Which of the following satisfies all the values of x such that |x+1| +

Question

Which of the following satisfies all the values of x such that  |x+1| + |x-2| < 7  ?

(A)  x > 3

(B)  x < 0

(C)  2 < x < 6

(D)  -3 < x < 4

(E)  -4 < x < 1

Charli08 · Accepted Answer

Since we have an absolute value we need to solve for 2 cases, positive and negative.
**REMEMBER: When dividing an inequality by a negative number the sign is flipped!

i) (x+1) + (x-2) <7

Solving we get x < 4

ii) Multiply each bracket by -1

(-x-1) + (-x+2) < 7

-2x + 1 < 7

x > -3

===> D

Markw1996 · Answer

To solve this one, we need to open up the absolute values and try all possibilities.

possibility 1: |x+1|+|x−2| both positive --> x+1 + x-2 < 7 --> 2x-1<7 -->x<4

possibility 2: |x+1|+|x−2| both negative --> -(x+1) - (x-2) <7 --> -2x <6 -->x>-3

possibility 3: |x+1| positive& |x−2| negative --> (x+1) + -(x-2) --> x cancel each other out . This not an option

Possibility 4: |x+1| negative & |x−2| positive --> -(x+1) + (x-2) --> x cancel each other out . This not an option

Hence we have the final solution of -3<x<4

BrushMyQuant · Answer

Given that |x+1| + |x-2| < 7 and we need to find the range for all possible values of x

Let's solve the problem using Substitution

We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)

(A) x > 3

Lets take x = 4 (which falls in this range of x > 3) and substitute in the equation |x+1| + |x-2| < 7
=> |4+1| + |4-2| < 7
=> |5| + |2| < 7
=> 5 + 2 < 7
=> 7 < 7 which is FALSE

(B) x < 0

Lets take x = -10 (which falls in this range of x < 0) and substitute in the equation |x+1| + |x-2| < 7
=> |-10+1| + |-10-2| < 7
=> |-9| + |-12| < 7
=> 9 + 12 < 7
=> 21 < 7 which is FALSE

(C) 2 < x < 6

We can again take x = 4 to prove this one FALSE

(D) -3 < x < 4

Lets take x = 0 (which falls in this range of -3 < x < 4) and substitute in the equation |x+1| + |x-2| < 7
=> |0+1| + |0-2| < 7
=> |0| + |-2| < 7
=> 0 + 2 < 7
=> 2 < 7 which is TRUE

(E) -4 < x < 1

Lets take x = -3 (which falls in this range of -4 < x < 1) and substitute in the equation |x+1| + |x-2| < 7
=> |-3+1| + |-3-2| < 7
=> |-2| + |-5| < 7
=> 2 + 5 < 7
=> 7 < 7 which is FALSE

So, Answer will be D
Hope it helps!

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lisahoang1301 · Answer

We solve piecewise at the absolute-value breakpoints x=−1 and x=2.

Region 1: x<−1 
∣x+1∣=−x−1,∣x−2∣=−x+2
Sum =(−x−1)+(−x+2)=−2x+1<7 ⇒ −2x<6 ⇒ x>−3
Combine with x<−1 gives −3<x<−1

Region 2: −1≤x<2 
∣x+1∣=x+1,∣x−2∣=−x+2 
Sum = (x+1)+(-x+2) = 3 < 7, which is true for every x in this interval. So the whole interval  Region 3: x≥2 
∣x+1∣=x+1,∣x−2∣=x−2
Sum =(x+1)+(x−2)=2x−1<7 ⇒ 2x<8 ⇒ x<4
Combine with x≥2 gives 2≤x<4

Combine all solutions: (−3,−1)∪ (D) −3<x<4

Which of the following satisfies all the values of x such that |x+1| +

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Which of the following satisfies all the values of x such that |x+1| +

Prep Toolkit

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