If y = |x – 1| and y = 3x + 3, then x must be between which of the

Question

If y = |x – 1| and y = 3x + 3, then x must be between which of the following values?

(A) 2 and 3
(B) 1 and 2
(C) 0 and 1
(D) –1 and 0
(E) –2 and –1

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

We should set the two equations for y equal and algebraically solve |x – 1| = 3x + 3 for x.

This requires two solutions: one for the case that x – 1 is positive, the other for the case that x –1 is negative:

x – 1 is positive:  (x – 1) = 3x + 3 --> x = -2.
INCORRECT: x – 1 = (–2) – 1 = –3
Therefore, the original assumption that x – 1 is positive does not hold.

x – 1 is negative:  –(x – 1) = 3x + 3 --> x = -1/2.
CORRECT: Therefore, the original assumption that x – 1 is negatives holds.

Thus, there is only one solution for x and y, which is y = 3/2 and x = -1/2.

The correct answer is D.

GSBae · Answer

Since we're finding the intersection points of two relatively easy functions to think about visually, the best way to approach this problem is with a graph.

https://i.imgur.com/3FIBDiR.png

The first equation should be a simple absolute value function with the bend at x = 1.

The second is a basic line with y intercept at y = 3 and x intercept at x = -1.

Based on these two functions, we see that they have to intersect somewhere between - 1 and 0 .

Answer: D

GMATinsight · Answer

y = |x – 1| confirms that y is non-negative value

i.e. 3x +3 must be non negative

i.e. 3x +3  >  0
i.e. 3x  >  -3
 i.e. x  >  -1

Now |x – 1| = 3x + 3
i.e.  + (x-1) = 3x + 3
i.e. x-1 = 3x + 3 
OR -x + 1 = 3x + 3

i.e. x = -4
OR x = -0.5

So we can conclude that x can't be -4 as it has to be greater than -1
hence x must be -0.5 only

Answer: Option  D

brandon7 · Answer

Bunuel, this is was the same thing I did, but I just had one question. Isn't it more accurate to say for the case where x>1, that since it leads to x=-2, this is an invalid solution because -2 is not >1, rather than saying -2 is invalid because -2 <0? Like when are testing solutions to be valid we are making sure that fall within the range, not just if they are positive or negative right?

Bunuel · Answer

Here is another way, which might help: |x – 1| = 3x + 3 When x - 1 < 0 , so when x < 1 we'll get -(x - 1) = 3x + 3 --> x = -\frac{1}{2} : keep because -\frac{1}{2} IS in the range x < 1 . When x - 1 \geq 0 , so when x \geq 1 we'll get x - 1 = 3x + 3 --> x = -2 : discard because -2 is NOT in the range x \geq 1 ). So, we got that |x – 1| = 3x + 3 has only one solutions x = -\frac{1}{2} . Answer: D. Hope it helps.

nitishmalagi · Answer

+1 kudos Bunuel. Nice expln

kaleem765 · Answer

Bunuel  ... this is how I approached this question. is it ok to do this way?

y=|x-1| so y=(x-1) or y=-(x-1)

First consider y=(x-1)
since y=3x+3 we have an equation 3x+3=x-1, solving this equation gives x=-2 which when plugged in the given equations does not hold. Plugging x=-2 in y=|x-1| gives y=3 and Plugging x=-2 in y=3x+3 gives y=-3 so x=-2 is not the desired solution.

Now consider y=-(x-1) 
Using the same approach, we get x=-1/2 which when plugged in the given equations gives y=3/2 in both cases. Plugging x=-1/2 in y=|x-1| gives y=3/2 and Plugging x=-1/2 in y=3x+3 gives y=3/2 so x=-1/2 is the desired solution. It lies within -1 and 0 so D is the right answer.

Bunuel · Answer

Yes, you can expand absolute value with positive sign and negative sing, solve and substitute back to check the validity of the roots.

ScottTargetTestPrep · Answer

Using substitution, we have:

|x – 1| = 3x + 3

For absolute value questions, we always have two cases: when (x - 1) is positive and when (x - 1) is negative:

Case 1: when (x - 1) is positive:

x - 1 = 3x + 3

-4 = 2x

-2 = x

When (x - 1) is negative:

-(x - 1) = 3x + 3

-x + 1 = 3x + 3

-2 = 4x

x = -1/2

We have two possible solutions: x = -2 and x = -½. However, x = -2 is not a solution, since it doesn’t satisfy the equation:

|-2 - 1| = 3(-2) + 3 ?

|-3| = -6 + 3 ?

3 = -3 ? → False

On the other hand, x = -1/2 is a solution, since it does satisfy the equation:

|-1/2 - 1| = 3(-1/2) + 3 ?

|-3/2| = -3/2 + 3 ?

3/2 = 3/2 ? → True

Thus, x is between -1 and 0.

Answer: D

BrushMyQuant · Answer

y = |x – 1|  and  y = 3x + 3,

Let's substitute the y = 3x + 3 in  y = |x – 1| , we get 
3x + 3 = |x – 1|
=> | x-1 | 3x + 3 ..(1)

Let's solve it using two methods

Method 1: Algebra

( To MASTER Absolute Value Problems,  watch this video  )

As we have |x-1| in the equation so we will have two cases 
 
   -   Case 1:  x - 1 ≥ 0 => x ≥ 1 
=> | x-1 | = x-1
=> x-1 = 3x + 3 (From (1))
=> 2x = -4
=> x = -2

But our condition was x ≥ 1
=>   NO SOLUTION   
  -   Case 2:  x ≤ 1 
=> | x-1 | = -(x-1)
=> -(x-1) = 3x + 3 (From (1))
=> -x + 1 = 3x + 3
=> 4x = -2
=> x = -2/4 = -0.5

But our condition was x ≤ 1, and -0.5 ≤ 1
=>  x = -0.5 is a  SOLUTION

Method 2: Substitution

| x-1 | = 3x + 3 
Let's pick values in the range of each option choice and see if it satisfies the above equation

(A) 2 and 3 
x = 2.5
=> | 2.5-1 | = 3*2.5 + 3 => 1.5 = 10.5 =>   NOT POSSIBLE

(B) 1 and 2 
x = 1.5
=> | 1.5-1 | = 3*1.5 + 3 => 0.5 = 7.5 =>   NOT POSSIBLE

(C) 0 and 1 
x = 0.5
=> | 0.5-1 | = 3*0.5 + 3 => 0.5 = 4.5 =>   NOT POSSIBLE

(D) -1 and 0 
x = -0.5
=> | -0.5-1 | = 3*-0.5 + 3 => 1.5 = 1.5 =>   POSSIBLE  
We don't need to solve further, but solving to complete the solution

(E) -2 and -1 
x = -1.5
=> | -1.5-1 | = 3*(-2.5) + 3 => 2.5 = -4.5 =>   NOT POSSIBLE

So,  Answer will be D 
Hope it helps!

To learn how to solve absolute value problems, watch the following video

https://www.youtube.com/watch?v=huFiBk8PaGY

work2 · Answer

how can x = -4? shouldn't it be x= -2?

Bunuel · Answer

You are right. The solution you quote has a typo.

From x - 1 = 3x + 3 we get x = -2.

However, x = -2 does  not  satisfy the original equation, because |x - 1| is positive while 3x + 3 would be negative.

The other case gives x = -0.5, which works.

So the final answer is still D.

If y = |x – 1| and y = 3x + 3, then x must be between which of the

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As we have \|x-1\| in the equation so we will have two cases
-Case 1: x - 1 ≥ 0 => x ≥ 1 => \| x-1 \| = x-1 => x-1 = 3x + 3 (From (1)) => 2x = -4 => x = -2 But our condition was x ≥ 1 => NO SOLUTION	-Case 2: x ≤ 1 => \| x-1 \| = -(x-1) => -(x-1) = 3x + 3 (From (1)) => -x + 1 = 3x + 3 => 4x = -2 => x = -2/4 = -0.5 But our condition was x ≤ 1, and -0.5 ≤ 1 => x = -0.5 is a SOLUTION

GMAT Problem Solving (PS) Questions

If y = |x – 1| and y = 3x + 3, then x must be between which of the

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