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Difficulty: Very hardAbsolute Value Equations and Inequalities

For all real numbers yy that satisfy the inequality ∣∣y−3∣−6∣≤2||y - 3| - 6| \leq 2, what is the sum of all possible integer values of yy?

  1. A
    −15-15
  2. 3030Answer
  3. C
    4545
  4. D
    5151
  5. E
    6565

Answer

The sum of all possible integer values of yy is 3030.
The compound inequality 4≤∣y−3∣≤84 \leq |y - 3| \leq 8 splits into two parts: ∣y−3∣≤8|y - 3| \leq 8 (which gives −5≤y≤11-5 \leq y \leq 11) and ∣y−3∣≥4|y - 3| \geq 4 (which gives y≤−1y \leq -1 or y≥7y \geq 7). The intersection of these intervals is [−5,−1]∪[7,11][-5, -1] \cup [7, 11]. Summing all the integers in these intervals gives (−5+−4+−3+−2+−1)+(7+8+9+10+11)=−15+45=30(-5 + -4 + -3 + -2 + -1) + (7 + 8 + 9 + 10 + 11) = -15 + 45 = 30.

Step-by-Step Solution

1
Set up the compound inequality representing the outer absolute value.
−2≤∣y−3∣−6≤2-2 \leq |y - 3| - 6 \leq 2
By definition, ∣u∣≤a|u| \leq a is equivalent to −a≤u≤a-a \leq u \leq a for a≥0a \geq 0.
2
Isolate the inner absolute value term by adding 66 to all parts of the inequality.
4≤∣y−3∣≤84 \leq |y - 3| \leq 8
Isolating the absolute value allows us to split the compound inequality into two separate cases.
3
Split the compound inequality into two separate inequalities and solve each one.
∣y−3∣≤8|y - 3| \leq 8 and ∣y−3∣≥4|y - 3| \geq 4
The expression ∣y−3∣|y - 3| must be simultaneously less than or equal to 88 and greater than or equal to 44.
4
Solve the first inequality, ∣y−3∣≤8|y - 3| \leq 8.
−5≤y≤11-5 \leq y \leq 11
Rewriting the inequality gives −8≤y−3≤8-8 \leq y - 3 \leq 8, and adding 33 to all parts yields the interval [−5,11][-5, 11].
5
Solve the second inequality, ∣y−3∣≥4|y - 3| \geq 4.
y≤−1y \leq -1 or y≥7y \geq 7
By definition, ∣u∣≥a|u| \geq a is equivalent to u≥au \geq a or u≤−au \leq -a. Thus, y−3≥4  ⟹  y≥7y - 3 \geq 4 \implies y \geq 7, and y−3≤−4  ⟹  y≤−1y - 3 \leq -4 \implies y \leq -1.
6
Determine the intersection of the two solution sets.
y∈[−5,−1]∪[7,11]y \in [-5, -1] \cup [7, 11]
The values of yy must lie within [−5,11][-5, 11] and also satisfy y≤−1y \leq -1 or y≥7y \geq 7.
7
Identify the integers within the final intervals and calculate their sum.
Sum = 3030
The integers in [−5,−1][-5, -1] are −5,−4,−3,−2,−1-5, -4, -3, -2, -1 (sum = −15-15). The integers in [7,11][7, 11] are 7,8,9,10,117, 8, 9, 10, 11 (sum = 4545). The total sum is −15+45=30-15 + 45 = 30.

Key Concept

Solving compound and nested absolute value inequalities

Alternative Method

Instead of solving the inequality algebraically, one can test the integer values around the critical points. Since yy must satisfy 4≤∣y−3∣≤84 \leq |y - 3| \leq 8, we can see that the distance of yy from 33 must be between 44 and 88 units. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the right of 33 are 7,8,9,10,117, 8, 9, 10, 11. The integers at distance 4,5,6,7,84, 5, 6, 7, 8 to the left of 33 are −1,−2,−3,−4,−5-1, -2, -3, -4, -5. Summing these ten integers yields 3030.
Estimated Time:3m 0s
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Topics

Absolute Value Equations and Inequalities

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