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

GMATinsight · Accepted 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

Bunuel · 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.

tjerkrintjema · Answer

I find this one hard, as these formulas should be equal to each other but when testing for values i get different answers on both y's

let's say x = 0,

0 - 1 = -1 = absolute 1

3*0 +3 = 3

So different values

----

x = 1

1-1 = 0 = absolute 0

3*1 + 3 = 6

Also different

---

x = -1

-1-1 = -2 absolute 2
3*-1 + 3 = 0

Also different

----

x = 2

2 -1 = 1 absolute 1
3*2 + 3 = 9

also different

-----

x = -2

-2-1 = -3 absolute 3
3*-2 +3 = -3

also different

---

x = 3

3-1 = 2 absolute 2
3*3 + 3 = 12

also different..

I guess I'm not understanding the question here, reading it wrong. But I would like some help here.

Noob23 · Answer

Hi tjerkrintjema,

I think you are missing the word "Between" in the question

The ans is "D"

x = -0.5

-0.5-1 = -1.5 absolute 1.5
-0.5*3 + 3 = 1.5

Hope this helps

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

SKas07 · Answer

So I solved for y initially to find the distance. One answer was -3 and the second for 3/2. Since distance from 1 cannot be negative, only 3/2 is valid.

But why then can it not be between 2 and 3 as well? Is that because it x as 5/2 needs to satisfy y=3x+3? That is the only reason I could think of. The possible answer I solved for was -1/2 and 5/2 (distance of 3/2 from 1 but only one of them is correct).

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

CyberStein · Answer

|x-1| = 3x + 3
therefore, x must be a negative number
Plug in -1.5 for x
2.5 =/ -4.5 + 3, eliminate E
plug in .5 for x
1.5 = 3(-.5) + 3 = 1.5

Therefore, the answer is (D) -1 and 0

ank4u · Answer

we can two cases x>1 and X<1 , for equation |x-1|=3x+3 for x>1,x=-2 cant be taken. For x<1, x=-1/2 ,hence and D

sasyaharry · Answer

A graph is the "best way to approach this problem" ? Not really.

Equating |x-1| to 3x+3 is quick and painless.

CAMANISHPARMAR · Answer

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

Princ · Answer

OA: D
There are two method
1st method
We have to find the intersection point of \(y=|x−1|\) and \(y=3x+3\),
if \(x<1\)
then \(y=|x−1|\) becomes \(y =-(x-1)\)
intersection point of \(y =-(x-1)\) and \(y=3x+3\) will be coordinate \((-\frac{1}{2},\frac{3}{2})\)

if \(x\geq{1}\)
then \(y=|x−1|\) becomes \(y =x-1\)
intersection point of \(y=x-1\) and \(y=3x+3\) will be coordinate \((-2,-3)\).
This cannot be possible as we are finding the solution for condition \(x\geq{1}\)

2nd method graphical method
intersection of \(y=|x−1|\) and \(y=3x+3\) can be seen at \((-\frac{1}{2},\frac{3}{2})\)

Attachment:

graphical method.PNG [ 36.54 KiB | Viewed 22948 times ]

from both method, it can be seen that x is falling between -1 and 0.

Bunuel · Answer

Merging topics. Please check the discussion above.

Kinshook · Answer

Given: 
1. y = |x – 1| and 
2. y = 3x + 3,

Asked: x must be between which of the following values?

y = |x-1| = 3x+3

Region 1: x>1
x -1 = 3x +3 
2x = -4 
x = -2 
NOT FEASIBLE

Region 2: x<=1
1-x = 3x + 3
4x = -2
x = -.5

IMO D

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

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

Prep Toolkit

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