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# Solving Inequalities

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Re: Solving Inequalities [#permalink]

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05 Jan 2017, 05:24
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Re: Solving Inequalities [#permalink]

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19 Aug 2017, 22:10
If 4<(7-x)/3, which of the following must be true?

I. 5<x
II. |x+3|>2
III. -(x+5) is positive

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 43358
Re: Solving Inequalities [#permalink]

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20 Aug 2017, 00:27
artisood17 wrote:
If 4<(7-x)/3, which of the following must be true?

I. 5<x
II. |x+3|>2
III. -(x+5) is positive

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III

Posted from my mobile device

Discussed here: https://gmatclub.com/forum/if-4-7-x-3-w ... 68681.html

Please follow the rules (https://gmatclub.com/forum/rules-for-po ... 33935.html) when posting a question. Thank you.
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Joined: 11 Sep 2017
Posts: 9
Re: Solving Inequalities [#permalink]

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11 Nov 2017, 18:44
cyberjadugar wrote:

Solving Inequalities

I was going through the posts on inequalities and found that many good concepts are explained here, but still people do have trouble solving the question using these concept.
In these posts, there were quadratic equations, curves, graphs and other mathematical stuff. With this post, I am trying to provide a simple method to solve such questions quickly. I won't be writing the concepts behind it.
Remember this is the same OLD concept, it's just presented differently.

Case 1: Multiplication

for example: $$(x-1)(x-2)(x-3)(x-7) \leq 0$$

To check the intervals in which this inequality holds true, we need to pick only one value from the number line.
Lets say x = 10, then (9)(8)(7)(3) > 0, in every alternate interval the sign would be + for the above expression

---(+)-----1---(-)---2---(+)---3-----------(-)----------7----(+)------

Thus, inequality would hold true in the intervals:
$$1 \leq x \leq 2$$
$$3 \leq x \leq 7$$,

Note that intervals are inclusive of 3 & 7

Case 2: Division

In case of division:
$$\frac{(x-1)(x-2)}{(x-3)(x-7)} \leq 0$$
Using the same approach as above;
$$1 \leq x \leq 2$$
$$3 < x < 7,$$ $$(x\neq3,$$ $$7)$$

Since (x-3)(x-7) is in denominator, its value can't be 0.

Following things to be kept in mind while using above method:
1. Cofficient of x should be positive: for ex - $$(x-a)(b-x)>0$$, can be written as $$(x-a)(x-b)<0$$
2. Even powers: for ex - $$(x-9)^2(x+3) \geq 0$$, $$(x-9)^2$$ is always greater than 0, so, it should be only considered to check the equality (=0)
3. Odd powers: $$(x-a)^3(x-b)^5>0$$, will be same as $$(x-a)(x-b)>0$$
4. Cancelling the common terms:
for ex - $$\frac {(x^2+x-6)(x-11)}{(x+3)} >0$$, it can be simplified as (x-2)(x-11)>0
or, ---(+)-----2---(-)-------------11----(+)------
thus $$x <2$$ or $$x>11$$, but since at x = -3 (in the original expression), we get undefined form, so, $$x \neq -3$$

A question for you:
For what values of x, does the following inequality holds true?
$$(x-a)(x-b)...(x-n)...(x-z) \geq 0$$, where {a, b, c,...} are integers.
[Reveal] Spoiler: Solution
The expression has (x-x), thus, it is always 0=0 for every value of x.

Reference post:
http://gmatclub.com/forum/inequalities-trick-91482.html

PS: I hope you find this post useful, please provide feedback to improve the quality of the post.

Thanks,

can anybody give example for this?

2. Even powers: for ex - (x−9)2(x+3)≥0(x−9)2(x+3)≥0, (x−9)2(x−9)2 is always greater than 0, so, it should be only considered to check the equality (=0)

thanks
Re: Solving Inequalities   [#permalink] 11 Nov 2017, 18:44

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# Solving Inequalities

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