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When more than two absolutes are present though, it becomes complicated to use the squaring approach. Another method seems to work better. I'll briefly solve your given problem using this other method.
|3x-2| <= |2x-5|
1. First you find the pivotal/critical points:
i.e.
a. you set 3x-2 = 0, so x = 2/3
b. you set 2x-5 = 0, so x = 5/2
2. If you take the 2 values, 2/3 and 5/2 on the number line, you get three domains
a. left of 2/3
b. b/w 2/3 and 5/2
c. right of 5/2
Next, you modify the intial inequation for these three domains, by taking out the modulus signs.
a. So, for x < 2/3 the inequation becomes,
-3x+2 <= -2x +5
(since both terms become negative for x < 2/3, you reverse their signs)
-3 <= x
b. For x b/w 2/3 and 5/2
3x -2 <= -2x + 5
(since 2x-5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign)
or x <= 7/5
c. For x > 5/2
3x-2 <= 2x-5
x <= -3 (not valid since x is assumed to be > 5/2 above)
Combining the above three results, you get
-3 <= x <= 7/5
------------------
Credit for this method goes to user "Fig" who has posted several problems/solutions using this approach. Please search and look at all those problems if you have any questions. Manhattan GMAT also uses this approach.
Last edited by oops on 13 Jul 2007, 21:09, edited 1 time in total.
When more than two absolutes are present though, it becomes complicated to use the squaring approach. Another method seems to work better. I'll briefly solve your given problem using this other method.
|3x-2| <= |2x-5|
1. First you find the pivotal/critical points: i.e. a. you set 3x-2 = 0, so x = 2/3 b. you set 2x-5 = 0, so x = 5/2
2. If you take the 2 values, 2/3 and 5/2 on the number line, you get three domains a. left of 2/3 b. b/w 2/3 and 5/2 c. right of 5/2
Next, you modify the intial inequation for these three domains, by taking out the modulus signs.
a. So, for x < 2/3 the inequation becomes, -3x+2 <= -2x +5 (since both terms become negative for x < 2/3, you reverse their signs) -3 <= x
b. For x b/w 2/3 and 5/2 3x -2 <= -2x + 5 (since 2x-5 becomes negative for x b/w 2/3 and 5/2, you reverse its sign) or x <7> 5/2 3x-2 <= 2x-5 x <3> 5/2 above)
Combining the above three results, you get -3 <= x <= 7/5 ------------------
Credit for this method goes to user "Fig" who has posted several problems/solutions using this approach. Please search and look at all those problems if you have any questions. Manhattan GMAT also uses this approach.
Well ... Finally, I should not post ... All is in your post :D
Gosh, Fig (master of absolutes!), didn't see you also responded to this mesg - we must've been typing it up simultaneously - credit goes to you - glad to see you back...