Problem 47
Question
You plan a trip that involves a 40 -mile bus ride and a train ride. The entire trip is 140 miles. The time (in hours) the bus travels is \(y_1=\frac{40}{x}\), where \(x\) is the average speed (in miles per hour) of the bus. The time (in hours) the train travels is \(y_2=\frac{100}{x+30}\). Write and simplify a model that shows the total time \(y\) of the trip.
Step-by-Step Solution
Verified Answer
The simplified model equation showing the total time of the trip is \(y = \frac{140x + 1200}{x^2 + 30x}\).
1Step 1: Understand the Problem
The total distance for the trip is 140 miles, with 40 miles covered by a bus ride and the remaining 100 miles covered by a train ride. The problem provides two equations that represent the time each mode of transportation takes, given their respective speeds. The task is to find a model equation that represents the total time of the trip.
2Step 2: Write the Total Time Equation
We know that time equals distance divided by speed for both the bus ride and the train ride. Therefore, the total time of the trip \(y\) can be found by summing up \(y_1\) and \(y_2\). We can therefore write the equation for the total time as \(y = y_1 + y_2\). Substituting \(y_1\) and \(y_2\) with the provided expressions, our equation becomes \(y = \frac{40}{x} + \frac{100}{x+30}\).
3Step 3: Simplify the Total Time Equation
To simplify this equation, find a common denominator for the two fractions. The common denominator is \(x(x+30)\). By rewriting the fractions with the common denominator, the equation becomes \(y = \frac{40(x+30) + 100x}{x(x+30)}\). Simplifying the numerator gives us \(y = \frac{40x + 1200 + 100x}{x^2 + 30x} = \frac{140x + 1200}{x^2 + 30x}\).
Key Concepts
Algebraic EquationsCommon DenominatorsRational Expressions
Algebraic Equations
Algebraic equations are fundamental in expressing relationships between different quantities. In our example, we look at a scenario involving a bus and train ride where algebra equations help represent the time taken for travel based on speed. Specifically, we use equations like \(y_1=\frac{40}{x}\) and \(y_2=\frac{100}{x+30}\) to denote the time spent in hours for the bus and the train rides, respectively. Here, \(x\) represents the average speed.
To find the total travel time \(y\), we create an equation that adds both individual times together: \(y = y_1 + y_2\). The beauty of algebra is in its ability to form such composite equations from individual parts and then simplify them to get a more comprehensive understanding of the entire system. When working on algebra equations, it's crucial to ensure that the operations performed on both sides of the equation maintain equality, leading to accurate solutions.
To find the total travel time \(y\), we create an equation that adds both individual times together: \(y = y_1 + y_2\). The beauty of algebra is in its ability to form such composite equations from individual parts and then simplify them to get a more comprehensive understanding of the entire system. When working on algebra equations, it's crucial to ensure that the operations performed on both sides of the equation maintain equality, leading to accurate solutions.
Common Denominators
When dealing with fractions, a common denominator is necessary to combine terms, as seen when simplifying algebra equations. The common denominator is the least common multiple of the denominators of separate fractions. In our exercise, we have denominators \(x\) and \(x+30\) that need to be combined. The least common multiple of these two expressions is their product, \(x(x+30)\), which becomes our common denominator.
Why do we need a common denominator? It allows us to perform addition or subtraction on fractions by ensuring the pieces we are adding are comparable—in our case, using a common denominator lets us sum the time spent on the bus and train into a single fraction. To simplify expressions efficiently, always find and use the smallest common denominator that enables you to combine terms without losing the integrity of the original expressions.
Why do we need a common denominator? It allows us to perform addition or subtraction on fractions by ensuring the pieces we are adding are comparable—in our case, using a common denominator lets us sum the time spent on the bus and train into a single fraction. To simplify expressions efficiently, always find and use the smallest common denominator that enables you to combine terms without losing the integrity of the original expressions.
Rational Expressions
Rational expressions are fractions where the numerator and denominator are polynomials. In our travel time equation, \(y = \frac{40}{x} + \frac{100}{x+30}\), we have two rational expressions that we need to combine. Just like numerical fractions, rational expressions require a common denominator before they can be added or subtracted.
To simplify a rational expression, one seeks to combine like terms and reduce the expression to its lowest terms, if possible. To reinforce understanding, let's revisit our travel time equation after finding the common denominator: \(y = \frac{40(x+30) + 100x}{x(x+30)}\), which simplifies to \(y = \frac{140x + 1200}{x^2 + 30x}\). The numerator here represents the sum of time in terms of speed, and the denominator represents the product of different speed scenarios affecting the bus and the train. The simplification of rational expressions is critical in algebra, as it makes equations more manageable and solutions clearer to interpret.
To simplify a rational expression, one seeks to combine like terms and reduce the expression to its lowest terms, if possible. To reinforce understanding, let's revisit our travel time equation after finding the common denominator: \(y = \frac{40(x+30) + 100x}{x(x+30)}\), which simplifies to \(y = \frac{140x + 1200}{x^2 + 30x}\). The numerator here represents the sum of time in terms of speed, and the denominator represents the product of different speed scenarios affecting the bus and the train. The simplification of rational expressions is critical in algebra, as it makes equations more manageable and solutions clearer to interpret.
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Problem 46
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