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GRE Prep Club Team Member
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If y = |x – 1| and y = 3x + 3, then x must be between which of the
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23 Apr 2021, 15:32
23 Apr 2021, 15:32
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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
(A) 2 and 3
(B) 1 and 2
(C) 0 and 1
(D) –1 and 0
(E) –2 and –1
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Schools: XLRI Jamshedpur, India - Class of 2014
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If y = |x 1| and y = 3x + 3, then x must be between which of the
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16 Jul 2021, 10:38
16 Jul 2021, 10:38
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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
General Discussion
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Re: If y = |x – 1| and y = 3x + 3, then x must be between which of the
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25 Apr 2021, 01:53
25 Apr 2021, 01:53
1
GeminiHeat wrote:
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
(A) 2 and 3
(B) 1 and 2
(C) 0 and 1
(D) –1 and 0
(E) –2 and –1
Let's equate both the equations;
\(|x - 1| = 3x + 3\)
Case I:
\(x - 1 = 3x + 3\)
\(-2x = 4\)
\(x= -2\)
Case II:
\(-(x - 1) = 3x + 3\)
\(-x + 1 = 3x + 3\)
\(-4x = 2\)
\(x = \frac{-1}{2}\)
If we plug in both the values of \(x\) to check if it satisfies the equation or not [Which we must always do]
We can see that \(x = -2\) cannot be the solution.
\(|-2 - 1| = 3(-2) + 3\)
\(|-3| = -6 + 3\)
\(3 ≠ -3\)
So only \(x = \frac{-1}{2}\) can be the solution which lies between \(0\) and \(-1\)
Hence, option D
