Chapter 6.3, Problem 6STP

A hotel is offering the two weekend specials described below.

Plan 1: 3 nights and 4 meals for $233.

Plan 2: 3 nights and 3 meals for $226.50.

For accounting purposes, the hotel will record income for the stay and the meals separately. So the cost per night for each special must be the same, and the cost per meal for each special must be the same.

a. What is a system of equations for the situation?

b. Solve the system. What is the cost per night and the cost per meal?

To determine

To write the system of equations for the given situation

Answer to Problem 6STP

  $\begin{array}{l}3x+4y=233\\ 3x+3y=226.5\end{array}$

Explanation of Solution

Given:

An hotel offers the 2 plans, i.e.,

Plan 1: 3 nights and 4 meals for $233

Plan 2: 3 nights and 3 meals for $226.50

Formula Used:

  $\text{Total Cost = (No}\text{. of nights}\times \text{Cost of per night) + (No}\text{. of meals }\times \text{Cost of per meal)}$

Calculation:

Let,

The Cost of per night be $ $x$

The Cost of per meal be $ $y$

Given,

Plan 1: 3 nights and 4 meals for $233

Therefore,

$3x+4y=233$

Also,

Plan 2: 3 nights and 3 meals for $226.50

Therefore,

$3x+3y=226.5$

Conclusion:

Hence, the required system of equations is:

  $3x+4y=233$ and $3x+3y=226.5$

To determine

To solve the system of equation obtained in the previous part

Answer to Problem 6STP

Cost of per night is $ 69

Cost of per meal is $ 6.5

Explanation of Solution

The obtained system of equations is:

  $\begin{array}{l}3x+4y=233\\ 3x+3y=226.5\end{array}$

Calculation:

Let,

$3x+4y=233-----(i)$

  $3x+3y=226.5------(ii)$

Now, subtract equation (ii) and ( i ),

$\begin{array}{l}\Rightarrow (3x+4y)-(3x+3y)=233-226.5\\ \Rightarrow y=6.5\end{array}$

Substituting the value of $y$ in equation ( i ),

  $\begin{array}{l}\Rightarrow 3x+4\times 6.5=233\\ \Rightarrow 3x=207\\ \Rightarrow x=69\end{array}$

Conclusion:

Hence, cost of per night is $ 69 and cost of per meal is $ 6.5.