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Difficulty: MediumAbsolute Value Equations and Inequalities

What is the complete solution set for the inequality ∣4−3v∣≥7|4 - 3v| \ge 7?

  1. A
    v≥−1v \ge -1 or v≤113v \le \frac{11}{3}
  2. B
    −1≤v≤113-1 \le v \le \frac{11}{3}
  3. v≤−1v \le -1 or v≥113v \ge \frac{11}{3}Answer
  4. D
    v≤−1v \le -1
  5. E
    v≤−7v \le -7 or v≥7v \ge 7

Answer

The solution set is the union of two open-ended intervals, representing all values of vv such that vv is less than or equal to −1-1 or vv is greater than or equal to 113\frac{11}{3}.
To solve the absolute value inequality ∣4−3v∣≥7|4 - 3v| \ge 7, we rewrite it as two separate inequalities: 4−3v≥74 - 3v \ge 7 or 4−3v≤−74 - 3v \le -7. For the first case, subtracting 4 from both sides of 4−3v≥74 - 3v \ge 7 results in −3v≥3-3v \ge 3. Dividing by −3-3 and reversing the inequality sign gives v≤−1v \le -1. For the second case, subtracting 4 from both sides of 4−3v≤−74 - 3v \le -7 results in −3v≤−11-3v \le -11. Dividing by −3-3 and reversing the inequality sign gives v≥113v \ge \frac{11}{3}. Combining these gives the correct solution set: v≤−1v \le -1 or v≥113v \ge \frac{11}{3}.

Step-by-Step Solution

1
Set up two separate linear inequalities based on the absolute value inequality template.
4−3v≥74 - 3v \ge 7 or 4−3v≤−74 - 3v \le -7
An absolute value inequality of the form ∣A∣≥B|A| \ge B where B>0B > 0 is equivalent to the compound statement A≥BA \ge B or A≤−BA \le -B.
2
Solve the first inequality: 4−3v≥74 - 3v \ge 7. Subtract 4 from both sides, then divide by −3-3 and reverse the inequality sign.
v≤−1v \le -1
Subtracting 4 yields −3v≥3-3v \ge 3. Dividing by the negative number −3-3 requires reversing the direction of the inequality sign.
3
Solve the second inequality: 4−3v≤−74 - 3v \le -7. Subtract 4 from both sides, then divide by −3-3 and reverse the inequality sign.
v≥113v \ge \frac{11}{3}
Subtracting 4 yields −3v≤−11-3v \le -11. Dividing by the negative number −3-3 requires reversing the direction of the inequality sign.
4
Combine the individual solutions into a single compound statement.
v≤−1v \le -1 or v≥113v \ge \frac{11}{3}
The solution to a 'greater than or equal to' absolute value inequality is the union of the individual solutions.

Key Concept

Solving absolute value inequalities by splitting them into two linear cases and correctly reversing the inequality sign when dividing by a negative number.

Alternative Method

Alternatively, test test-values from each interval. For instance, choosing v=0v = 0 (which is in the middle interval) yields ∣4−0∣=4≥7|4 - 0| = 4 \ge 7, which is false. Choosing v=−2v = -2 (which is in the left interval) yields ∣4−3(−2)∣=∣10∣=10≥7|4 - 3(-2)| = |10| = 10 \ge 7, which is true. Choosing v=4v = 4 (which is in the right interval since 4>1134 > \frac{11}{3}) yields ∣4−12∣=∣−8∣=8≥7|4 - 12| = |-8| = 8 \ge 7, which is true. This confirms the solution set must cover v≤−1v \le -1 and v≥113v \ge \frac{11}{3}.
Estimated Time:1m 30s
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Topics

Absolute Value Equations and Inequalities

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