Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

I think its quicker just to square both sides. 30 sec you have the answer and little room for mistakes.
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The question is

Is |x| < 1 ?

Asking whether x falls between 1 and -1, exclusive.

if |x|<1
x<1; 0.5, 0.2,0
and
x>-1; -0.5,-0.2

or -1<x<1

Moment you write x=1 or x=-1; the |x|<1 becomes false.


(1) |x + 1| = 2|x – 1|

For modulus on both sides:

Case I.
solve the equation with no sign change

so +(x+1) = +2(x-1)
x+1=2x-2
x=3. Not between -1 and 1.

Case II.
solve the equation with sign change on one side. Either RHS or LHS. Let's do the sign change on LHS

so -(x+1) = +2(x-1)
-x-1=2x-2
3x=1.
x=1/3
x is between -1 and 1.

CaseII(b):
Even if we did sign change on RHS, we would have gotten the same result. Let's try
+(x+1) = -2(x-1)
x+1=-2x+2
3x=1
x=1/3. Same result as before.

So; sign change should be done for either LHS or RHS.

Now, we have two solutions for x; 1/3(between -1 and 1), 3(not between -1 and 1)

Not sufficient.


###Also please substitute these factors of x into the main equation and check whether the factors indeed satisfy the equation. because say if 3 didn't satisfy the equation and 1/3 does. The statement would be sufficient.###


(2) |x – 3| > 0

Here modulus only on LHS;
So, try this with both signs;

+(x-3) > 0
x-3>0
x>3

and

-(x-3) > 0
-x+3>0
-x>-3
x<3

Here x can be either less than 3 or more than 3. Not 3.

However; this doesn't tell us definitively whether x lies between -1 and 1.

Not Sufficient.

Using both:

We know x can be either (1/3 or 3) by 1st statement. Second statement tells us that x can not be 3.

The only value thus left is 1/3.

Sufficient.

Ans: "C"
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[quote="Bunuel"][quote="mn2010"]
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}.

Hi Bunuel:

Can you please explain why, in B above, we must NOT change the sign in |x+1| leaving it as x+1, and must change the sign in 2|x-1| making it 2(-x+1)?

Thank you.

Yes. You should really try the links I've provided in the above posts.



No that's not the point. When you have absolute values on both sides of the equation expansion can be either + + (or which is the same - - because x + 1 = 2(x - 1) is the same as -(x + 1) = -2(x - 1)) OR + - (or which is the same - + because x + 1 = -2(x - 1) is the same as -(x + 1)=2(x - 1)) so basically only two options.

So as I've written in my solution: "alternately you could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: x + 1 = 2(x - 1)) and then with different sign (for example LHS positive and RHS negative x + 1 = 2(-x + 1)), solve for x both equations, and finally check whether the solutions satisfy |x + 1| = 2|x - 1|."

Here are two more links which might helps to understand this:
absolute-value-ds-100357.html
some-inequalities-questions-93760.html

Hope it helps.
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If (2) were |x – 3| = 0 then yes, we would have that x=3 and in this case this statement would be sufficient, as we could answer NO to the question whether -1<x<1.
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(1) |x + 1| = 2|x – 1|

This has 2 cases.. X>0 and X<0
If X>0 , then X+1 = 2(x-1)
If X<0 , then X+1 = -2(x-1)

Solving these equations we get X= 3 or X= 1/3. Since we have YES and NO situation => Not sufficient

(2) |x – 3| > 0

Solving this equation , we get x>3 or X<3, in either cases, X<> 3. So not sufficient.

(1) + (2) ==> X= 1/3 . Since X<> 3.

So the answer is (c).

Regards,
Mustu

We have to consider 4 cases totally..

1) + +
2) + -
3) - +
4) - -

If u analyse more closely, then u will find that case 1 and 4 ( - and - cancels out) are the same and cases 2 and 3 are the same( + and - 0r - and + are the same). Hope this is clear.

Regards,
Mustu

Exactly. So 1) gives us two options, but we don't know which x is. 2) tells us that x does not equal 3, so we know that x = 1/3
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My Debrief

Thanks 144144

This is a wonderful method

statement 1 :
(X+1)^2 = 4 (x-1)^2

this translated to a quadratic equation

X^2 +6X - 1 = 0 clearly there are 2 answers so not sufficient

Statement 2 translates to X not eqaul to 3

combining both, we have our solution

Kudos to you