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converting Inequality to absolute form

Expert replies
by gmatdriller » Sat Mar 03, 2012 11:29 pm
permit me to ask this somewhat dumb question:
how do i translate a number line inequality to an expression in absolute form?
For example, how to convert the attached number line diagram diagram to its absolute form:

Inequality: -3 <= x < 7
Absolute form: ???

Thanks.
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Source: — Problem Solving |

by shashank.ism » Sat Mar 03, 2012 11:46 pm
absolute form ... Hmmm.... ok Go this way..
First look at the endpoints. Negative -3 and 7 are 10 units apart. Half of 10 is 5. So I want to adjust this inequality so that it relates to -5 and 5, instead of to -3 and 7. To accomplish this, I will adjust the ends by subtracting 2 from all both sides

-3 <= x <= 7
-3 - 2 <= x - 2 <= 7-2
-5 <= x - 2 <= 5
--> |x-2| <=5

There it is you got your answer... cheers :)
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by krusta80 » Sun Mar 04, 2012 12:34 am
gmatdriller wrote:permit me to ask this somewhat dumb question:
how do i translate a number line inequality to an expression in absolute form?
For example, how to convert the attached number line diagram diagram to its absolute form:

Inequality: -3 <= x < 7
Absolute form: ???

Thanks.
So, to generalize, for all inequalities with a < 0 and b > 0:

a < x < b

We can convert to |x - (b+a)/2| < (b-a)/2, when |a| <= |b|
and |x + (b+a)/2| < (b-a)/2, when |a| > |b|
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by gmatdriller » Sun Mar 04, 2012 1:42 am
Thanks for the posts.
for one, i can convert single segment inequality to its absolute form.
I could not do same for two segment inequalities (that is quadratic forms).

this is similar to the OR (|x-a| > b) situation
Is this knowledge necessary for the gmat test?
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by shashank.ism » Sun Mar 04, 2012 2:08 am
krusta80 wrote:
gmatdriller wrote:permit me to ask this somewhat dumb question:
how do i translate a number line inequality to an expression in absolute form?
For example, how to convert the attached number line diagram diagram to its absolute form:

Inequality: -3 <= x < 7
Absolute form: ???

Thanks.
So, to generalize, for all inequalities with a < 0 and b > 0:

a < x < b

We can convert to |x - (b+a)/2| < (b-a)/2, when |a| <= |b|
and |x + (b+a)/2| < (b-a)/2, when |a| > |b|

Yes you are correct krusta... anyway we don't need any generalization... just check problem and solve it in seconds....
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by shashank.ism » Sun Mar 04, 2012 2:15 am
gmatdriller wrote:Thanks for the posts.
for one, i can convert single segment inequality to its absolute form.
I could not do same for two segment inequalities (that is quadratic forms).

this is similar to the OR (|x-a| > b) situation
Is this knowledge necessary for the gmat test?
Yup there is a Thanks button for that :) ....

In quadratic form u get 2 soultions. and there is an OR so if it gets continous result.... u can do it.. like x<3 or x>24
otherwise not
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