5.2 Direct vs. Partial Variations & Word Problems

5.2 Direct vs. Partial Variations & Word Problems

In the Ontario Mathematics Curriculum, linear relations are categorized into two primary structural types based on their initial values: Direct Variation and Partial Variation. Distinguishing between these two variation types is essential for understanding real-world proportional reasoning, consumer applications, and algebraic modeling on the Ontario Mathematics Proficiency Test (MPT).

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1. Direct Variation ($y = kx$)

A direct variation describes a mathematical relationship between two variables where one is a constant multiple of the other. The dependent variable $y$ varies directly as the independent variable $x$.

Algebraic Representation

$y = kx$

Where:

• $k$ is the constant of variation (also called the constant of proportionality or unit rate).
• $k = \frac{y}{x}$ for any non-zero point $(x, y)$ on the line.
• The $y$-intercept is $b = 0$.

Key Graphical Characteristics

1. Passes through the Origin: The graph always passes through $(0,0)$. When $x = 0$, $y = 0$.
2. Straight Line: The graph is a continuous straight line with constant slope $m = k$.
3. Proportional Ratio: The ratio $\frac{y}{x}$ remains constant for every point on the graph.

Real-World Examples

• Hourly Wage (without base pay): Earnings $E = 17.20h$, where $h$ is hours worked.
• Distance at Constant Speed: Distance $d = 90t$, where $t$ is time in hours driving at $90 \text{ km/h}$.
• Mass and Volume: Mass $m = \rho V$, where $\rho$ is density.

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2. Partial Variation ($y = mx + b$, $b \neq 0$)

A partial variation describes a relationship between two variables in which the dependent variable $y$ is equal to a fixed initial amount plus a variable amount that depends directly on $x$.

Algebraic Representation

$y = mx + b \quad \text{where } b \neq 0$

Where:

• $b$ is the fixed component or initial value ($y$-intercept at $(0, b)$).
• $m$ is the variable component or rate of change (slope).

Key Graphical Characteristics

1. Does NOT Pass through the Origin: The graph intersects the $y$-axis at $(0, b)$ where $b \neq 0$.
2. Straight Line: The graph is a straight line with slope $m$.
3. Non-Constant Ratio: The ratio $\frac{y}{x}$ changes at different points along the line due to the presence of the fixed constant $b$.

Real-World Examples

• Taxi Fare: Total fare $F = 4.50 + 1.85d$, where $4.50 is the base flag-drop fee and $1.85 is the cost per kilometer $d$.
• Equipment Rental: Cost $C = 50 + 15h$, where $50 is the security deposit/admin fee and $15 is the hourly rental rate.
• Sales Compensation: Total salary $S = 2000 + 0.05s$, where $2,000 is base salary and 5% is commission on sales $s$.

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3. Direct vs. Partial Variation Comparison Matrix

graph TD
    A["Linear Relation Modeling"] --> B["Initial Value b = 0?"]
    B -->|"Yes: Passes through (0,0)"| C["Direct Variation: y = kx"]
    B -->|"No: y-intercept (0,b) with b ≠ 0"| D["Partial Variation: y = mx + b"]

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4. Solving Real-World Comparison & Break-Even Problems

A major expectation on the Ontario MPT is evaluating two competing pricing schemes (one direct variation and one partial variation, or two partial variations) to find the break-even point (point of intersection).

Systematic Problem-Solving Protocol

1. Define Variables: State the independent variable ($x$) and dependent variable ($y$) with units.
2. Formulate Linear Equations: Express Option 1 ($y_1$) and Option 2 ($y_2$) in slope-intercept form.
3. Set Up Algebraic Equivalence: Set $y_1 = y_2$ to find the break-even value of $x$.
4. Solve for $x$: Isolate $x$ using inverse operations.
5. Interpret Results: Evaluate which option is more cost-effective above or below the threshold value of $x$.

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5. Step-by-Step Worked Examples

Worked Example 1: Direct Variation Calculation

Problem: A school fundraiser sells custom t-shirts. The total cost of purchasing 25 t-shirts is $350. Assuming cost varies directly with the number of shirts:

1. Determine the constant of variation $k$.
2. Write the direct variation equation.
3. Calculate the cost to purchase 80 t-shirts.

Solution:

• Step 1: Calculate constant of variation $k$. $k = \frac{C}{n} = \frac{350}{25} = 14$ The cost is $14 per shirt.

• Step 2: Write the equation. $C(n) = 14n$

• Step 3: Calculate cost for 80 shirts. $C(80) = 14 \times 80 = 1120$ The total cost for 80 shirts is $1,120.

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Worked Example 2: Comparative Break-Even Analysis (Direct vs. Partial)

Problem: A gym member considers two payment plans:

• Plan A (Direct Variation): Pay no monthly membership fee, but pay $12 per fitness class attended.
• Plan B (Partial Variation): Pay a fixed monthly fee of $40, plus $4 per fitness class attended.

1. Write linear equations for the total monthly cost of Plan A ($C_A$) and Plan B ($C_B$) for $n$ classes.
2. Calculate the exact number of classes where both plans cost the same.
3. Determine which plan is cheaper if a member attends 12 classes per month.

Solution:

• Step 1: Write linear equations. $C_A = 12n$ $C_B = 4n + 40$

• Step 2: Find the break-even point ($C_A = C_B$). $12n = 4n + 40$ $12n - 4n = 40$ $8n = 40$ $n = \frac{40}{8} = 5 \text{ classes}$ At 5 classes per month, both plans cost exactly $60.

• Step 3: Evaluate for 12 classes.

  • Plan A: $C_A(12) = 12(12) = 144$ ($144)
  • Plan B: $C_B(12) = 4(12) + 40 = 48 + 40 = 88$ ($88)

  Conclusion: For 12 classes, Plan B is cheaper by $56.

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6. Ontario Curriculum Context & Pedagogical Guidance

In Grade 9 (MTH1W), teachers use direct and partial variations to transition students from proportional reasoning (multiplicative strategies in Grades 7–8) to formal linear modeling. When constructing assessment items, ensure students recognize that a graph representing a partial variation will never cross the origin unless the fixed fee $b = 0$, at which point it collapses into a direct variation.