Problem 65 - Precalculus

Question

An airline is deciding which meals to buy from its provider. If the airline orders the same number of meals of types I and II totalling 150 meals, the cost is \(\$ 1275 ;\) if they order \(60 \%\) of type I and \(40 \%\) of type II, the cost is \(\$ 1260 .\) What is the cost of each type of meal?

Step-by-Step Solution

Answer

The cost of a type I meal is $8, and the cost of a type II meal is $9.

Step 1: Define Variables

Let's define \( x \) as the cost of a type I meal and \( y \) as the cost of a type II meal. We need to find values for \( x \) and \( y \).

Step 2: Set Up Equations

Using the given information, we can set up two equations. 1. For the first scenario: If they order the same number of meals \( n \) for both types totaling 150 meals, then: \[ 2n = 150 \Rightarrow n = 75 \] The total cost equation becomes: \[ 75x + 75y = 1275. \]2. For the second scenario: 60% and 40% of 150 meals are ordered: \[ 0.6 \times 150 = 90 \text{ meals of type I}, \ 0.4 \times 150 = 60 \text{ meals of type II} \] The total cost equation is: \[ 90x + 60y = 1260. \]

Step 3: Simplify the Equations

Let's simplify both equations:1. \[ 75x + 75y = 1275 \Rightarrow x + y = 17 \] (dividing by 75)2. \[ 90x + 60y = 1260 \Rightarrow 3x + 2y = 42 \] (dividing by 30)

Step 4: Solve the System of Equations

Now solve the equations:1. \( x + y = 17 \)2. \( 3x + 2y = 42 \)To solve by substitution or elimination, let's use substitution with the first equation:\[ y = 17 - x \]Substitute into the second equation:\[ 3x + 2(17 - x) = 42 \]\[ 3x + 34 - 2x = 42 \]\[ x = 8 \]Substitute \( x = 8 \) back into the first equation:\[ 8 + y = 17 \Rightarrow y = 9 \]

Step 5: Verify Solution

Check our solutions by substituting \( x = 8 \) and \( y = 9 \) back into the original condition:1. \( 75x + 75y = 1275 \rightarrow 75(8) + 75(9) = 1275 \), holds true.2. \( 90x + 60y = 1260 \rightarrow 90(8) + 60(9) = 1260 \), holds true.

Key Concepts

Understanding Linear Equations

Linear equations are foundational in the world of mathematics. They are equations of the first order, meaning the variables are not raised to any power greater than one. Linear equations often take the form of \( ax + by = c \), where \(a, b,\) and \(c\) are constants, and \(x\) and \(y\) are variables.

When dealing with problems, linear equations help determine relationships between variables. In the context of our airline meal example, linear equations allow you to set constraints for the cost of meals. You end up with two distinct equations.
This is crucial as we often have multiple unknowns—like the cost of type I and II meals—and need to solve them within a certain framework, such as a budget.

Linear equations can be manipulated by basic operations like addition, subtraction, multiplication, or division, to solve for the unknowns. They lay the groundwork for solving systems of equations, a vital step when you have more than one equation with two or more variables.

Applying the Substitution Method

The substitution method is a strategic approach for solving systems of equations, especially when one equation is easily solvable for one variable. The idea is to express one variable in terms of another and then substitute this expression into another equation.

Using our exercise example, we first manipulate the equation \( x + y = 17 \) by expressing \(y\) in terms of \(x\): \( y = 17 - x \). This new expression for \( y \) is then substituted into the second equation, \( 3x + 2y = 42 \).

• This substitution effectively reduces the system from two variables to one, making it easier to solve.
• After substituting, you get \( 3x + 2(17-x) = 42 \), which further simplifies to \( x = 8 \).

Once \(x\) is found, it's simple to find \(y\) using the first equation: plug \(x = 8\) back into \( y = 17 - x \), yielding \( y = 9 \).

This method simplifies solving, especially useful when one equation can be easily manipulated into an expression for one variable.

Mastering the Elimination Method

The elimination method offers another way to solve systems of equations. Unlike substitution, elimination focuses on removing one variable by combining the equations.

We start by aligning terms for both equations. Consider the simplified equations from our exercise:

• \(x + y = 17\)
• \(3x + 2y = 42\)

To eliminate a variable, you manipulate the equations such that adding or subtracting them cancels out one of the variables. For instance, if we multiply the first equation by 2, we get \(2x + 2y = 34\).

Subtracting this from the second equation:

(\(3x + 2y = 42\) - \(2x + 2y = 34\)) gives \(x = 8\).

• Finding \(x\) this way allows its solution without solving for \(y\) first.
• Then simply substitute back to find the other variable.

Elimination is efficient for many equations and balances the workload by avoiding substitution, especially when coefficients are conveniently cancellable.