2.2 Proportions & Direct/Inverse Variation

Defining and Solving Proportions

A proportion is a mathematical statement establishing that two ratios are strictly equal. It is expressed in the standard fractional form:

$\frac{a}{b} = \frac{c}{d} \quad (b \ne 0, \; d \ne 0)$

In classical terminology, $a$ and $d$ are called the extremes, while $b$ and $c$ are called the means.

The Cross-Multiplication Property

The fundamental theorem of proportions states that for any valid proportion, the product of the extremes equals the product of the means:

$\frac{a}{b} = \frac{c}{d} \iff a \cdot d = b \cdot c$

This property allows for the immediate conversion of rational equations into linear or polynomial equations.

Solving Linear Proportions with Binomial Expressions

On the ACCUPLACER test, proportions frequently involve algebraic binomials in the numerators or denominators. When applying cross-multiplication, always enclose binomial terms in parentheses to distribute coefficients correctly.

$\frac{2x + 5}{3} = \frac{5x - 2}{4}$ $4(2x + 5) = 3(5x - 2)$ $8x + 20 = 15x - 6$ $20 + 6 = 15x - 8x$ $26 = 7x \implies x = \frac{26}{7}$

Worked Example 1: Scale Factor & Cartography

On an architectural blueprint for a commercial medical clinic, a scale of $0.75\text{ inches} = 12\text{ feet}$ is used. If the actual length of the surgical recovery ward is $68\text{ feet}$, what is the corresponding length on the blueprint in inches?

1. Set up a proportion matching blueprint units to real-world units: $\frac{0.75\text{ in}}{12\text{ ft}} = \frac{x\text{ in}}{68\text{ ft}}$
2. Apply cross-multiplication: $12 \cdot x = 0.75 \cdot 68$ $12x = 51$
3. Solve for $x$: $x = \frac{51}{12} = 4.25\text{ inches} \;\left(4\frac{1}{4}\text{ inches}\right)$

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Direct Variation ($y = kx$)

Two variables $x$ and $y$ exhibit direct variation (or are said to be directly proportional) if a constant ratio exists between them for all non-zero values.

Mathematical Formulation

$y = kx \quad \text{or} \quad \frac{y}{x} = k$

where $k$ is a non-zero constant called the constant of proportionality (or constant of variation).

Geometric and Functional Characteristics

• Graph: The graph of a direct variation relationship is a straight line passing directly through the origin $(0, 0)$ with slope equal to $k$.
• Behavior: If $x$ is multiplied by a factor $c$, $y$ is multiplied by the exact same factor $c$. If $x$ doubles, $y$ doubles; if $x$ is halved, $y$ is halved.
• Proportional Equivalence: For any two paired data points $(x_1, y_1)$ and $(x_2, y_2)$: $\frac{y_1}{x_1} = \frac{y_2}{x_2}$

(Exam Warning: The linear equation $y = 3x + 4$ is NOT direct variation because when $x = 0$, $y = 4 \ne 0$. A true direct variation line must have a $y$-intercept of $0$.)

Worked Example 2: Direct Variation in Physics

Under Hooke's Law, the distance $d$ that an elastic spring stretches varies directly with the force $F$ applied to it. When a force of $45\text{ Newtons}$ is applied, the spring stretches $7.2\text{ centimeters}$. How far will the spring stretch when a force of $75\text{ Newtons}$ is applied?

1. Find the constant of proportionality $k$: $k = \frac{d_1}{F_1} = \frac{7.2\text{ cm}}{45\text{ N}} = 0.16\text{ cm/N}$
2. Write the specific direct variation equation: $d = 0.16 F$
3. Substitute the new force $F = 75\text{ N}$: $d = 0.16 \times 75 = 12.0\text{ centimeters}$

Alternatively, solve directly via proportion: $\frac{7.2}{45} = \frac{d_2}{75} \implies 45 d_2 = 7.2 \times 75 = 540 \implies d_2 = \frac{540}{45} = 12.0\text{ cm}$

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Inverse Variation ($y = \frac{k}{x}$ or $x \cdot y = k$)

Two variables $x$ and $y$ exhibit inverse variation (or are said to be inversely proportional) if their mathematical product remains constant for all values.

Mathematical Formulation

$y = \frac{k}{x} \quad \text{or} \quad x \cdot y = k$

where $k$ is the non-zero constant of variation.

Geometric and Functional Characteristics

• Graph: The graph of an inverse variation relationship is a rectangular hyperbola asymptotic to the $x$-axis and $y$-axis (lying in Quadrants I and III if $k > 0$).
• Reciprocal Scaling: If $x$ is multiplied by a constant factor $c$, $y$ is divided by $c$ (multiplied by $\frac{1}{c}$). If $x$ triples, $y$ decreases to $\frac{1}{3}$ of its original value.
• Product Invariance Equation: For any two paired data points $(x_1, y_1)$ and $(x_2, y_2)$: $x_1 \cdot y_1 = x_2 \cdot y_2$

Real-World Applications of Inverse Variation

• Speed vs. Travel Time: For a fixed distance $D$, speed $r$ and time $t$ vary inversely: $r \cdot t = D$.
• Workforce vs. Project Duration: For a fixed total workload, number of workers $W$ and completion hours $H$ vary inversely: $W \cdot H = \text{Constant Worker-Hours}$.
• Boyle's Law in Chemistry: Pressure $P$ and volume $V$ of an ideal gas at constant temperature satisfy $P \cdot V = k$.
• Gear and Pulley Mechanics: Gear teeth count $N$ and rotational speed $S$ satisfy $N_1 S_1 = N_2 S_2$.

Worked Example 3: Workforce Scheduling (Worker-Hours)

A landscaping company determines that a crew of $6$ technicians can clear and sod a corporate park in $14\text{ hours}$. If the client requires the project to be completed in exactly $8\text{ hours}$, how many technicians must be assigned to the crew, assuming all technicians work at the same constant rate?

1. Identify the constant of variation $k$ (total required worker-hours): $k = W_1 \cdot H_1 = 6\text{ technicians} \times 14\text{ hours} = 84\text{ worker-hours}$
2. Set up the product invariance equation with new time $H_2 = 8\text{ hours}$: $W_2 \cdot H_2 = 84$ $W_2 \cdot 8 = 84$
3. Solve for $W_2$: $W_2 = \frac{84}{8} = 10.5\text{ technicians}$
4. Interpret practical constraints: Since partial technicians cannot be hired, the company must deploy at least $11$ technicians to finish within the 8-hour deadline (or exactly $10.5$ if fractional labor-hours are modeled).

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Comparison of Direct vs. Inverse Variation

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Joint and Combined Variation

Many physical and economic models involve three or more variables simultaneously.

Definitions and Formulas

1. Joint Variation ($z$ varies jointly as $x$ and $y$): $z = k \cdot x \cdot y$ Example: The volume of a cylinder $V = \pi r^2 h$ varies jointly as $h$ and the square of $r$.
2. Combined Variation ($z$ varies directly as $x$ and inversely as $y$): $z = \frac{k \cdot x}{y}$ Example: Electrical resistance $R = \frac{\rho L}{A}$ varies directly with length $L$ and inversely with cross-sectional area $A$.

Systematic Two-Step Solving Procedure

• Step 1: Substitute initial baseline data $(x_0, y_0, z_0)$ into the model to solve explicitly for the constant $k$.
• Step 2: Re-write the general equation with the calculated numerical value of $k$, then substitute the new given variables to solve for the target unknown.

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Collaborative Work-Rate Problems

Work-rate problems represent a high-yield application of rational proportions on the ACCUPLACER test. In collaborative work problems, each entity's rate is expressed as the fraction of the total job completed per unit of time.

$\text{Work Done} = \text{Rate} \times \text{Time} \implies \text{Rate} = \frac{1}{\text{Individual Time } t}$

The Combined Rate Formula

When multiple entities work together simultaneously without interfering with one another, their individual working rates are additive:

$\text{Rate}_{\text{combined}} = \text{Rate}_1 + \text{Rate}_2 + \dots + \text{Rate}_n$ $\frac{1}{t_{\text{together}}} = \frac{1}{t_1} + \frac{1}{t_2}$

Solving for $t_{\text{together}}$ yields the Product-over-Sum Shortcut for two workers: $t_{\text{together}} = \frac{t_1 \cdot t_2}{t_1 + t_2}$

Worked Example 4: Collaborative Drainage Pipes

A commercial storage reservoir has two drainage pipes. Pipe A can completely empty the full reservoir in $6\text{ hours}$ when operating alone. Pipe B can empty the reservoir in $9\text{ hours}$ when operating alone. If both pipes are opened simultaneously, how long will it take to drain the entire reservoir?

1. Express individual rates per hour: $\text{Rate}_A = \frac{1}{6}\text{ of reservoir/hr}, \quad \text{Rate}_B = \frac{1}{9}\text{ of reservoir/hr}$
2. Sum rates to find combined rate: $\text{Rate}_{\text{together}} = \frac{1}{6} + \frac{1}{9} = \frac{3}{18} + \frac{2}{18} = \frac{5}{18}\text{ of reservoir/hr}$
3. Invert the combined rate to find total time: $t_{\text{together}} = \frac{18}{5}\text{ hours} = 3.6\text{ hours} = 3\text{ hours and } 36\text{ minutes}$

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Common Pitfalls & ACCUPLACER Exam Traps

1. Applying Direct Variation Logic to Inverse Problems: If $4$ machines build an order in $12\text{ hours}$, $8$ machines will NOT take $24\text{ hours}$. More machines take LESS time: $4 \times 12 = 48 \implies 8 \times t = 48 \implies t = 6\text{ hours}$.
2. Averaging Speeds Directly (Harmonic Mean Trap): If a delivery truck drives to a destination at $40\text{ mph}$ and returns along the identical route at $60\text{ mph}$, the average speed for the round trip is NOT $\frac{40 + 60}{2} = 50\text{ mph}$. Because more time is spent at the slower speed, the true average speed is $\frac{\text{Total Distance}}{\text{Total Time}} = \frac{2D}{\frac{D}{40} + \frac{D}{60}} = \frac{2}{\frac{5}{120}} = \frac{240}{5} = 48\text{ mph}$.
3. Omitting Parentheses During Cross-Multiplication: In $\frac{x+2}{4} = \frac{x-1}{3}$, forgetting parentheses leads to the common error $3x + 2 = 4x - 1$, rather than the correct $3(x + 2) = 4(x - 1) \implies 3x + 6 = 4x - 4$.
4. Linear Equations with Non-Zero Intercepts: Never classify $y = mx + b$ ($b \ne 0$) as direct variation. While linear, it fails the proportionality requirement $\frac{y}{x} = \text{constant}$.