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Farina
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 21 Feb 2020, 02:30
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ScottTargetTestPrep
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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Greatly explained. Thanks a lot
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 25 Feb 2020, 08:22
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ScottTargetTestPrep
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

Greatly explained. Thanks a lot
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Sure thing!
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 29 Feb 2020, 10:55
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ScottTargetTestPrep
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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

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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Hi, your solution is simple to understand. I just have one confusion, this is a very basic question related to mod. When we have to remove the mod sign, we put +ve and +ve sign to the equation which is on other side. Like if y = |x-1| then it should be +- y = x-1 and then +- (3x+3) = x-1 . Because that's what we have been doing in modulus questions
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 04 Mar 2020, 13:17
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Hi, your solution is simple to understand. I just have one confusion, this is a very basic question related to mod. When we have to remove the mod sign, we put +ve and +ve sign to the equation which is on other side. Like if y = |x-1| then it should be +- y = x-1 and then +- (3x+3) = x-1 . Because that's what we have been doing in modulus questions
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You can put the ± sign either to the right or to the left of the equation. It will not make a difference.

±y = x - 1 means the following: either y = x - 1 or -y = x - 1

Multiplying -y = x - 1 by -1, we get y = -(x - 1). So, we get "either y = x - 1 or y = -(x - 1)". This is equivalent to saying y = ±(x - 1).
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 28 Aug 2022, 15:32
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Bunuel
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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

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.
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This seems like the best and easiest way to do it. Key term here is BETWEEN. If x = -1/2 and -2 then according to options x falls BETWEEN 0 and -1 right? Yes.
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If y = |x – 1| and y = 3x + 3, then x must be between which of the 17 Oct 2024, 09:16
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\(y = |x – 1|\) 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

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If y = |x – 1| and y = 3x + 3, then x must be between which of the 11 Nov 2025, 10:30
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