If y = |x – 1| and y = 3x + 3, then x must be between which of the : Problem Solving (PS)

tjerkrintjema

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04 Jun 2015, 05:18

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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

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04 Jun 2015, 06:35

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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

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25 Jul 2016, 06:43

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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.

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

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09 Aug 2017, 16:46

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Bunuel

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

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.

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

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Bunuel

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

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.

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

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02 Sep 2017, 05:04

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brandon7

Bunuel

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

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.

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?

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.

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23 Sep 2017, 09:57

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+1 kudos Bunuel. Nice expln

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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

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kaleem765

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.

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

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04 Oct 2017, 16:36

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Bunuel

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

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

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Bunuel

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

|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

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Bunuel

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

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

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eaze

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.

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

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

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

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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

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19 May 2018, 09:51

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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 19830 times ]

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

Bunuel

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19 May 2018, 14:21

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CAMANISHPARMAR

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

Merging topics. Please check the discussion above.

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Bunuel

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

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

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