Problem 30 - Introductory and Intermediate Algebra for College Students 4th

Question

A car rental agency charges 180 dollar per week plus 0.25 dollar per mile to rent a car. How many miles can you travel in one week for 395 dollar? (GRAPH CAN NOT COPY)

Step-by-Step Solution

Answer

You can travel 860 miles in one week for 395 dollars.

Step 1: Understand and Set Up the Equation

The total cost is equal to the weekly cost plus the cost per mile multiplied by the distance traveled. Since we know the total cost is 395 dollars, the weekly cost is 180 dollars, and the cost per mile is 0.25 dollars. This can be represented as follows: \[395 = 180 + 0.25m\], where 'm' represents the number of miles.

Step 2: Solve the Equation for 'm'

In the equation \[395 = 180 + 0.25m, \] we want to isolate 'm'. We can do that by first subtracting the constant, which in this case is 180, from both sides of the equation. Hence, \[395 - 180 = 180 - 180 + 0.25m\] which translates to \[215 = 0.25m.\] Then, we can solve for 'm' by dividing both sides of the equation by the coefficient of m which is 0.25 which gives, \[m = \frac{215}{0.25}.\]

Step 3: Compute the Value of 'm'

The last step is computing \[m= \frac{215}{0.25}.\] The division leads to 'm' being equal to 860.

Key Concepts

Algebraic Equations

Understanding algebraic equations is crucial for solving various mathematical problems. An algebraic equation is a statement of equality between two expressions that contain variables and constants. In particular, linear equations are the simplest form of algebraic equations, involving only first-degree variables—meaning the variable is not raised to any power higher than one.

For example, the equation given in the car rental problem, \[395 = 180 + 0.25m\], is a linear equation where \(m\) represents the number of miles driven and the numbers represent the cost associated with the car rental. By manipulating such equations with basic operations—addition, subtraction, multiplication, and division—we can solve for the unknown variable. The goal is always to isolate the variable on one side to find its value.

Word Problems in Algebra

Word problems can seem daunting, but they are just real-world scenarios described with numbers and unknowns, framed within a story. To solve word problems in algebra:\(

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• Begin by carefully reading the problem to understand what is being asked.
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• \)Identify the known quantities and the unknown variables.
• Translate the story into a mathematical equation using these variables.
• Solve the equation using algebraic methods to find the value of the unknown.
• Lastly, interpret the solution in the context of the problem to ensure it makes sense.
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\)In the car rental exercise, it's important to translate the verbal description into the linear equation. The use of symbols like \(m\) to represent miles is crucial in transitioning from a word problem to an algebraic equation, which can then be manipulated and solved.

Systems of Linear Equations

While the car rental problem involves a single linear equation, many real-world problems require working with systems of linear equations, where two or more linear equations are solved together to find a common solution for their variables.

In general, a system of linear equations can be written in the form:\(ax + by = c\) and \(dx + ey = f\), where \(x\) and \(y\) are the variables, and \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\) are constants. Solutions to these systems can be found graphically by plotting the lines and finding their intersection, or algebraically using methods such as substitution or elimination.

These systems are all around us, from calculating simultaneous financial transactions to predicting traffic patterns. Thus, understanding how to solve them is an essential skill for solving complex problems in various fields.