If \(y = |x – 1|\) and \(y = 3x + 3,\) then \(x\) must be between which of the following values?

[VJesus12]

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

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT

[Jay@ManhattanReview]

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

[spoiler]OA=D[/spoiler]

Source: Manhattan GMAT

Since \(y = |x – 1|\), we have \(x – 1\) is either positive or negative.

Case 1: Say \(x – 1\) is positive

Thus, \(3x + 3 = x – 1 => x = – 2\). But this is not possible since at \(x=-2\), we have \(x – 1=-2-1=-3,\) a negative value.

Case 2: Say \(x – 1\) is negative

Thus, \(3x + 3 = -x + 1 => x = -1/2\). This is possible since at \(x= -1/2\), we have \(x - 1=-1/2-1=-3/2,\) a negative value.

So, \(x\) lies between – 1 and 0.

The correct answer: D

Hope this helps!

-Jay
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[deloitte247]

Given that y = |x-1| and y = 3x + 3
Then y = 3x + 3 where y = |x-1|
|x-1| = 3x + 3

Now the absolute value or modulus of a real number (x - 1) which is denoted by |x - 1| is the positive value of (x - 1) i.e |x - 1| - (x - 1) if x is positive or greater than or equals 0 and |x - 1| = - (x - 1) if x is negative or less than 0
=> |x - 1| = 3x + 3
$$\pm\left(x-1\right)=3x+3$$


For the positive
+(x - 1) = 3x + 3
x - 1 = 3x + 3
-3x + x = 3 + 1
-2x = 4
x = -2
Since (x - 1) is expected to be positive in this case, let's plug in the value of x to check
x - 1 = -2 -1 = -3; x is not equal to 3

For the negative
-(x - 1) = 3x + 3
-x - 1 = 3x + 3
-3x - x = 3 - 1
-4x/-4 = 2/4
x = -1/2
Since (x - 1) is expected to be negative in this case, let's plug in the value of x to check
x - 1 = -1/2 - 1 = -3/2; x is equals -1/2 as it yields a negative value as expected

So, the value of x is between - 1 and 0

Answer = D

[Scott@TargetTestPrep]

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

[spoiler]OA=D[/spoiler]

Solution:

We can set |x - 1| = 3x + 3 and solve for x. However, we can also rewrite the equation as |x - 1| - 3x - 3 = 0 and let f(x) = |x - 1| - 3x - 3. The argument now is: since f(x) is a continuous function and the values of f(a) and f(b) have opposite signs (i.e., one is positive and the other is negative), then there must be a value of x between a and b such that f(x) = 0.

With this in mind, let’s check the answer choices and let’s start with C first. If it works, we’ve found our answer. If it doesn’t, we can still see that we should move forward or backward.

f(0) = |0 - 1| - 3(0) - 3 = 1 - 0 - 3 = -2

f(1) = |1 - 1| - 3(1) - 3 = 0 - 3 - 3 = -6

We see both values are negative and there are no opposite signs. C is not the correct answer. Now let’s move on to D (notice that since we want a positive sign, the term -3x would be positive when x = -1).

f(-1) = |-1 - 1| - 3(-1) - 3 = 2 + 3 - 3 = 2

We see that f(0) = -2 and f(-1) = 2 have opposite signs in their values, there must be a value of x between 0 and -1 such that f(x) = 0 or |x - 1| = 3x + 3.

Alternate Solution:

Since y = |x - 1|, it must be true that y = x - 1 or y = 1 - x.

If y = x - 1, then solving x - 1 = 3x + 3; we obtain 2x = -4, which is equivalent to x = -2. If x = -2; on the other hand, |x - 1| will not equal x - 1. Indeed, x = -2 produce different values for y when we substitute it in the equalities y = |x - 1| and y = 3x + 3.

If y = 1 - x, then solving 1 - x = 3x + 3; we obtain 4x = -2, which is equivalent to x = -1/2. Substituting x = -1/2, we get |-1/2 - 1| = 3(-1/2) + 3 = 3/2. Thus, the only possible value for x is -1/2. We see that x is between -1 and 0.

Answer: D