Section 4.4: Basic Algebra and Problem Solving

Basic Algebra and Problem Solving

Algebra is a powerful mathematical tool that allows police officers to solve problems involving unknown values. Whether determining the speed of a vehicle prior to an accident, calculating the distribution of personnel across stations, or solving financial and operational equations, algebraic reasoning is an everyday requirement. The Victoria Police Entrance Examination (VPOL) assesses your ability to set up and solve basic linear equations, translate word problems into algebraic statements, and substitute values into physical formulas.

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Solving Basic Linear Equations

A linear equation is a mathematical statement asserting that two algebraic expressions are equal. It contains a variable (an unknown value represented by a letter, such as $x$), a coefficient (a number multiplying the variable), and constants (fixed numbers).

The Principles of Solving Equations

To solve a linear equation, you must isolate the variable on one side of the equal sign. The fundamental rule is: whatever operation you perform on one side of the equation, you must perform on the other side to maintain equality.

• Addition and Subtraction: Invert addition with subtraction, and vice versa.
• Multiplication and Division: Invert multiplication with division, and vice versa.

Step-by-Step Examples

• One-Step Equation: $x + 14 = 30$ Subtract 14 from both sides: $x = 30 - 14 \Rightarrow x = 16$
• Two-Step Equation: $4x - 9 = 27$ Add 9 to both sides to isolate the term with the variable: $4x = 27 + 9 \Rightarrow 4x = 36$ Divide both sides by 4 to solve for $x$: $x = \frac{36}{4} \Rightarrow x = 9$
• Multi-Step Equation with Brackets: $3(x + 5) = 24$ Expand the brackets (or divide both sides by 3 first): $x + 5 = \frac{24}{3} \Rightarrow x + 5 = 8$ Subtract 5 from both sides: $x = 8 - 5 \Rightarrow x = 3$

To build confidence for the VPOL exam, practice isolating variables in different positions. For example, if you are given the formula $A = \frac{1}{2}bh$ and asked to solve for the height $h$, you must multiply both sides by 2 to get $2A = bh$, and then divide by $b$ to isolate $h$: $h = \frac{2A}{b}$. Being able to rearrange equations fluidly ensures you can tackle any formula-based question, regardless of which variable is unknown. Common mistakes include performing operations out of order or forgetting to apply a change to both sides of the equation, which results in incorrect values.

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Translating Word Problems into Equations

Many algebra questions on the VPOL will not be presented as equations; instead, they will be written as word problems. You must translate the written scenario into a mathematical equation.

Key Translation Terms

• "Sum", "increased by", "more than", "total": Indicates addition (+).
• "Difference", "decreased by", "less than": Indicates subtraction (-).
• "Product", "times", "of": Indicates multiplication ($\times$).
• "Quotient", "shared between", "divided by": Indicates division ($\div$).
• "Is", "equals", "results in": Indicates the equal sign (=).

Worked Example

• Scenario: A local police division has a total of 56 officers on duty. The number of constables on patrol is 4 more than triple the number of sergeants. How many sergeants are on duty?
• Method:
  1. Define the variables: Let $S$ represent the number of sergeants.
  2. Express the number of constables in terms of $S$: Constables $= 3S + 4$.
  3. Set up the equation using the total number of officers: $S + \text{Constables} = 56$ $S + (3S + 4) = 56$
  4. Combine like terms: $4S + 4 = 56$
  5. Solve for $S$: $4S = 52 \Rightarrow S = \frac{52}{4} \Rightarrow S = 13$
  6. Conclusion: There are 13 sergeants on duty. (We can check: Constables $= 3(13) + 4 = 39 + 4 = 43$. Total officers $= 13 + 43 = 56$, which is correct).

When translating word problems, pay close attention to commas and phrasing. For instance, "twice the sum of $x$ and 5" is written as $2(x + 5)$, whereas "the sum of twice $x$ and 5" is written as $2x + 5$. The placement of parentheses completely alters the order of operations, and selecting the wrong algebraic model will lead to incorrect solutions. Writing down each step and naming the variables clearly on your working paper is the best way to prevent these errors.

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Interpreting and Substituting into Formulas

Policing often relies on established formulas to estimate variables from physical evidence, such as vehicle speeds from skid marks or alcohol dissipation rates.

The Speed-Distance-Time Formula

A fundamental formula in traffic operations is: $\text{Distance} = \text{Speed} \times \text{Time}$ By rearranging the variables, we get: $\text{Speed} = \frac{\text{Distance}}{\text{Time}} \quad \text{and} \quad \text{Time} = \frac{\text{Distance}}{\text{Speed}}$

• Worked Example: A suspect vehicle is tracked by highway patrol traveling at a constant speed of 110 km/h. How many kilometers will the vehicle travel in 18 minutes?
  1. Convert minutes to hours: $18\text{ minutes} \div 60\text{ minutes/hour} = 0.3\text{ hours}$.
  2. Substitute values into the distance formula: $\text{Distance} = 110\text{ km/h} \times 0.3\text{ h} = 33\text{ kilometers}$

Accident Reconstruction Formulas

Forensic collision investigators use the skid distance and the road's coefficient of friction (drag factor) to estimate speed: $s = \sqrt{254 \cdot d \cdot f}$ where $s$ is the speed in km/h, $d$ is the skid distance in meters, and $f$ is the coefficient of friction.

• Worked Example: After a collision, a vehicle leaves skid marks measuring 40 meters. The coefficient of friction of the wet bitumen is 0.5. Calculate the estimated speed of the vehicle.
  1. Substitute the values into the formula: $s = \sqrt{254 \times 40 \times 0.5}$
  2. Simplify: $s = \sqrt{254 \times 20} = \sqrt{5080}$
  3. Calculate the square root: $s \approx 71.27\text{ km/h}$
  4. Conclusion: The vehicle was traveling at approximately 71 km/h when the brakes were applied.