4.3 Linear Inequalities & Word Problem Translations

4.3 Linear Inequalities & Word Problem Translations

While an equation expresses an exact equality between two quantities, an inequality describes an order or boundary relationship between expressions. In real-world scenarios—such as budgeting limits, speed boundaries, or minimum height requirements—situations are governed by constraints rather than single exact values. On the Praxis 5003 exam, candidates must master solving linear inequalities, graphing solution sets on number lines, and accurately translating complex verbal statements into algebraic models.

Inequality Symbols & Graphic Representations

There are four primary inequality symbols used in algebra, classified as either strict or non-strict:

Graphing Solutions on a Number Line

• Open Circle ($\u25CB$): Indicates that the boundary value is not included in the solution set (used for strict inequalities $<$ and $>$).
• Closed/Solid Circle ($\u25CF$): Indicates that the boundary value is included in the solution set (used for non-strict inequalities $\le$ and $\ge$).
• Ray / Shading Direction: Points to the left for values smaller than the boundary ($<$ or $\le$) and to the right for values larger than the boundary ($>$ or $\ge$).

Solving Linear Inequalities & The Crucial Negative Flip Rule

Solving linear inequalities follows the exact same procedures as solving linear equations—using inverse operations to isolate the variable—with one vital exception:

[!IMPORTANT] The Negative Reversal Rule: When multiplying or dividing both sides of an inequality by a negative number, you MUST reverse (flip) the direction of the inequality symbol.

Mathematical Justification for the Reversal Rule

Consider the true numerical inequality:
$2 < 5$

• If we add or subtract numbers (positive or negative), the order is preserved:
  $2 - 3 < 5 - 3 \implies -1 < 2 \quad (\text{True})$
• If we multiply both sides by positive $2$:
  $2(2) < 5(2) \implies 4 < 10 \quad (\text{True})$
• But if we multiply both sides by negative $-1$ without flipping the sign:
  $2(-1) < 5(-1) \implies -2 < -5 \quad (\text{FALSE!})$ On a number line, $-2$ lies to the right of $-5$, meaning $-2$ is greater than $-5$. Therefore, to maintain truth, we must flip the sign:
  $-2 > -5 \quad (\text{True})$

Step-by-Step Worked Example 1: Solving a Multi-Step Inequality

Problem: Solve $-4x + 9 > 29$ and graph the solution.

Solution:

1. Subtract $9$ from both sides: $-4x + 9 - 9 > 29 - 9$ $-4x > 20$

2. Divide both sides by $-4$ AND flip the inequality sign from $>$ to $<$: $\frac{-4x}{-4} < \frac{20}{-4}$ $x < -5$

3. Graphing the solution:

   • Place an open circle at $-5$.
   • Draw a shaded ray pointing to the left (towards negative infinity).

Final Answer: $x < -5$

Translating English Phrases into Algebraic Expressions & Inequalities

Accurate translation from natural language to mathematical symbols is a primary focus of Praxis 5003 word problems. Pay close attention to subtle phrasing differences.

[!WARNING] Syntax Order Trap: "$8$ less than $n$" translates to $n - 8$, NOT $8 - n$. The phrase "less than" acts as a turn-around phrase, placing the starting quantity ($n$) first.

Step-by-Step Word Problem Worked Examples

Step-by-Step Worked Example 2: Budgeting Inequality Word Problem

Problem: A school club is renting a charter bus for a field trip. The rental company charges a flat base fee of $$150$ plus $$12$ per student. If the club has a total budget of at most $$450$ for transportation, what is the maximum number of students that can attend the trip?

Solution:

1. Define the variable:

   • Let $s$ = number of students attending.

2. Translate "at most $$450$" to algebraic inequality: $\text{Base Fee} + \text{Per-Student Fee} \le \text{Budget}$ $150 + 12s \le 450$

3. Solve the inequality for $s$:

   • Subtract $150$: $12s \le 450 - 150 \implies 12s \le 300$
   • Divide by $12$: $s \le \frac{300}{12} \implies s \le 25$

4. Interpret in context:

   • The maximum number of students that can attend without exceeding the $$450$ budget is $25$.

Final Answer: $25$ students

Step-by-Step Worked Example 3: Comparative Word Problem Translation

Problem: Maya's age is $3$ years less than twice her brother Leo's age. If the sum of their ages is at least $27$ years, what is the minimum possible age of Leo?

Solution:

1. Define variables:

   • Let $L$ = Leo's age.
   • Maya's age = $2L - 3$.

2. Set up the inequality for their sum: $\text{Leo's Age} + \text{Maya's Age} \ge 27$ $L + (2L - 3) \ge 27$

3. Combine like terms and solve for $L$: $3L - 3 \ge 27$ $3L \ge 30 \implies L \ge 10$

4. Interpret:

   • The minimum possible age for Leo is $10$ years old.

Final Answer: $10$ years old

Praxis Candidate Traps & Elementary Misconceptions

1. Forgetting to flip the inequality sign: This is the single most common error on the Praxis test. Whenever you divide or multiply an inequality by a negative coefficient, immediately flip the inequality sign.
2. Confusing 'at most' with 'greater than': Candidates often associate the word 'most' with greater than ($>$). Remind yourself that if you have at most $$20$, you have $$20$ or less ($\u2264 20$).