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Louis14
I've been honing in my conceptual clarity vis-a-vis Inequalities and Absolute Values, and I wanted to clarify two things:

1. If, say, we have 1/|x| <= x, can we multiply both sides by |x| and make it 1 <= x*|x|? Can we do this when we have an "=" sign instead of an inequality sign?

2. If, say, we have x+1/x-y = 0. Can we multiply both sides by (x-y) and make the equation x+1=0?

Thanks!

It may help you to use substitute some small integers (1,2, -2, 0,-4) and play around with them in those formulas you just mentioned.
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chetan2u
I've been honing in my conceptual clarity vis-a-vis Inequalities and Absolute Values, and I wanted to clarify two things:

1. If, say, we have 1/|x| <= x, can we multiply both sides by |x| and make it 1 <= x*|x|? Can we do this when we have an "=" sign instead of an inequality sign?
Yes, you can. You can multiply any 'thing' on both sides without changing the inequality sign till the time you are sure that that 'thing' is positive.
I get that but don’t you think the value of |x| could be 0 as well?

2. If, say, we have x+1/x-y = 0. Can we multiply both sides by (x-y) and make the equation x+1=0?
The answer is again in affirmative, yes. We cannot have the denominator as 0, otherwise the fraction becomes undefined. So, the NUMERATOR has to be 0, that is x+1=0.
So, sir, are you suggesting that denominator can never be zero? As in, say we have a*b=b*c. Can we divide both side by b, assuming that denominator can never be 0?

Thanks

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If the question says that \(\frac{x+1}{x-y} = 0\), then are confirming that \(x-y\neq 0\)
But a*b=b*c is multiplication with b, and so b can be 0. The division by b is being done by you, but the test makers does not tell you that.
Say it is given that \(\frac{a}{b}=2\) or \(\frac{a}{b}=\frac{c}{b}\). Now the test makers/creators have given the division by b to you so b will not be 0, and most of the times they will mention it too..
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chetan2u
I've been honing in my conceptual clarity vis-a-vis Inequalities and Absolute Values, and I wanted to clarify two things:

1. If, say, we have 1/|x| <= x, can we multiply both sides by |x| and make it 1 <= x*|x|? Can we do this when we have an "=" sign instead of an inequality sign?
Yes, you can. You can multiply any 'thing' on both sides without changing the inequality sign till the time you are sure that that 'thing' is positive.
I get that but don’t you think the value of |x| could be 0 as well?


On the GMAT, if they start a problem by giving you something like 1/|x| <= x, then they'll also state that x cannot equal 0. In general, the GMAT tries to make it clear that the values in a problem won't be 'undefined' (for instance, like a fraction divided by 0). My impression is that they do this to avoid getting involved in mathematical debates! :)

Quote:

2. If, say, we have x+1/x-y = 0. Can we multiply both sides by (x-y) and make the equation x+1=0?

Similarly, if you see this on the GMAT, the problem will almost certainly specify that x and y are not equal. In that case, x-y won't equal 0, and you can safely multiply.

Quote:
So, sir, are you suggesting that denominator can never be zero? As in, say we have a*b=b*c. Can we divide both side by b, assuming that denominator can never be 0?

You definitely can't do that, though. That's because there's nothing here to suggest that b can't be 0. In the previous two problems, the GMAT would probably tell you that the denominator of the fraction isn't 0, because if it was, they would have written a problem with an undefined fraction in it, that you couldn't really figure out. But in this problem, a*b = b*c is a perfectly valid equation regardless of whether b=0. So, you can't assume that b doesn't equal 0 unless they tell you that.

To solve something that looks like this, here's what you need to do instead of dividing by b:

a*b = b*c
a*b - b*c = 0
b(a - c) = 0

EITHER b = 0, or a-c = 0 (or both)
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