Absolute Confusion!!!

[vittalgmat]

Hi,
I would like the experts to weigh in on this. (Ron, Ian, Stuart, Stacey??)
I will ask my question first.
What method is applicable to what class of problems?

As I understand, there are two ways to algebraically solve absolute equations (apart from the conceptual distance numberline method). The following are the two ways I found.

Method 1) Whenever u see problem of the type
Is |x| < 1 ?
(1) |x + 1| = 2|x - 1|

Solving the two cases a) (x +1 ) = 2(x-1) and
b) (x +1) = - 2(x -1)
are sufficient.
Ref: see Ron Purewal's explanation here:
https://www.manhattangmat.com/forums/abs ... t3648.html


Method 2) For solving problems of the type, we need to use a critical point approach

If y = /x + 7/ + /2 - x/, is y = 9?
(1) x < 2
(2) x > -7
(Ref: see the MGMAT Tutorial on Absolute values Slide 61
https://www.manhattangmat.com/tutorials/ ... -value.cfm )


The critical point approach is illustrated in the same tutorial. This link below also explains the critical point approach but as Ron mentioned, critical point approach is a overkill for that problem in the link below. It could have been solved by Method 1.

https://www.beatthegmat.com/tricky-inequ ... 22180.html

Thank you very much

[sanju09]

Hope this link helps...

https://www.scribd.com/doc/7058655/Math- ... Inequality



...but never trouble trouble till trouble troubles you

[vittalgmat]

Sanju,
This is the same as the one mentioned in my original post!!!
https://www.manhattangmat.com/tutorials/ ... -value.cfm

Thanks

[sanju09]

Sanju,
This is the same as the one mentioned in my original post!!!
https://www.manhattangmat.com/tutorials/ ... -value.cfm

Thanks

so what's the real hitch man?

[vittalgmat]

not sufficient.
I want to know how to figure out what to use when?
So whenever I see a problem of type | something | = | something else| then use method 1
ELSE I use method 2 ?

Is this a safe conclusion ?

thanks

[sanju09]

I am sorry vittal as I haven't seen either of it so curiously, so I will try to explain my point on the particular problems later on; but for now, a quick tip from my experience, "ALGEBRA is the final resort to crack all LOGICS in QUICK time", other approaches are too welcome on "first come (to hit brain) first serve basis" only if that works for you in a given situation. Take care and greet me as this one is my 700th post on this forum; which I have noted just now, honestly...

[vittalgmat]

Thanks Sanju for all your efforts
I have PMed Ron Purewal. Hopefull he will answer soon.

Btw u still have 60 more post to go before equating the # of posts with Gmat score.

thanks

[sanju09]

Thank you so much and congratulations vittal for still being "Hopefull" here, since all of us are only "Hopeful" here; but you are a bit mo+re than us.......just kidding...sorry....have fun2

[vittalgmat]

Sanju,
That is called "unintended typo" or "printer's devil". But the typo probably reflected my true inner nature

have a nice day

[sanju09]

Sanju,
That is called "unintended typo" or "printer's devil". But the typo probably reflected my true inner nature

have a nice day

so nice of you vittal, thank you

[lunarpower]

Method 2) For solving problems of the type, we need to use a critical point approach

If y = /x + 7/ + /2 - x/, is y = 9?
(1) x < 2
(2) x > -7
(Ref: see the MGMAT Tutorial on Absolute values Slide 61
https://www.manhattangmat.com/tutorials/ ... -value.cfm )

yeah, some sort of "critical point" approach is going to be the best way to hack away at a problem like that one. no doubt.

since these are INEQUALITIES, though, you can also use the backup method of TESTING EXTREMES.

i.e.
statement 1:
TEST EXTREMES
test x = 1.9 (i.e. very close to 2) --> y = |8.9| + |0.1| --> y = 9 YES
test x = -1000 (the other extreme) --> y = |-993| + |1002| --> y is not 9 NO
insufficient.

statement 2:
TEST EXTREMES
test x = -6.9 (i.e. very close to -7) --> y = |0.1| + |8.9| --> y = 9 YES
test x = 1000 (the other extreme) --> y = |1007| + |-998| --> y is not 9 NO
insufficient.

together:
TEST EXTREMES
test x = -6.9 (i.e. very close to -7) --> y = |0.1| + |8.9| --> y = 9 YES
test x = 1.9 (i.e. very close to 2) --> y = |8.9| + |0.1| --> y = 9 YES
test other values in the middle, if you're still skeptical
you'll get all YES's
SUFFICIENT

(c)