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Difficulty: HardLinear Inequalities in One Variable

For a constant aa, the inequality 5x−a(3−2x)≥45x - a(3 - 2x) \ge 4 has a solution set of the form x≤dx \le d, where dd is a constant. Which of the following must be true about the value of aa?

  1. a<−52a < -\frac{5}{2}Answer
  2. B
    a>−52a > -\frac{5}{2}
  3. C
    a>52a > \frac{5}{2}
  4. D
    a<52a < \frac{5}{2}

Answer

The value of aa must satisfy a<−52a < -\frac{5}{2}.
To find the correct range for aa, we first expand the inequality 5x−a(3−2x)≥45x - a(3 - 2x) \ge 4 using the distributive property, which yields 5x−3a+2ax≥45x - 3a + 2ax \ge 4. Grouping the xx terms gives (5+2a)x≥3a+4(5 + 2a)x \ge 3a + 4. The problem states that the solution set is of the form x≤dx \le d. Because the inequality sign flipped from greater-than-or-equal-to (≥\ge) to less-than-or-equal-to (≤\le), the coefficient of xx must be negative. Setting the coefficient 5+2a<05 + 2a < 0 and solving for aa gives a<−52a < -\frac{5}{2}.

Step-by-Step Solution

1
Expand the inequality to separate the terms.
5x−3a+2ax≥45x - 3a + 2ax \ge 4
Apply the distributive property to the term −a(3−2x)-a(3 - 2x), paying close attention to the signs: −a×3=−3a-a \times 3 = -3a and −a×(−2x)=2ax-a \times (-2x) = 2ax.
2
Group and factor the terms containing xx on the left side, and move the constant terms to the right side.
(5+2a)x≥3a+4(5 + 2a)x \ge 3a + 4
Factor out xx from the terms 5x5x and 2ax2ax to isolate the variable, and add 3a3a to both sides of the inequality.
3
Analyze the relationship between the coefficient of xx and the inequality sign of the solution set.
5+2a<05 + 2a < 0
The original inequality has a greater-than-or-equal-to sign (≥\ge), but the given solution set is of the form x≤dx \le d (less-than-or-equal-to). For the inequality sign to reverse when dividing both sides by the coefficient of xx, the coefficient (5+2a)(5 + 2a) must be negative.
4
Solve the inequality for aa.
a<−52a < -\frac{5}{2}
Subtract 5 from both sides to get 2a<−52a < -5, then divide both sides by 2.

Key Concept

Solving linear inequalities in one variable with symbolic coefficients and applying the inequality sign reversal rule when multiplying or dividing by a negative value.
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Topics

Linear Inequalities in One Variable

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