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Chapter 3: Problem 21

Two days at Busch Gardens (Tampa Bay) and 3 days at Universal Studios Florida(Orlando) cost \(\$ 510,\) while 4 days at Busch Gardens and 2 days at UniversalStudios cost \(\$ 580 .\) (Prices are based on single-day admissions.) What wasthe cost per day for each park?

### Short Answer

Expert verified

Cost per day at Busch Gardens: \(90, Cost per day at Universal Studios: \)110.

## Step by step solution

01

## - Define variables

Let the cost per day at Busch Gardens be denoted as \( x \) and the cost per day at Universal Studios be denoted as \( y \). This gives us the following system of equations based on the problem statement.

02

## - Write the system of equations

Based on the given costs:1. For 2 days at Busch Gardens and 3 days at Universal Studios: \( 2x + 3y = 510 \)2. For 4 days at Busch Gardens and 2 days at Universal Studios: \( 4x + 2y = 580 \).

03

## - Simplify the second equation

Divide the second equation by 2 to make it simpler: \( 2x + y = 290 \). Now the system of equations is: 1. \( 2x + 3y = 510 \)2. \( 2x + y = 290 \).

04

## - Eliminate one variable

Subtract the second equation from the first equation to eliminate \( 2x \):\( (2x + 3y) - (2x + y) = 510 - 290 \). Simplifying gives \( 2y = 220 \), or \ y = 110 \.

05

## - Solve for the remaining variable

Substitute \( y = 110 \) back into the second equation: \( 2x + 110 = 290 \). Simplifying gives \( 2x = 180 \), or \ x = 90 \.

06

## - State the final answer

The cost per day at Busch Gardens is \( x = 90 \) and the cost per day at Universal Studios is \( y = 110 \).

## Key Concepts

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

###### algebra

Algebra is the branch of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It allows for the solving of problems with unknown values by setting up equations based on given information. In the problem we addressed, algebra was essential to form and solve the system of equations for determining the daily costs of tickets to Busch Gardens and Universal Studios. Knowing algebra helps you manipulate and solve equations to find those unknowns, such as the cost per day at each park. This involves operations like addition, subtraction, multiplication, and division, as well as using techniques like substitution and elimination.

###### linear equations

Linear equations are equations of the first degree, meaning they involve only the sum of terms, each of which is the product of a constant and a variable raised to the first power. In our problem, we dealt with two linear equations:

1. \(2x + 3y = 510\)

2. \(4x + 2y = 580\)

Linear equations often represent real-world problems where relationships between quantities can be described in a straight line when graphed. The goal is to find the values of the variables that satisfy all equations in the system. It’s important to understand how to manipulate these equations using algebraic techniques.

###### substitution method

The substitution method for solving systems of equations involves solving one equation for one variable and then substituting this solution into the other equation. In our example, we simplified the second equation first to get:

\(2x + y = 290\).

By solving for y in terms of x, or x in terms of y, we can replace this value in the other equation. However, in our problem, we simplified and chose to use elimination primarily. The substitution method can be very efficient, particularly when one of the equations is simple to solve for one of the variables.

###### elimination method

The elimination method involves adding or subtracting equations to eliminate one of the variables. This makes it easier to solve for the remaining variable. In our problem:

We subtracted the simplified second equation \(2x + y = 290\) from the first equation \(2x + 3y = 510\):

\[(2x + 3y) - (2x + y) = 510 - 290\]

Simplifying, we get \(2y = 220\).

By solving \(2y = 220\), we found that \(y = 110\).

We then substituted \(y = 110\) back into \(2x + y = 290\) to find \(x = 90\).

This method is particularly useful when the coefficients of one variable are easily made the same or opposites.

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