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leonidas
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mod(x + 1) = 2 mod(x - 1)

When one solves this with 4 different conditions, one gets the value of x=3 and x=1/3. Is there a shorter way of identifying this equation and solving it only twice (instead of trying all 4 conditions) and get the two values?

I am trying to find some time saving strategies.
I tried the 4 conditions as:
(x+1) , (x-1)
-(x+1), -(x-1)
-(x+1), (x-1)
(x+1), -(x-1)

Hope I am making this clearer.



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In many cases, the fastest way to analyze these equations is to interpret the absolute values as distances on the number line. Recall that |a-b| is the distance between a and b on the number line. So,

|x + 1| = 2 |x - 1|
|x-(-1)| = 2 |x-1|

which just says that x is twice as far from -1 as it is from 1. Clearly there will be one point between -1 and 1 which satisfies this (x = 1/3), and one point to the right of 1 (x=3).
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leonidas
mod(x + 1) = 2 mod(x - 1)

When one solves this with 4 different conditions, one gets the value of x=3 and x=1/3. Is there a shorter way of identifying this equation and solving it only twice (instead of trying all 4 conditions) and get the two values?

I am trying to find some time saving strategies.
I tried the 4 conditions as:
(x+1) , (x-1)
-(x+1), -(x-1)
-(x+1), (x-1)
(x+1), -(x-1)

Hope I am making this clearer.
Show more

essentially these 4 conditions are in fact the same as 2 conditions:

1: if lx+1l and lx-1l both are +ve;
x+1 = 2x-2
x = 3

2: if lx+1l and lx-1l both are -ve;
-x-1 = -2x+2
x = 3


1 & 2 yeild the same value because if we change the side of 2, it becomes 1. lets do it again with if lx+1l and lx-1l both are -ve;
-x-1 = -2x+2
change the side
2x - 2 = x+1 ........ it is the same as 1.

x = 3


3: if lx+1l is +ve and lx-1l is -ve;
x+1 = -2x+2
x = 1/3

4: if lx+1l is -ve and lx-1l is +ve;
-x-1 = 2x-2
x = 1/3

the same applies to 3 and 4 too.

so do only twice i.e one for +ve and another for -ve. 8-)
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can you post the full question ?

Thanks
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leonidas
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IanStewart- Thats one of the best ways of dealing with absolute value problems. I need more practice to think on those lines.
GMAT Tiger- I agree, I should try the equations just twice instead of all 4 conditions. This will save time.

Thanks guys....

Here is the complete problem:

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
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leonidas

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
Show more

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

So answer is C.
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leonidas

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

So answer is C.
Show more

Squaring works or solving 2 equations works just fine.
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amitdgr
leonidas

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

So answer is C.
Show more

I like this one too. So, whenever, there is an absolute vaue on either side, one of the best approaches is to square?
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amitdgr
leonidas

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I like Ian's approach. Definitely faster.

(1)
There is another technique. Square on both sides. This can be applied where there are chances of eliminating the Modulus like we have here in our statement 1

so |x + 1| = 2|x - 1| becomes (x+1)² = 4(x-1)²

x²+1+2x = 4x²+4-8x
3x²-10x+3=0
(3x-1)(x-3)=0

so x can be 1/3 or 3. so (1) is insufficient

(2)
|x-3|≠0 means x is not 3. But this does not tell what x is. So this is also insufficient.

(1) & (2)
(1) says x =1/3 or 3 and (2) says x≠3. So combining both gives x=1/3

So answer is C.
Show more

Good approach. I never thought of this. This really makes calculation fool proof.



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