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Archit3110
Bunuel
Is x equal to –1?

(1) |3x – 2| + |3x + 2| = 6
(2) |x – 1| + |3x + 2| = 5

solving for #1
|3x – 2| + |3x + 2| = 6
gives us x=1 and -1
insufficeint
#2|x – 1| + |3x + 2| = 5
x=1
so x is not = -1
IMO B
Can you please explain it in little detail?
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sourabhsashwat
Archit3110
Bunuel
Is x equal to –1?

(1) |3x – 2| + |3x + 2| = 6
(2) |x – 1| + |3x + 2| = 5

solving for #1
|3x – 2| + |3x + 2| = 6
gives us x=1 and -1
insufficeint
#2|x – 1| + |3x + 2| = 5
x=1
so x is not = -1
IMO B
Can you please explain it in little detail?


When you will Solve For 1:

There Will be 4 Cases:
1) 3x-2 + 3x+2=6 which gives x=1
2)-3x+2 +3x+2=6 Since x will cancel out so no values of x
3)3x-2-3x-2=6 Again x will cancel out
4)-3x+2-3x-2=6 which gives x=-1

When you will solve For 2:

Again, there will be 4 cases:
1)x-1+3x+2=5 which gives x=1
2)-x+1+3x+2 which gives x=1
3)x-1-3x-2=5 which gives x=-4
4)-x+1-3x-2=5 which gives x=-3/2

B Definitely Doesn't gives us value x=-1, therefore the answer is B

Whenever you open the mod. Take 2 cases : |x|=+x
|x|=-x
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can you solve this equation for -+ ....this case cannot exist if we use the line method?
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