6.3 Linear Inequalities: Solving, Graphing on Number Lines, & Word Problems

6.3 Linear Inequalities: Solving, Graphing on Number Lines, & Word Problems

While linear equations determine exact, specific numerical values where expressions are equal, real-world constraints—such as staying within an operating budget, maintaining a minimum speed, or observing maximum load limits—are governed by inequalities. On the TABE assessment, you must be able to solve linear inequalities, reverse direction signs appropriately, graph continuous solutions on number lines, and evaluate compound relationships.

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1. Inequality Symbols & The Concept of Continuous Solution Sets

An inequality compares two algebraic expressions using one of five relational symbols:

Discrete vs. Continuous Solution Sets

Unlike the equation $x = 3$ (which contains exactly one discrete point), the inequality $x > 3$ contains an infinite continuum of real numbers ($3.0001, 3.5, 4, 100, 10^{12}$). Because an infinite set of numbers cannot be listed individually, solutions are represented graphically on a number line or via algebraic interval notation.

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2. Graphing Inequalities on a 1D Number Line

Graphing an inequality on a number line requires two decisions:

1. Boundary Point (Circle Type):
   • Use an Open Circle ($\circ$) for strict inequalities ($<$ or $>$) to show the endpoint is excluded.
   • Use a Closed / Solid Circle ($\bullet$) for inclusive inequalities ($\le$ or $\ge$) to show the endpoint is included.
2. Shading Direction (Ray):
   • Shade to the Right (toward $+\infty$) when the variable is greater than the boundary value ($x > c$ or $x \ge c$).
   • Shade to the Left (toward $-\infty$) when the variable is less than the boundary value ($x < c$ or $x \le c$).

Graph of x > 2 (Open Circle at 2, Shaded Right):
<---+---+---+---(o)=======>
   -1   0   1    2   3   4

Graph of x <= -1 (Closed Circle at -1, Shaded Left):
<=======*---+---+---+---+--->
   -3  -2  -1   0   1   2   3

3. The Golden Rule of Inequalities: Multiplying or Dividing by a Negative

Solving a linear inequality follows the exact same inverse operations as solving a linear equation, with one vital, fundamental exception:

Why Does the Inequality Symbol Reverse?

Consider the true numerical inequality: $2 < 6$.

• If we multiply both sides by positive $2$: $2(2) < 6(2) \implies 4 < 12$ (Still True).
• If we multiply both sides by negative $1$: $2(-1) \text{ vs } 6(-1) \implies -2 \text{ vs } -6$. On a number line, $-2$ is to the right of $-6$, which means $-2 > -6$! Multiplying by a negative reflects quantities across zero, completely inverting their relative order.

Step-by-Step Worked Demonstration

Problem: Solve $-4x + 15 \le 39$.

$$\begin{aligned} \text{Step 1 (Undo Addition):} &\quad -4x + 15 - 15 \le 39 - 15 \\ &\quad -4x \le 24 \\ \text{Step 2 (Divide by } -4 \text{ and FLIP sign):} &\quad \frac{-4x}{-4} \ge \frac{24}{-4} \quad (\le \text{ flips to } \ge) \\ &\quad x \ge -6 \end{aligned}$$

Graphical Representation: A solid closed circle at $-6$ with shading extending infinitely to the right.

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4. Multi-Step Linear Inequalities

Example: Solve $3(2x - 5) - 4x < 7x + 9$.

$$\begin{aligned} \text{Step 1 (Distribute):} &\quad 6x - 15 - 4x < 7x + 9 \\ \text{Step 2 (Combine Like Terms):} &\quad 2x - 15 < 7x + 9 \\ \text{Step 3 (Collect Variables on Left):} &\quad 2x - 7x - 15 < 9 \\ &\quad -5x - 15 < 9 \\ \text{Step 4 (Undo Subtraction):} &\quad -5x < 24 \\ \text{Step 5 (Divide by } -5 \text{ and FLIP):} &\quad x > -\frac{24}{5} \quad (x > -4.8) \end{aligned}$$

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5. Compound Inequalities: "AND" vs. "OR"

A compound inequality connects two individual inequalities with either the conjunction "AND" or the disjunction "OR".

Compound Inequalities:
├── 1. "AND" Conjunction (Intersection ∩) ──> a <= x <= b (Bounded Overlapping Segment)
└── 2. "OR" Disjunction (Union ∪) ──────────> x < a OR x > b (Two Diverging Rays)

1. "AND" Compound Inequalities (Intersection $\cap$)

In an "AND" inequality, both conditions must be true simultaneously. It is typically written as a continuous double-ended inequality: $a < x < b$.

Worked Example: Solve $-8 \le 2x + 4 < 14$.

• To solve, perform inverse operations to all three parts simultaneously:

$$\begin{aligned} \text{Subtract 4 from all 3 parts:} &\quad -8 - 4 \le 2x + 4 - 4 < 14 - 4 \\ &\quad -12 \le 2x < 10 \\ \text{Divide all 3 parts by 2:} &\quad \frac{-12}{2} \le \frac{2x}{2} < \frac{10}{2} \\ &\quad -6 \le x < 5 \end{aligned}$$

Graph: A closed circle at $-6$, an open circle at $5$, and the segment between them shaded.

2. "OR" Compound Inequalities (Union $\cup$)

In an "OR" inequality, a value of $x$ is a solution if it satisfies at least one of the individual inequalities. The two parts are solved separately.

Worked Example: Solve $4x - 3 < -15 \quad \text{or} \quad 2x + 7 \ge 17$.

• Left inequality: $4x < -12 \implies x < -3$.
• Right inequality: $2x \ge 10 \implies x \ge 5$.
• Combined Solution: $x < -3 \quad \text{or} \quad x \ge 5$. Graph: An open circle at $-3$ shaded to the left, and a solid circle at $5$ shaded to the right.

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6. Real-World Constraints & Word Problems

Workplace Problem 1: Freight Elevator Maximum Load

Problem: A warehouse cargo elevator has a safe working load limit of $2,500\text{ lbs}$. An operator weighing $190\text{ lbs}$ must transport pallets of engine parts weighing $140\text{ lbs}$ each. What is the maximum number of pallets $p$ that can be safely loaded?

1. Set Up Inequality: $140p + 190 \le 2500$
2. Solve: $$\begin{aligned} 140p &\le 2500 - 190 \\ 140p &\le 2310 \\ p &\le \frac{2310}{140} = 16.5 \end{aligned}$$
3. Practical Interpretation: Because you cannot load a partial pallet without exceeding safe capacity, round down to the nearest integer: $16$ pallets maximum.

Academic Target Problem 2: Average Score Target

Problem: A nursing student needs an overall average of at least $84$ across four pharmacology exams to pass. On the first three exams, the student scored $79, 88,$ and $81$. What score $s$ must be earned on the fourth exam?

$$\begin{aligned} \frac{79 + 88 + 81 + s}{4} &\ge 84 \\ \frac{248 + s}{4} &\ge 84 \\ 248 + s &\ge 84 \times 4 \\ 248 + s &\ge 336 \\ s &\ge 336 - 248 \\ s &\ge 88 \end{aligned}$$

Conclusion: The student must score at least $88$ on the final exam.