Problem 68 - Algebra A Combined Function

Question

The second best-selling car is the Volkswagen Golf. A driver of this car took a day trip around the California coastline driving at two different speeds. He drove 70 miles at a slower speed and 300 miles at a speed 40 miles per hour faster. If the time spent driving the faster speed was twice that spent at the slower speed, find the two speeds during the trip. (Source: Top Ten of Everything)

Step-by-Step Solution

Answer

The two speeds are 35 mph and 75 mph.

Step 1: Define Variables

Let's define the variables needed for this problem. Let \( x \) be the slower speed in miles per hour. The faster speed is then \( x + 40 \) miles per hour.

Step 2: Formulate Time Equations

The time spent traveling is distance divided by speed. For the slower speed, the time is \( \frac{70}{x} \). For the faster speed, the time is \( \frac{300}{x + 40} \). It's given that the time at the faster speed is twice the time at the slower speed, so we have the equation: \[ \frac{300}{x + 40} = 2 \cdot \frac{70}{x} \].

Step 3: Solve the Equation

Start by cross-multiplying in the equation from Step 2: \( 300x = 2 \cdot 70(x + 40) \). Simplify to get: \[ 300x = 140x + 5600 \]. Move terms to one side to solve for \( x \): \[ 160x = 5600 \].

Step 4: Find the Slower Speed

Divide both sides of the equation by 160 to isolate \( x \): \[ x = \frac{5600}{160} = 35 \]. The slower speed is 35 miles per hour.

Step 5: Find the Faster Speed

Since the faster speed was \( x + 40 \), substitute \( x = 35 \): \[ 35 + 40 = 75 \]. The faster speed is 75 miles per hour.

Key Concepts

linear equation

Linear equations are fundamental in solving algebra word problems, and they're especially useful in managing real-life scenarios involving relationships between quantities. A linear equation is an algebraic equation of the form \( ax + b = c \). In this particular problem, the linear equation helps us determine the speeds for the trip based on given conditions.How Linear Equations Work in Word Problems
In our example, we started with an equation that compared time traveled at different speeds. When the times are given in terms of variables, we form equations that link those variables. By solving the linear equation, we find surface-level relationships that satisfy all the conditions of the problem.

• Translate word problems into algebraic expressions.
• Identify relationships between unknowns.
• Solve for values using algebraic manipulation.

These steps are crucial for breaking down the problem and finding the solution.Linear equations often require moving terms from one side of an equation to another, combining like terms, and simplifying the expressions to isolate the variable. Once simplified, we can find the precise numeric values that solve the equation.

variable definition

Defining variables is a critical first step in solving algebra word problems. In this scenario, we had to identify the speeds at which the car traveled at different points. We defined a variable \( x \) to represent the slower speed, which allowed us to subsequently express the faster speed as \( x + 40 \).Why Variables are Important
Variables act as placeholders for unknowns we need to solve for. By defining them clearly at the start, we make it easier to set up our equations. Here’s how to think about defining variables:

• Analyze the problem to identify the unknowns.
• Choose symbols that will represent these unknowns efficiently.
• Express other quantities in terms of these variables.

For instance, setting \( x \) as the slower speed allows the formulation of other conditions (e.g., the faster speed is \( x + 40 \)). This makes it easier to relate different parts of the problem to each other.

rate and time problems

Rate and time problems are a classic type of algebra word problem that often involve finding speeds and travel times. In these scenarios, understanding the basic formula \( \text{Distance} = \text{Rate} \times \text{Time} \) is key.Applying the Formula
In our car trip example, this formula is the backbone of solving the problem. For each part of the journey, we know the distance and need to determine the time involved:

• For the slow speed: \( \text{Time} = \frac{70}{x} \).
• For the fast speed: \( \text{Time} = \frac{300}{x + 40} \).

Given that time at the faster speed was twice the slower speed's time, the corresponding linear equation was formed: \( \frac{300}{x + 40} = 2 \cdot \frac{70}{x} \).Understanding Rates and Times

• Time goes inversely as speed; increase speed means less time to cover the same distance.
• Consistent units are essential; ensure distances and speeds are compatible.

These problems help build intuition about how movement and time relate, which is applicable in diverse scenarios.