We are looking -5<=x<=3
a) x<= 3 and x >= -3 so this would be -3<=x<=3 (not right)
b)x<=5 and x>= -5 so this would be -5<=x<=5 (not right)
c)x<=5 and x>=-1 so this would be -1<=x<=5 (not right)
d)x<=5 and x>=-3 so this would be -3<=x<=5 (not right)
e)x<=3 and x>=-5 so this would be -5<=x<=3 (this is what we are looking for)
E
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Sorry I am a bit confused with how you get these numbers. Can you plz provide some help.. thanksdmateer25 wrote:We are looking -5<=x<=3
a) x<= 3 and x >= -3 so this would be -3<=x<=3 (not right)
b)x<=5 and x>= -5 so this would be -5<=x<=5 (not right)
c)x<=5 and x>=-1 so this would be -1<=x<=5 (not right)
d)x<=5 and x>=-3 so this would be -3<=x<=5 (not right)
e)x<=3 and x>=-5 so this would be -5<=x<=3 (this is what we are looking for)
E
-
GMATGuruNY
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The correct answer choice must correspond to the range shown in the picture: from -5 to 3, inclusive.
Thus, the correct answer must work for EVERY value between -5 and 3, inclusive -- and for no values outside this range.
Plug x=-5 into all of the answer choices.
Eliminate any answer choice in which x=-5 doesn't work.
Answer choice A: |x|≤3
|-5|≤3.
5≤3.
Doesn't work. Eliminate A.
Answer choice B: |x|≤5
|-5|≤5
5≤5.
This works. Hold onto B.
Answer choice C: |x-2|≤3
|-5-2|≤3
7≤3.
Doesn't work. Eliminate C.
Answer choice D: |x-1|≤4
|-5-1|≤4
6≤4.
Doesn't work. Eliminate D.
Answer choice E: |x+1|≤4
|-5+1|≤4
4≤4.
This works. Hold onto E.
Only B and E remain.
Answer choice B -- |x|≤5 -- includes every value from -5 to 5, but the needed range is only from -5 to 3.
Eliminate B.
The correct answer is E.
Algebraically:
Absolute value means distance from 0.
Answer choice A: |x|≤3
|x|≤3 means that the value of x cannot be more than 3 places from 0.
Thus, x can be any value between -3 and 3, inclusive:
-3≤x≤3
Answer choice B: |x|≤5
|x|≤5 means that the value of x cannot be more than 5 places from 0.
Thus, x can be any value between -5 and 5, inclusive:
-5≤x≤5.
Answer choice C: |x-2|≤3
|x-2|≤3 means that the value of x-2 cannot be more than 3 places from 0.
Thus, x-2 can be any value between -3 and 3, inclusive:
-3≤x-2≤3
-1≤x≤5 (Solve for x by adding 2 to each part of the inequality.)
Answer choice D: |x-1|≤4
|x-1|≤4 means that the value of x-1 cannot be more than 4 places from 0.
Thus, x-1 can be any value between -4 and 4, inclusive:
-4≤x-1≤4
-3≤x≤5 (Solve for x by adding 1 to each part of the inequality.)
Answer choice E: |x+1|≤4
|x+1|≤4 means that the value of x+1 cannot be more than 4 places from 0.
Thus, x+1 can be any value between -4 and 4, inclusive:
-4≤x+1≤4
-5≤x≤3 (Solve for x by subtracting 1 from each part of the inequality.)
Plugging values into the answer choices seems easier and faster.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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For any modulus equality, you can break that into as follows:
|x-1|<=4 can be broken into -> -4<=(x-1)<=4
Solve all the 5 like this and you get an E as answer
))
Cheers
Ami/-
|x-1|<=4 can be broken into -> -4<=(x-1)<=4
Solve all the 5 like this and you get an E as answer
))Cheers
Ami/-
