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Not a question but an intermediary step

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by OptimusPrep » Mon Apr 25, 2016 7:58 pm
sushantsahaji wrote:Is |2x+5|>= -3 a valid statement?

It is mentioned in one of the questions answer.
If valid, why? Can |2x+5| be equal to -3?

Regards,

Sushant
I am having difficulty in understanding your question. I would be better if you posted the entire question, but let me try.

Case 1:
x >= - 5/2 - (i)
In this case, modulus will open with a positive sign.
2x + 5 >= -3
2x >= -8
x >= -4 (ii)
From (i) and (ii), x >= -5/2

Case 2:
x < -5/2 - (iii)
In this case, modulus will open with a negative sign.
2x + 5 < -3
2x < -8
x < -4 - (iv)
From (iii) and (iv), x < -4
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by sushantsahaji » Tue Apr 26, 2016 2:34 am
Hi,

Solve |2x+5|+4>=1 is the original question. Question is not tough to solve. But I have a doubt in second step,

|2x+5|>=-3. (|2x+5|can never be equal to -3.)

So, is the above statement valid?
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by GMATGuruNY » Tue Apr 26, 2016 4:13 am
sushantsahaji wrote:Hi,

Solve |2x+5|+4≥1 is the original question.
Strange problem.
Since an absolute value cannot be negative, |2x+5| ≥ 0.
Thus, the least possible value of |2x+5| + 4 is 4.
Since |2x+5| + 4 ≥ 4, it is not possible that |2x+5| + 4 = 1.
That said, any value of x will satisfy the inequality in red.

I would ignore this problem.
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by sushantsahaji » Wed Apr 27, 2016 3:23 am
Thank you Mitch.
Source of the problem: https://www.regentsprep.org/regents/math ... actice.htm
Practice website suggested on Gmatprepnow.
GMATGuruNY wrote:
sushantsahaji wrote:Hi,

Solve |2x+5|+4≥1 is the original question.
Strange problem.
Since an absolute value cannot be negative, |2x+5| ≥ 0.
Thus, the least possible value of |2x+5| + 4 is 4.
Since |2x+5| + 4 ≥ 4, it is not possible that |2x+5| + 4 = 1.
That said, any value of x will satisfy the inequality in red.

I would ignore this problem.
What is the source?
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