What is the value of x? (1) |2x − 1| = 3x + 6 (2) x^2 = 1

Question

What is the value of x?

(1) |2x − 1| = 3x + 6
(2) x^2 = 1

Mo2men · Accepted Answer

Hi Max,

When you deal with such statement, you need to add extra step. You need to check the solutions you get in the original equation.

x =-7, substitute in original equation: |2x − 1| = 3x + 6

|-14 − 1| = -21 + 6......15=-15.........so, x=-7 is not viable solution.

x=-1
|- 3| = 3........3=3..............so x=-1 is viable solution

So we have only one answer.

Wereheretotakeover · Answer

Hi,

please correct me if I'm wrong.

Statement 1:
|2x − 1| = 3x + 6

equals:
2x-1 = 3x+6
x=-7

2x-1 = -(3x+6)
x=1

Statement 1 alone is not sufficient because x could either be 1 or -7

Statement 2:
x²=1

x could either be 1 or -1

Statement 2 alone is not sufficient

(C): Statement 1 and 2 combined are sufficient because x must equal 1

I hope my answer is right

Greetings
Max

Nunuboy1994 · Answer

Have to be careful with the algebra here

Stmnt 1

2x- 1 = 3x + 6 
 -1= x + 6
 x = -7

-l2x -1l = 3x + 6
 -2x + 1 = 3x +6 
 1= 5x +6
 -5 =5x
 x = -1

Be careful- only negative one satisfies the equation so x= -7 cannot be a value of X

Suff

Stmnt 2

X^2= 1
x= -1 , 1

Insuff

A

pikolo2510 · Answer

Hey Bunuel

I need some help. I'm going wrong somewhere but I can't identify where. Below are the steps :-
(I'm trying to follow the same steps you used here in this post https://gmatclub.com/forum/what-is-x-12 ... l#p1037498)

Statement I: -
|2x − 1| = 3x + 6
LHS is non-negative, therefore RHS should alos be non-negative

3x+6>=0
x>= -2

2x − 1 = 3x + 6
Solving the above equation, we get
x= -7
We should discard this value as -7 < -2. For RHS to be positive we need something >= -2

How do I proceed ahead after this step? I want to use the conceptual method you used in the post I mentioned above. Please help. Thanks

Bunuel · Answer

You cannot use this method here becasue not for all values of x which are >= -2, the expression in the modulus (2x − 1) is positive.

In the problem you refer to we have |x| = 3x – 2. RHS must be >= 0, so x >= 2/3. Now, if x >= 2/3, then |x| = x, so we can write x = 3x - 2, which is not the case for the problem above.

pikolo2510 · Answer

Got it, Thanks a ton Bunuel!

GMATYoda · Answer

For the first statement you need to use the two case approach for absolute values.

|2x−1|=3x+6 means that: 3x+6 could equal 2x−1, in which case: 
x=−7
or, 3x+6 could equal −(2x−1) in which case: 
3x+6=−2x+1, so 
5x=−5
and therefore:

x=−1
So with two possible values it would be very tempting to say that statement 1 is not sufficient, then recognize that while statement 2 is clearly not sufficient on its own, it eliminates x=−7as a possibility when you use the two statements together. But wait!

If you return to your work from statement 1 to plug your solutions back in for a quick logic test, you'll see that −7 is an extraneous solution: |2(−7)−1|=3(−7)+6
means that:

|−15|=−15

Which does not work, since the absolute value on the left means that the left-hand side will be POSITIVE 15, while the right is stuck at NEGATIVE 15. This is known as an extraneous solution, and is why the process for solving absolute values always includes the step "plug your solutions back into the equation to verify that they are valid." Here, since −7 is invalid, statement 1 guarantees that x=−1 is the sole solution, and statement 1 is therefore sufficient. The correct answer is A.

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What is the value of x? (1) |2x − 1| = 3x + 6 (2) x^2 = 1

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