OG 130

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

dmateer25 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

Sorry I am a bit confused with how you get these numbers. Can you plz provide some help.. thanks

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.

E

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/-

GMATGuruNY wrote:
... text removed...

Thanks a lot GMATGuruNY! Very well explained. Kudos++

I am more confident with the 2nd method