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Difficulty: Very hardSolving Linear Inequalities

A logistics company determines that its daily operating cost, CC (in dollars), for a delivery truck satisfies the inequality a(2C−3)3+54≤C−a2−78\frac{a(2C - 3)}{3} + \frac{5}{4} \leq \frac{C - a}{2} - \frac{7}{8}, where aa is a constant regional fuel efficiency parameter such that a<−2a < -2. Which of the following represents the range of possible operating costs CC?

  1. C≥51−12a12−16aC \geq \frac{51 - 12a}{12 - 16a}Answer
  2. B
    C≤51−12a12−16aC \leq \frac{51 - 12a}{12 - 16a}
  3. C
    C≥4a+8928−16aC \geq \frac{4a + 89}{28 - 16a}
  4. D
    C≥228a+2112−16aC \geq \frac{228a + 21}{12 - 16a}
  5. E
    C≤36a+5116a+12C \leq \frac{36a + 51}{16a + 12}

Answer

The range of possible operating costs is C≥51−12a12−16aC \geq \frac{51 - 12a}{12 - 16a}.
To solve the inequality, we first eliminate the denominators by multiplying the entire inequality by 24, resulting in 8a(2C−3)+30≤12(C−a)−218a(2C - 3) + 30 \leq 12(C - a) - 21. Expanding both sides and gathering all terms with CC on the left gives (16a−12)C≤12a−51(16a - 12)C \leq 12a - 51. Because a<−2a < -2, the coefficient 16a−1216a - 12 is negative. Dividing by a negative number reverses the inequality direction, giving C≥12a−5116a−12C \geq \frac{12a - 51}{16a - 12}. Multiplying the numerator and denominator by −1-1 yields the correct solution.

Step-by-Step Solution

1
Clear the denominators by multiplying all terms by the least common multiple of 3, 4, 2, and 8, which is 24.
8a(2C−3)+30≤12(C−a)−218a(2C - 3) + 30 \leq 12(C - a) - 21
Eliminating fractions simplifies the algebraic manipulation of the linear inequality.
2
Expand both sides of the inequality.
16aC−24a+30≤12C−12a−2116aC - 24a + 30 \leq 12C - 12a - 21
Distributing terms allows grouping the variable CC and the constants.
3
Isolate the terms containing CC on the left side and all other terms on the right side.
16aC−12C≤12a−5116aC - 12C \leq 12a - 51
Grouping like terms is necessary to solve for CC.
4
Factor out CC on the left side.
(16a−12)C≤12a−51(16a - 12)C \leq 12a - 51
This isolates the variable CC with a single coefficient.
5
Determine the sign of the coefficient (16a−12)(16a - 12) based on the condition a<−2a < -2.
Since a<−2a < -2, we have 16a<−3216a < -32, which implies 16a−12<−4416a - 12 < -44. Thus, the coefficient is negative.
Knowing whether the coefficient is positive or negative determines whether the inequality sign must flip upon division.
6
Divide both sides by (16a−12)(16a - 12) and reverse the inequality sign.
C≥12a−5116a−12C \geq \frac{12a - 51}{16a - 12}
Dividing an inequality by a negative number requires reversing the direction of the inequality sign.
7
Simplify the resulting fraction by multiplying the numerator and denominator by −1-1.
C≥51−12a12−16aC \geq \frac{51 - 12a}{12 - 16a}
This yields the simplified final expression matching the target choice.

Key Concept

Solving linear inequalities involving fractions and variable parameters, with strict application of the inequality sign-flip rule when dividing by a negative algebraic term.
Estimated Time:3m 0s
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Topics

Solving Linear Inequalities

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