6.3 Linear Modeling & Inequality Graphs

Constructing Linear Models from Real-World Scenarios

In quantitative reasoning and intermediate algebra, a linear model is an equation that describes a constant rate of change between two real-world variables. The general mathematical structure of a linear model is expressed in slope-intercept or function notation:

$y = mx + b \quad \text{or} \quad f(x) = mx + b$

Linear Model Breakdown: y = mx + b

  Total Output Value (y) = [Variable Unit Rate (m)] · [Quantity (x)] + [Fixed Base Amount (b)]
       ↑                                ↑                                   ↑
  Dependent Variable             Slope / Marginal Rate                 y-Intercept / Initial Value
  (e.g., Total Bill $)           (e.g., $ per mile)                    (e.g., Base Service Fee $)

1. Anatomy of Model Parameters

• Independent Variable ($x$): The input quantity that varies freely (e.g., time, distance driven, units produced, hours worked).
• Dependent Variable ($y$ or $f(x)$): The output quantity whose value depends on $x$ (e.g., total cost, remaining volume, temperature, account balance).
• Slope ($m = \frac{\Delta y}{\Delta x}$): The marginal rate of change or unit rate. It specifies how much $y$ increases (if $m > 0$) or decreases (if $m < 0$) for each one-unit increase in $x$.
• $y$-Intercept ($(0, b)$): The initial value, baseline starting amount, fixed fee, or setup cost before any activity occurs ($x = 0$).

2. Contextual Interpretation of Intercepts and Domain Restrictions

• The $y$-Intercept ($x = 0$): Represents the starting condition of the system. For a cellular plan with equation $C(g) = 15g + 45$, the $y$-intercept is $45, which is the flat monthly base charge regardless of data usage.
• The $x$-Intercept ($y = 0$): Represents the point where the dependent variable reaches zero. For an airplane descending according to $h(t) = -600t + 24000$, setting $h(t) = 0$ yields $t = 40\text{ minutes}$, representing the landing time.
• Applied Domain and Range Constraints: In physical applications, variables are frequently constrained to non-negative real numbers ($x \ge 0, y \ge 0$). Models must be interpreted strictly within their valid operational domain.

---

Applied Linear Modeling Archetypes

---

Graphing Two-Variable Linear Inequalities

A linear inequality in two variables relates two expressions using inequality symbols ($<, \le, >, \ge$). The graph of a linear inequality is not just a single line, but an entire half-plane containing an infinite set of ordered pairs $(x, y)$ that make the inequality true.

$\text{Standard Forms: } Ax + By < C, \quad Ax + By \le C, \quad Ax + By > C, \quad Ax + By \ge C$ $\text{Slope-Intercept Forms: } y < mx + b, \quad y \le mx + b, \quad y > mx + b, \quad y \ge mx + b$

Boundary Line Styles & Half-Plane Shading:

      Strict: < or >                             Inclusive: ≤ or ≥
  (Dashed Boundary Line)                       (Solid Boundary Line)
Points ON line are NOT solutions           Points ON line ARE solutions

           y                                          y
        /  │  /                                    /  │  /
       /   │ /                                    /   │ /
      - - -┼- - - (Dashed)                       ─────┼───── (Solid)
     /     │                                    /     │
    /      │                                   /      │

---

The Systematic 3-Step Graphing Algorithm

Step 1: Graph the Boundary Line

Replace the inequality symbol with an equal sign ($=$) to obtain the equation of the boundary line: $Ax + By = C \quad \text{or} \quad y = mx + b$

• If the original inequality contains $<$ or $>$ (strict inequalities), draw a dashed (broken) line to indicate that points lying exactly on the boundary are excluded from the solution set.
• If the original inequality contains $\le$ or $\ge$ (inclusive inequalities), draw a solid line to indicate that points on the boundary are included in the solution set.

Step 2: Determine the Shaded Half-Plane

Choose one of two reliable methods to decide which side of the boundary line to shade:

• Method A: The Test Point Method (Universal & Foolproof)

  1. Select any test coordinate $(x_0, y_0)$ that does NOT lie on the boundary line. The origin $(0, 0)$ is the fastest and easiest choice whenever the line does not pass through $(0,0)$.
  2. Substitute $(x_0, y_0)$ into the original inequality statement.
  3. If the resulting statement is TRUE, shade the entire half-plane containing the test point.
  4. If the resulting statement is FALSE, shade the opposite half-plane on the other side of the boundary line.

• Method B: The Slope-Intercept Inspection Rule (When $y$ is Isolated) When an inequality is solved explicitly for $y$ with a positive coefficient:

  • $y > mx + b$ or $y \ge mx + b$: Shade ABOVE the boundary line (higher $y$-values).
  • $y < mx + b$ or $y \le mx + b$: Shade BELOW the boundary line (lower $y$-values).

Step 3: Verify with an Arbitrary Solution Coordinate

Pick a coordinate clearly situated in the shaded region and confirm it satisfies the original inequality.

---

The Critical Negative Coefficient Trap in Inequalities

When converting a standard form inequality with a negative $y$-coefficient into slope-intercept form, you must reverse the inequality direction when multiplying or dividing by a negative number:

$-3y \le 6x - 12 \implies \frac{-3y}{-3} \ge \frac{6x}{-3} - \frac{12}{-3} \implies y \ge -2x + 4$

Failing to flip the symbol leads to shading the wrong half-plane and selecting incorrect solution sets.

---

Step-by-Step Worked Examples

Worked Example 1: Constructing and Evaluating a Linear Cost Model

A commercial equipment rental company charges a flat insurance and preparation fee of $\$65.00$ plus an hourly operating rate of $\$28.50$ per hour for an industrial wood chipper. A landscaping contractor has a maximum project budget of $\$350.00$.

1. Construct the linear model expressing total cost $C(h)$ as a function of rental hours $h$: $C(h) = 28.50h + 65.00$

   • $m = +28.50$: Marginal cost per rental hour ($/hour).
   • $b = 65.00$: Fixed initial base fee ($).

2. Determine the maximum whole number of hours the contractor can rent the machine without exceeding the budget: $28.50h + 65.00 \le 350.00$ $28.50h \le 350.00 - 65.00$ $28.50h \le 285.00$ $h \le \frac{285.00}{28.50} = 10$ The contractor can rent the chipper for up to $10$ full hours.

---

Worked Example 2: Graphing a Linear Inequality with the Test Point Method

Graph the linear inequality: $4x - 3y > 12$.

Step 1: Determine the boundary line equation and line style

• Boundary equation: $4x - 3y = 12$.
• The inequality uses "$>$" (strict inequality) $\implies$ Dashed boundary line.

Step 2: Find the intercepts of the boundary line

• $x$-intercept (let $y = 0$): $4x - 3(0) = 12 \implies 4x = 12 \implies x = 3 \implies (3, 0)$.
• $y$-intercept (let $x = 0$): $4(0) - 3y = 12 \implies -3y = 12 \implies y = -4 \implies (0, -4)$.

Step 3: Test the origin $(0, 0)$ Substitute $(0, 0)$ into the original inequality $4x - 3y > 12$: $4(0) - 3(0) > 12 \implies 0 > 12 \quad (\mathbf{FALSE})$ Because the statement is false, the origin $(0, 0)$ is NOT a solution. Therefore, shade the half-plane that does not contain $(0, 0)$ (the region below and to the right of the dashed boundary line).

Step 4: Alternative Verification by Isolating $y$ $4x - 3y > 12 \implies -3y > -4x + 12$ Divide by $-3$ and reverse the inequality sign: $y < \frac{4}{3}x - 4$ The isolated form $y < \frac{4}{3}x - 4$ indicates shading below the dashed line, confirming the test point result.

---

Worked Example 3: Applied Straight-Line Asset Depletion

A municipal transit agency operates an electric bus fleet. A newly installed quick-charge battery bank with a total capacity of $450\text{ kilowatt-hours (kWh)}$ discharges at a constant rate of $37.5\text{ kWh per hour}$ during transit operations.

1. Formulate the battery capacity model $E(t)$ as a function of operational hours $t$: $E(t) = -37.5t + 450$
2. Interpret the slope and intercepts:
   • Slope $m = -37.5\text{ kWh/hour}$: The battery loses $37.5\text{ kWh}$ of energy each hour.
   • $y$-intercept $(0, 450)$: The initial full battery charge is $450\text{ kWh}$.
   • $x$-intercept (exhaustion time): Set $E(t) = 0$: $-37.5t + 450 = 0 \implies 37.5t = 450 \implies t = \frac{450}{37.5} = 12\text{ hours}$ The battery is completely depleted after exactly $12$ hours of continuous transit service.

---

Common Pitfalls & ACCUPLACER Exam Traps

1. Failing to Reverse Inequality Signs When Dividing by Negatives: In $-5y \ge 15x + 20$, dividing by $-5$ without flipping yields $y \ge -3x - 4$ (wrong). The correct result is $y \le -3x - 4$.
2. Confusing Solid vs. Dashed Boundary Lines: Drawing a solid line for strict inequalities ($<, >$) or a dashed line for inclusive inequalities ($\le, \ge$) results in point-loss. Solid lines include the boundary ($=$); dashed lines exclude it.
3. Blindly Shading Based on Standard Form Signs: Seeing "$\le$" in $2x - 5y \le 10$ and assuming you should shade "below" is a classic trap. Because the coefficient of $y$ is negative ($-5$), dividing by $-5$ turns the inequality into $y \ge \frac{2}{5}x - 2$, which requires shading ABOVE the line. Always use the test-point method or isolate $y$ properly.
4. Testing a Point that Lies on the Boundary Line: If the line passes through $(0,0)$, testing $(0,0)$ gives $0 = 0$ or $0 \ge 0$, which does not determine which half-plane to shade. Always choose a point distinctly off the boundary (such as $(0, 1)$ or $(1, 0)$).