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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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(A) 2 and 3
(B) 1 and 2
(C) 0 and 1
(D) –1 and 0
(E) –2 and –1
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Bunuel
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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GSBae
Joined: 23 May 2013
Last visit: 07 Mar 2025
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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
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
General Discussion
Bunuel
y = |x – 1| confirms that y is non-negative value
i.e. 3x +3 must be non negative [because y = 3x + 3]
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
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Bunuel
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?
brandon7
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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nitishmalagi
Joined: 17 Mar 2017
Last visit: 10 Apr 2018
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Concentration: Strategy, International Business
Posts: 13
+1 kudos Bunuel. Nice expln
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.
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.
kaleem765
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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Bunuel
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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\(y = |x – 1|\) 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 )
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
_________________
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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All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
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how can x = -4? shouldn't it be x= -2?
GMATinsight
work2
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.
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