Question

Three Nissans, two Fords, and four Chevrolets can be rented for $$\$ 106$$ per day. At the same rates two Nissans, four Fords, and three Chevrolets cost $$\$ 107$$ per day, whereas four Nissans, three Fords, and two Chevrolets cost $$\$ 102$$ per day. Find the rental rates for all three kinds of cars.

Short answer

Nissan: $32, Ford: $1, Chevrolet: $2 per day.

Step-by-step solution

Define Variables

Let \( x \) be the cost to rent a Nissan, \( y \) be the cost to rent a Ford, and \( z \) be the cost to rent a Chevrolet per day. We need to solve for \( x, y, \) and \( z \).

Set Up Equations

Based on the given information, set up the following system of equations:\[ 3x + 2y + 4z = 106 \] \[ 2x + 4y + 3z = 107 \] \[ 4x + 3y + 2z = 102 \]

Use the Elimination Method on the First Two Equations

Multiply the first equation by 4 and the second equation by 3 to eliminate \( z \). This results in: \[ 12x + 8y + 16z = 424 \] \[ 6x + 12y + 9z = 321 \] Subtract the second from the first: \[ 6x - 4y + 7z = 103 \]

Use the Elimination Method on the First and Third Equations

Multiply the first equation by 2 and the third equation by 3 to eliminate \( z \). This results in: \[ 6x + 4y + 8z = 212 \] \[ 12x + 9y + 6z = 306 \] Subtract the first from the second: \[ 6x + 5y - 2z = 94 \]

Solve for y and z

Now solve the two new equations for \( y \) and \( z \):\[ 6x - 4y + 7z = 103 \] \[ 6x + 5y - 2z = 94 \] Eliminate \( x \) by subtracting the first modified equation from the second : \[ 9y - 9z = -9 \]. Simplifying gives: \[ y - z = -1 \], or \[ y = z - 1 \].

Substitute y in Terms of z

Using \( y = z - 1 \) substitute into the first equation \( 3x + 2(z - 1) + 4z = 106 \): \[ 3x + 2z - 2 + 4z = 106 \] Simplifying results in: \[ 3x + 6z = 108 \].

Solve for x and z

From \( 3x + 6z = 108 \), we get \( 3x = 108 - 6z \), or \( x = 36 - 2z \). Substitute back into one of the original equations, for example the first one: \[ 3(36 - 2z) + 2(z - 1) + 4z = 106 \]. Simplifying gives \[ 108 - 6z + 2z - 2 + 4z = 106 \]. Further simplifying: \( 108 - 2 = 106 \). Hence, \( z = 2 \).

Find Values of x, y, z

Use \( z = 2 \) in \( y = z - 1 \) to get \( y = 1 \). Then, substitute \( z = 2 \) back in \( x = 36 - 2z \) to find \( x = 32 \). So, the rental costs are: Nissan \( x = 32 \), Ford \( y = 1 \), and Chevrolet \( z = 2 \).

Key concepts

Linear equations

Linear equations are the foundation of many mathematical concepts such as systems of equations. These equations typically involve variables that are multiplied by coefficients and are set equal to a constant. Here, your main goal is to find the values of the variables that make the equation true. For example, in our car rental problem, the costs of renting different cars are represented by the variables \( x \), \( y \), and \( z \). Together, they form the system of equations:

• \( 3x + 2y + 4z = 106 \)
• \( 2x + 4y + 3z = 107 \)
• \( 4x + 3y + 2z = 102 \)

Solving these equations means finding consistent values of \( x \), \( y \), and \( z \) that satisfy all three equations simultaneously.
The solutions to linear equations like these can be approached using various methods, including substitution and elimination, which we'll explore next.

Substitution method

The substitution method is a strategy for solving systems of equations where you express one variable in terms of another, and then substitute this expression into the other equations. This helps reduce the number of variables, making it easier to solve the system step by step.
For instance, when using substitution in our given problem, we already found part of our solution through elimination. We discovered that \( y = z - 1 \). This means, wherever you see \( y \) in the equations, you can replace it with \( z - 1 \).
This substitution simplifies equations and focuses your efforts on fewer unknowns. As shown, substituting directly in the equation \( 3x + 2(z - 1) + 4z = 106 \), we rearrange and simplify to find a simpler equation involving only \( x \) and \( z \).
With substitution, patience and careful calculation are key as mistakes in basic arithmetic can lead to errors further down the solution path.

Elimination method

The elimination method is another effective way to solve systems of linear equations. It involves adding or subtracting equations to eliminate one of the variables, thereby simplifying the system.
In the context of our car rental problem, elimination was used in several steps: we combined equations strategically to remove one of the variables from consideration. For example, we manipulated the equations to eliminate \( z \) first:
1. Multiplied the first equation by 4 and the second one by 32. Subtracted the second result from the first
This process resulted in a new equation that no longer contained the \( z \) variable.
Then, by repeating similar operations with other pairs of equations, we managed to eliminate another variable, ultimately landing us in a position to solve for one of the variables directly.
Elimination is efficient but requires careful arithmetic and can sometimes involve dealing with larger numbers, which need precision during calculations to avoid any mistakes.