Problem 43 - Precalculus with Limits

Question

Nutrition Two cheeseburgers and one small order of French fries from a fast- food restaurant contain a total of 830 calories. Three cheeseburgers and two small orders of French fries contain a total of 1360 calories. Find the caloric content of each item.

Step-by-Step Solution

Answer

The caloric content of a cheeseburger is \(c\) calories and a small order of French fries is \(f\) calories.

Step 1: Identify the Equations

From the problem, the equations are: \n 2c + f = 830 and 3c + 2f = 1360 \n Where c represents the calorific content of a cheeseburger and f represents the calorific content of french fries.

Step 2: Solve the system of equations

The equations can be solved by different methods such as substitution, elimination or matrix method. We'll use elimination here. First, multiply the first equation by 2 and the second by 1 and subtract the second from the first. This gives: \n 2*(2c + f) - 1*(3c + 2f) = 2*830 - 1*1360 . Simplifying this gives the value of c.

Step 3: Find the Value of f

Substitute the value of c into the first equation to get the value of f. That is: 2c + f = 830. With the derived value of c, calculate for f.

Step 4: Verify the solution

Substitute the found values of c and f into both equations to verify if they satisfy the equations. If yes, then the solution is correct.

Key Concepts

Linear Equations

Linear equations are fundamental tools in algebra that represent a straight line when graphed on a Cartesian plane. They typically come in the form of \( ax + by = c \), where \( a \), \( b \), and \( c \), are constants, and \( x \) and \( y \) are the variables we want to solve for. In real-world scenarios, like our nutrition problem, linear equations model relationships between quantities; calories in our cheeseburgers \( (c) \) and fries \( (f) \) in this case. Setting up the correct equations is crucial to finding a solution—each equation in our problem represents different combinations of the items adding up to a total calorie count.

Elimination Method

The elimination method is an algebraic strategy used to find the values of variables that satisfy a system of linear equations. Unlike the substitution method, which isolates one variable before substituting it into the other equations, elimination involves adding or subtracting the equations to cancel out one of the variables. For example, in our food problem, by doubling the first equation and subtracting the second, we eliminated the \( f \) variable, allowing us to solve for \( c \) directly. This method can be particularly effective when the coefficients of one of the variables in the different equations are opposites or can easily be made into opposites.

Algebraic Problem-Solving

Solving algebraic problems typically involves understanding the problem, translating it into mathematical language (equations), performing algebraic manipulations (like the elimination method), and finding the solution. When solving for the nutrition problem, we first defined our variables and translated the word problem into a system of equations. Then we applied the elimination method, which was an algebraic operation, to deduce the caloric content of a cheeseburger \( (c) \) and then used substitution to find the calories in French fries \( (f) \)—exemplifying structured problem-solving in algebra.

Verifying Solutions

After calculating the values of variables, it's essential to verify that these values correctly solve the initial system of equations. This ensures the solution is not only mathematically valid but practical as well, confirming that any assumptions made during solving were correct. In the nutrition problem, we plug in the calculated values of \( c \) and \( f \) into the original equations to check for accuracy. If our values satisfy both equations, we can confidently state that our solution is correct. Verifying solutions also helps in detecting any errors we might have made during the process of solving.