maihuna
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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
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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aj5105
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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."
"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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DanaJ
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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.
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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kanha81
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Nice! However, my question is why can't we solve it as: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
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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