Problem 56 - Algebra and Trigonometry Real Mathematics, Real People

Question

On the first part of a 336 -mile trip, a salesperson averaged 58 miles per hour. The salesperson averaged only 52 miles per hour on the last part of the trip because of an increased volume of traffic. The total time of the trip was 6 hours. Find the amount of time at each of the two speeds.

Step-by-Step Solution

Answer

The salesperson travels at 58 mph for 4 hours and at 52 mph for 2 hours.

Step 1: Identify variables and equations

We know that the total distance of the trip is 336 miles, and that it took 6 hours. Denote the time spent at 58 mph as \( t1 \) and the time spent at 52 mph as \( t2 \). From here, two equations can be formed using formula of distance and total time: \n1) \( 58t1 + 52t2 = 336 \)2) \( t1 + t2 = 6 \)

Step 2: Solve the system of equations

Begin by multiplying the second equation by 52 to prepare for substitution: \( 52t1 + 52t2 = 312 \). Now, we subtract the second equation from the first one: \( 6t1 = 24 \), which upon simplifying, yields \( t1 = 4 \) hours.

Step 3: Find the second variable

After obtaining \( t1 = 4 \) hours, substitute this value into the second equation from Step 1: \( 4 + t2 = 6 \). Simplify this equation to find that \( t2 = 2 \) hours.

Key Concepts

Algebra Word Problems

Solving algebra word problems involves converting real-world situations into mathematical statements using variables and symbols. The first critical step is to understand and translate the given information from the exercise. In our case, we're dealing with a trip divided into two parts with different average speeds. We start by identifying our unknowns—we want to find out the time spent at each speed. With a total distance of 336 miles and a total time of 6 hours, the problem mirrors practical scenarios we may encounter, such as planning travel times.

We break down the task by labeling our unknowns with variables, which in this case are the times at the two different speeds, expressed as t1 and t2. The key is then to express the rates (speeds) and distances in algebraic equations that reflect the relationship between them. Once the information is translated and equations are set up, the stage is set for using algebraic methods to find the solutions.

Distance Rate Time Problems

Distance, rate (speed), and time problems are classic applications of algebra in real-world contexts. They are based on the fundamental formula distance = rate × time. This relationship is essential in formulating our equations for the problem at hand. We're given two phases of travel—each with a known rate but unknown time.

For the first part of the journey at 58 miles per hour, the distance covered can be expressed as 58 × t1. Similarly, the second part at 52 miles per hour gives us 52 × t2. The total distance is the sum of the two, leading to our first equation. Furthermore, since the total time is known (6 hours), we also have a straightforward equation representing time: t1 + t2 = 6. Solving this set of equations gives us the individual travel times that contribute to the overall journey.

Solving Algebraic Equations

Once we have our equations, the next phase is to solve the system, which consists of two algebraic equations with two unknowns (t1 and t2). There are various methods to solve such systems, including substitution, elimination, and graphing. We chose the elimination method in the solution to efficiently subtract one equation from the other to solve for one variable at a time.

Step 1: To eliminate t2, we multiplied the second equation by 52 (the speed during t2), aligning it with t2's coefficient in the distance equation. This created two equations with the same coefficient for t2, making it easy to subtract and find t1.

Step 2: After finding t1, we substituted its value back into one of the original equations to solve for t2. The process of first eliminating one variable and then substituting to find the other is a systematic approach that leads to the final solution of algebraic word problems involving two variables.