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Difficulty: Very hardReal Numbers, Number Line, and Absolute Value

If xx and yy are real numbers such that ∣2x−5∣≤9|2x - 5| \le 9 and ∣y+3∣≤4|y + 3| \le 4, what is the maximum possible value of ∣x−2y+1∣|x - 2y + 1|?

  1. 2222Answer
  2. B
    2020
  3. C
    1414
  4. D
    1313
  5. E
    66

Answer

The maximum possible value of ∣x−2y+1∣|x - 2y + 1| is 22.
The correct answer is 22 because solving the absolute value inequalities gives −2≤x≤7-2 \le x \le 7 and −7≤y≤1-7 \le y \le 1. To maximize x−2y+1x - 2y + 1, we take the largest possible value of xx (77), the smallest possible value of yy (−7-7, making −2y=14-2y = 14), and add 11, obtaining 7+14+1=227 + 14 + 1 = 22. Since the lower bound of x−2y+1x - 2y + 1 is −3-3, the maximum absolute value is ∣22∣=22|22| = 22.

Step-by-Step Solution

1
Find the range of possible values for xx from the inequality ∣2x−5∣≤9|2x - 5| \le 9.
−9≤2x−5≤9  ⟹  −4≤2x≤14  ⟹  −2≤x≤7-9 \le 2x - 5 \le 9 \implies -4 \le 2x \le 14 \implies -2 \le x \le 7.
An absolute value inequality ∣A∣≤B|A| \le B unwraps to −B≤A≤B-B \le A \le B.
2
Find the range of possible values for yy from the inequality ∣y+3∣≤4|y + 3| \le 4.
−4≤y+3≤4  ⟹  −7≤y≤1-4 \le y + 3 \le 4 \implies -7 \le y \le 1.
Unwrapping the absolute value inequality gives the upper and lower bounds for yy on the real number line.
3
Determine the range of possible values for −2y-2y.
Multiplying −7≤y≤1-7 \le y \le 1 by −2-2 and reversing the inequality signs yields −2(1)≤−2y≤−2(−7)  ⟹  −2≤−2y≤14-2(1) \le -2y \le -2(-7) \implies -2 \le -2y \le 14.
Multiplying an inequality by a negative number reverses the direction of the inequality signs.
4
Combine the ranges to find the minimum and maximum bounds for z=x−2y+1z = x - 2y + 1.
Minimum z=(−2)+(−2)+1=−3z = (-2) + (-2) + 1 = -3; Maximum z=7+14+1=22z = 7 + 14 + 1 = 22. Thus, −3≤x−2y+1≤22-3 \le x - 2y + 1 \le 22.
Adding the individual minimums gives the absolute minimum, and adding the individual maximums gives the absolute maximum.
5
Calculate the maximum value of the absolute value ∣x−2y+1∣|x - 2y + 1|.
max(∣x−2y+1∣)=max(∣−3∣,∣22∣)=22\\max(|x - 2y + 1|) = \\max(|-3|, |22|) = 22.
The absolute value of a quantity ranging from −3-3 to 2222 reaches its maximum distance from zero at 2222.

Key Concept

Properties of Real Numbers, Number Line Inequalities, and Absolute Value Operations
Estimated Time:2m 30s
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Topics

Real Numbers, Number Line, and Absolute Value

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