Problem 22 - Algebra and Trigonometry with Analytic Geometry

Question

A salesperson purchased an automobile that was advertised as averaging \(25 \mathrm{mi} / \mathrm{gal}\) in the city and \(40 \mathrm{mi} / \mathrm{gal}\) on the highway. A recent sales trip that covered 1800 miles required 51 gallons of gasoline. Assuming that the advertised mileage estimates were correct, how many miles were driven in the city?

Step-by-Step Solution

Answer

400 miles were driven in the city.

Step 1: Define Variables

Let \( x \) be the miles driven in the city and \( y \) be the miles driven on the highway. Thus, the total miles driven is \( x + y = 1800 \).

Step 2: Set Up Mileage Equations

According to the problem, city mileage is \(25 \text{ mi/gal} \) and highway mileage is \(40 \text{ mi/gal} \). We can write the gasoline consumption equations as \( \frac{x}{25} + \frac{y}{40} = 51 \).

Step 3: Solve the System of Equations

We have two equations: \( x + y = 1800 \) and \( \frac{x}{25} + \frac{y}{40} = 51 \). First, express \( y \) from the first equation: \( y = 1800 - x \). Substitute \( y \) in the second equation: \( \frac{x}{25} + \frac{1800-x}{40} = 51 \).

Step 4: Eliminate Fractions

Multiply the entire equation by 200 (LCM of 25 and 40) to eliminate fractions: \( 8x + 5(1800-x) = 10200 \).

Step 5: Simplify the Equation

Expand and simplify the equation: \( 8x + 9000 - 5x = 10200 \), which simplifies to \( 3x = 1200 \).

Step 6: Solve for \( x \)

Divide both sides by 3: \( x = 400 \). This means 400 miles were driven in the city.

Key Concepts

Understanding System of Equations

In mathematics, a system of equations is a set of two or more equations that share the same set of variables. The core goal is to find values for these variables that satisfy all equations simultaneously. By using systems of equations, we can solve complex word problems that involve multiple conditions.
In our exercise, two equations were derived: one for the total mileage and another for gasoline consumption, both involving the variables for city and highway miles driven (\(x\) and \(y\) ). Here is a brief overview of the equations:

• \(x + y = 1800\) represents the total miles driven.
• \( \frac{x}{25} + \frac{y}{40} = 51\) involves the total gasoline used based on mileage in city and highway.

These equations make up a system because they describe the same scenario but from different perspectives. Solving systems of equations like these can help break down the problem into manageable parts. Utilizing methods like substitution or elimination allows us to find the solution efficiently.

Mileage Calculation Breakdown

Mileage calculation is crucial when dealing with problems that involve distances and gasoline consumption. In this exercise, the mileage estimates are provided in miles per gallon (mi/gal), a typical measure of efficiency for vehicles. These estimates allow us to calculate total gasoline usage on different routes, making it a practical application of high school algebra.

• The city mileage is \(25 \text{ mi/gal} \).
• The highway mileage is \(40 \text{ mi/gal} \).

By knowing these values, we can set up accurate models to represent how gas consumption varies between city and highway driving. Once we have the total miles and the total gallons of gasoline used, we formulate equations to solve for the actual miles driven in each condition.

The Process of Algebraic Substitution

Algebraic substitution is a handy technique for solving a system of equations, where one equation is solved for one variable, and this expression is substituted into another equation. This method is particularly useful when one of the variables is straightforward to isolate.
In this exercise, substitution begins with the equation for total miles:

• From \(x + y = 1800\), we isolate \(y\) as \(y = 1800 - x\).

This expression for \(y\) is then substituted into the gasoline equation:

• \( \frac{x}{25} + \frac{1800 - x}{40} = 51\).

This substitution creates an equation with a single variable, \(x\), which can then be solved through arithmetic processes. Substitution reduces the complexity of solving multiple-variable equations by focusing on one variable at a time.