Problem 34 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. A small company has budgeted \(\$ 6,000\) per month to lease vehicles. On this budget, the company can lease 12 cars and 4 trucks each month, or 8 cars and 6 trucks. Find the monthly cost to lease a car and to lease a truck.

Step-by-Step Solution

Answer

The monthly cost to lease a car is \ \$300\ \ , and to lease a truck is \ \$600\.

Step 1 - Define the Variables

Let’s define the variables: Let \( c \) represent the monthly cost to lease a car and \( t \) represent the monthly cost to lease a truck.

Step 2 - Set Up the Equations

We are given two scenarios: 1. 12 cars and 4 trucks cost \$6,000 per month. This gives the equation \( 12c + 4t = 6000 \).2. 8 cars and 6 trucks also cost \$6,000 per month. This gives the equation \( 8c + 6t = 6000 \).

Step 3 - Simplify the Equations

Simplify both of the equations: 1. Divide the first equation by 4: \( 3c + t = 1500 \).2. Divide the second equation by 2: \( 4c + 3t = 3000 \).

Step 4 - Use the Subtraction Method

Multiply the simplified first equation by 3: \( 9c + 3t = 4500 \). Then subtract the second simplified equation (4c + 3t = 3000) from this new equation: \[ 9c + 3t - (4c + 3t) = 4500 - 3000 \]. This simplifies to \( 5c = 1500 \).

Step 5 - Solve for Cost of Leasing a Car

Solve for \( c \): \[ 5c = 1500 \] \[ c = \frac {1500}{5} = 300 \]. Therefore, the monthly cost to lease a car is \ \$300\ \.

Step 6 - Solve for Cost of Leasing a Truck

Substitute \( c = 300 \) back into the first simplified equation: \[ 3(300) + t = 1500 \] \[ 900 + t = 1500 \] \[ t = 1500 - 900 = 600 \]. Therefore, the monthly cost to lease a truck is \ \$600\ \.

Key Concepts

Defining Variables

Understanding and defining variables is crucial when solving algebraic equations, as it helps to represent unknown values in a clear and concise manner.
In this problem, we are dealing with two types of vehicle leases: cars and trucks. We need to find their monthly costs, so let's assign each a variable:

• Let c represent the monthly cost to lease a car.
• Let t represent the monthly cost to lease a truck.

By defining these variables, we set the foundation for setting up our equations and ultimately solving the problem.
Using easy-to-remember letters for variables, like c for cars and t for trucks, also helps keep track of what each variable represents throughout the calculation.

Setting Up Equations

Once we have our variables defined, the next step is to convert the given conditions of the problem into algebraic equations. This uses the provided information to form mathematical sentences.
We have two scenarios:

• In the first scenario, leasing 12 cars and 4 trucks costs \(6,000. This can be translated to the equation: \( 12c + 4t = 6000 \)
• In the second scenario, leasing 8 cars and 6 trucks also costs \)6,000. This translates to the equation: \( 8c + 6t = 6000 \)

By setting up these two equations, we have a system of equations that we can solve to find the cost of leasing a car and a truck.

Simplifying Equations

Simplifying the equations is an important step as it makes solving them more manageable. Simplification often involves dividing all terms of the equation by a common factor.
In the first equation:\( 12c + 4t = 6000 \)
We divide each term by 4:
\( 3c + t = 1500 \)
In the second equation:\( 8c + 6t = 6000 \)
We divide each term by 2:
\( 4c + 3t = 3000 \)
Now, our simplified equations are easier to work with: \( 3c + t = 1500 \) and \( 4c + 3t = 3000 \)
This reduction can significantly lessen computational load and clarify the algebraic process.

Subtraction Method in Algebra

The subtraction method (or elimination method) is useful for eliminating one variable, making it easier to solve the system of equations.
First, we need to align our simplified equations for subtraction:
Multiply the first equation by 3:\( 9c + 3t = 4500 \)
Now subtract the second equation:
\( (9c + 3t) - (4c + 3t) = 4500 - 3000 \)
This simplifies to:
\( 5c = 1500 \)
This subtraction eliminates t, isolating c so that we can solve for it directly.

Solving for Variables

Now that we have isolated \(5c = 1500\), we solve for \(c\) by dividing both sides of the equation by 5:
\( c = \frac{1500}{5} = 300 \)
Therefore, the monthly cost to lease a car is \(300.
Next, we substitute \(c = 300\) back into one of our simplified equations, preferably the simpler one:
\(3c + t = 1500\)
This gives us:

\(3(300) + t = 1500 \)
\(900 + t = 1500 \)
\(t = 1500 - 900 = 600 \)
Therefore, the monthly cost to lease a truck is \)600. By following these steps methodically, we are able to determine both unknown costs accurately.