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Chapter 4: Problem 42

California 1 South is a historic highway that stretches 123 miles along the coast from Monterey to Morro Bay. Suppose that two cars start driving this highway, one from each town. They meet after 3 hours. Find the rate of each car if one car travels 1 mile per hour faster than the other car.

Short Answer

Expert verified
The slower car travels at 20 mph, and the faster car travels at 21 mph.

Step by step solution

01

Define the Variables

Let's define the variables first: let the speed of the slower car be \( x \) miles per hour. Then, the speed of the faster car, which is 1 mile per hour faster, will be \( x + 1 \) miles per hour.
02

Setup the Equation

Since both cars meet after 3 hours and they together cover the total distance of 123 miles, we can express this as an equation: \( 3x + 3(x+1) = 123 \).
03

Simplify the Expression

Let's simplify the expression: \[3x + 3x + 3 = 123,\] combine like terms: \[6x + 3 = 123.\]
04

Solve for x

First, subtract 3 from both sides: \[6x = 120.\] Next, divide by 6 to solve for \( x \): \[x = 20.\] This means the slower car was traveling at 20 miles per hour.
05

Find the Rate of the Faster Car

Since the faster car is traveling 1 mile per hour more than the slower car, its speed is \( x + 1 = 21 \) miles per hour.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Equations
Linear equations are fundamental in algebra as they express relationships between variables in a proportional way. They often appear in the form of \( ax + b = c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. In our highway problem, both cars鈥� distances over time need to equal the total distance of 123 miles. Hence, we form a linear equation: \( 3x + 3(x+1) = 123 \). Here, both terms on the left side represent the distances each car travels in three hours. Breaking down complex problems into linear equations helps in simplifying the problem, making it easier to find the solution.
  • Linear Equations express relationships with constant rates
  • The equation \( 3x + 3(x+1) = 123 \) represents the combined distances covered by the two cars
  • They simplify problem-solving by creating a straightforward relationship
Variable Definition
Defining variables is crucial in solving problems mathematically. A variable is a symbol that stands in for an unknown quantity or a changing value. In the problem, we first assign a variable to represent a known unknown: the speed of the slower car. Let鈥檚 denote this as \( x \).
Since the faster car travels 1 mile per hour more, its speed is represented as \( x + 1 \).
This strategic setup helps us translate a real-world situation into a mathematical model, where each variable corresponds to a specific part of the problem.
  • Variables act as placeholders for unknowns
  • In our scenario, \( x \) represents the speed of the slower car
  • Choosing appropriate variables simplifies further solution processes
Equation Solving
Solving equations involves a series of manipulations that balance both sides to isolate and find the value of the unknown variable. In our problem, we begin with the equation \( 3x + 3(x+1) = 123 \). Simplifying this requires:
  • First, expanding the equation
  • Combining like terms to get \( 6x + 3 = 123 \)
  • Isolating \( x \) by subtracting 3, resulting in \( 6x = 120 \)
  • Finally, dividing by 6 to yield \( x = 20 \)
These steps, if followed precisely, lead us to the desired solution, in this case, the speed of the slower car being 20 miles per hour.
Problem Solving Steps
Effective problem solving in algebra often follows a structured approach, which can be replicated across various problems. The process is as follows:
First, clearly define the variables to represent unknowns (step 1). Next, set up the equation using the relationship defined by the problem (step 2). Simplify the expression by performing basic algebraic operations (step 3). Solve for the variable by isolating it through steps of subtraction and division (step 4). Finally, calculate any additional values needed (step 5).
  • Defining variables sets the foundation for solving
  • Constructing the equation models the problem mathematically
  • Simplifying lays the groundwork for easy calculation
  • Each step brings us closer to the solution
Using these steps systematically not only helps in finding the correct answer but also builds strong problem-solving skills in algebra.

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Most popular questions from this chapter

Use the system of linear equations below to answer the questions. \(\left\\{\begin{array}{l}x+y=5 \\ 3 x+3 y=b\end{array}\right.\) a. Find the value of \(b\) so that the system has an infinite number of solutions. b. Find a value of \(b\) so that there are no solutions to the system.

Write equivalent equations by multiplying both sides of each given equation by the given nonzero number. $$ -4 x+y=3 \text { by } 3 $$

The sales \(y\) of VHS movie format units (in billions of dollars) sold in the United States from 2000 to 2008 is given by \(y=-1.1 x+7.1,\) where \(x\) is the number of years after 2000 . The sales \(y\) of DVD movie format units (in billions of dollars) sold in the United States from 2000 to 2008 is given by \(y=1.9 x+4.7,\) where \(x\) is the number of years after 2000. (Source: EMedia Digital Studio Magazine) a. Use the substitution method to solve this system of equations. $$ \left\\{\begin{array}{l} y=-1.1 x+7.1 \\ y=1.9 x+4.7 \end{array}\right. $$ Round \(x\) to the nearest tenth and \(y\) to the nearest whole number. b. Explain the meaning of your answer to part (a). c. Sketch a graph of the system of equations. Write a sentence describing the trends in the popularity of these two types of movie formats. d. Use the VHS equation to find the sales of VHS units in 2007. Then explain your answer.

In recent years, the number of newspapers printed as morning editions has been increasing and the number of newspapers printed as evening editions has been decreasing. The number \(y\) of daily morning newspapers in existence from 1997 through 2007 is approximated by the equation \(146 x-10 y=-7086\), where \(x\) is the number of years since \(1997 .\) The number \(y\) of daily evening newspapers in existence from 1997 through 2007 is approximated by \(111 x+5 y=4058,\) where \(x\) is the number of years since 1997. (Source: Based on data from Newspaper Association of America) a. Use the addition method to solve this system of equations. $$ \left\\{\begin{array}{l} 146 x-10 y=-7086 \\ 111 x+5 y=4058 \end{array}\right. $$ (Round to the nearest whole number. Because of rounding, the \(y\) -value of your ordered pair solution may vary.) b. Interpret your solution from part (a). C. How many of each type of newspaper were in existence that year?

Solve each system of equations by the addition method. If a system contains fractions or decimals, you may want to first clear each equation of fractions or decimals.$$ \left\\{\begin{array}{l} 3 x+y=4 \\ 9 x+3 y=6 \end{array}\right. $$
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