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Need help with negative inequality

by ildude02 » Mon Jul 07, 2008 3:05 pm
If x is an integer, what's the value of X?

1)1/5 < 1/(x+1) < 1/2
2)(x-3)(x-4) = 0;

I have solved this question to find the answer, but I have a question with regards to how I solved statement 1 and wanted to hear your thoughts.

For statement 1, (x+1) can either be positve or negative, so if (x+1) is postive, we dont need to worry abt the reversing the inequalities, so we get, 1 < x < 4;

When we consider (x+1) being negative, this is where I wanted to get your input;

Can we just substitue (x+1) with -(x+1) and try to solve the inequality? as in; 1/5 < 1/-(x+1) < 1/2 . Is that OK? Or, do we need to reverse the inequality and leave the (x+1) as in; 1/5 > 1/(x+1) > 1/2; Solving both the possibilites obviously give different values.
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by wilderness » Tue Jul 08, 2008 2:45 am
Hi,

I think we do not need to take -ve into consideration for (1) because it says 1/(x+1) is between 1/5 and 1/2 i.e 1/(x+1) is between 0.2 and 0.5. So its not negative.

IMHO the best way to solve 1 is just invert the equations and change the signs (Is there any rule that forbids us from doing so ?? )
Then we have simply 5 > (x+1) > 2
hence 4 > x > 1

The answer is C.
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by Ian Stewart » Tue Jul 08, 2008 5:23 am
wilderness wrote: IMHO the best way to solve 1 is just invert the equations and change the signs (Is there any rule that forbids us from doing so ?? ) .
You can certainly do this if you know everything is positive. If x and y are positive, and you have the inequality:

1/x < 1/y

then you can multiply both sides by xy to get:

y < x

We don't need to worry about whether to reverse the inequality, provided we know that x and y are both positive. The same is true if x and y are both negative: then xy is positive, so we can multiply by xy without reversing the inequality. You do need to be careful, however, when x and y have opposite signs. Then, when you multiply by xy, you are multiplying by a negative, and you must reverse the inequality. If, say:

1/x < 1/y

then x could be negative, and y positive, and we have, multiplying by xy and reversing the inequality:

y > x

So yes, there is a rule one can follow here- but you need to be very careful if you don't know the signs of x and y.
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by wilderness » Tue Jul 08, 2008 6:30 am
Thanks. You just keep amazing me with your tips.
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by ildude02 » Tue Jul 08, 2008 7:11 am
I missed it, you are gith that we dont need to consider (x+1) < 0 since it will not satify the given inequality. Thanks.

But my main question was, if we do need to consider one, can we just substitue the variable x+1 with -(x+1) and continue solving the inequality to find the upper an dtjhe lower limit? Appreciate your response.
wilderness wrote:Hi,

I think we do not need to take -ve into consideration for (1) because it says 1/(x+1) is between 1/5 and 1/2 i.e 1/(x+1) is between 0.2 and 0.5. So its not negative.

IMHO the best way to solve 1 is just invert the equations and change the signs (Is there any rule that forbids us from doing so ?? )
Then we have simply 5 > (x+1) > 2
hence 4 > x > 1

The answer is C.
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by AleksandrM » Tue Jul 08, 2008 9:32 am
Hey, I ended up with:

1) 10/(x + 1) - 5 < 0

2) x = 3 x = 4

Both: If x = 4, then 1) does not hold true. If x = 3, then 1 holds true.

And so I went with C.

Is my approach correct?
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by jay2007 » Tue Jul 08, 2008 9:36 am
Solving the first in equality i get the values of x as 3 or 4

1/5 < 1/(x+1) means, x=1,2,3,4,0
1/(x+1) < 1/2 means, x=3,4,5,..... or -2, -3, -4....

So, x should be 3 or 4

Solving the second equation i get the values of x as 3 or 4.

So, my answer is E.


What is the OA?
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by AleksandrM » Tue Jul 08, 2008 9:50 am
jay,

For the first one, you cannot cross-multiply because you do not know the sign of x, therefore you do not know whether to "flip" the inequality sign or not.
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by zakir » Tue Jul 08, 2008 9:15 pm
I would do this as-
From 1st equation:
(x+1)<5, so x <4>2, so x>1
x can be either 2 or 3

From 2nd equation:
x=3, or x=4

Combining 1st and 2nd, x=3.

So the correct answer is C.

(x is not negative, as that would make 1st equation untrue)
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by lunarpower » Thu Jul 10, 2008 10:48 pm
AleksandrM wrote:Hey, I ended up with:

1) 10/(x + 1) - 5 < 0

2) x = 3 x = 4

Both: If x = 4, then 1) does not hold true. If x = 3, then 1 holds true.

And so I went with C.

Is my approach correct?
nope; you left off the left side of the double inequality in statement 1. your version (10/(x+1) - 5 < 0) is a rearrangement of 1/(x+1) < 1/2; you've completely neglected the fact that, per the original problem statement, 1/(x+1) is also greater than 1/5.
this observation totally changes the game, because a number that's stuck between 1/5 and 1/2 must be positive; with just the listed half of the inequality, you can't determine whether 1/(x+1) is positive or negative.

because everything in the double inequality is positive, you can just go ahead and take reciprocals of everything. (RULE: if everything is positive or everything is negative, you can take reciprocals of everything and flip all the inequality signs around.)
this yields 5 > x+1 > 2, or, more traditionally, 2 < x+1 < 5.
therefore, 1 < x < 4.

your rephrase of (2) is correct.

therefore, the answer is still c, although not quite for the reasons you've listed.
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