Problem 23 - Intermediate Algebra

Question

Two bottles of catsup, 2 jars of peanut butter, and 1 jar of pickles cost $$\$ 7.78$$. Three bottles of catsup, 4 jars of peanut butter, and 2 jars of pickles cost $$\$ 14.34$$. Four bottles of catsup, 3 jars of peanut butter, and 5 jars of pickles cost $$\$ 19.19$$. Find the cost per bottle of catsup, the cost per jar of peanut butter, and the cost per jar of pickles.

Step-by-Step Solution

Answer

Catsup costs $1.22 per bottle, peanut butter $1.77 per jar, and pickles $1.80 per jar.

Step 1: Define the Variables

Assume the cost per bottle of catsup is \( x \), the cost per jar of peanut butter is \( y \), and the cost per jar of pickles is \( z \).

Step 2: Set Up the Equations

The problem gives us three equations based on the costs and the items:1. \( 2x + 2y + z = 7.78 \)2. \( 3x + 4y + 2z = 14.34 \)3. \( 4x + 3y + 5z = 19.19 \)

Step 3: Use Elimination Method

We'll solve these equations using the elimination method. First, use equations 1 and 2 to eliminate \( z \) by multiplying equation 1 by 2 and subtracting it from equation 2:\(2(2x + 2y + z) = 2(7.78) \rightarrow 4x + 4y + 2z = 15.56 \)Subtract from equation 2:\( (3x + 4y + 2z) - (4x + 4y + 2z) = 14.34 - 15.56 \)\(-x = -1.22 \rightarrow x = 1.22\)

Step 4: Substitute to Solve for Another Variable

Substitute \( x = 1.22 \) back into equation 1:\( 2(1.22) + 2y + z = 7.78 \rightarrow 2.44 + 2y + z = 7.78 \rightarrow 2y + z = 5.34 \).

Step 5: Reduce the System of Equations

Use equation 3 with \( x = 1.22 \):\( 4(1.22) + 3y + 5z = 19.19 \rightarrow 4.88 + 3y + 5z = 19.19 \rightarrow 3y + 5z = 14.31 \).

Step 6: Eliminate Another Variable

Using the reduced equations:1. \( 2y + z = 5.34 \)2. \( 3y + 5z = 14.31 \)Multiply equation 1 by 5 and subtract from equation 2 to eliminate \( z \):\( 5(2y + z) = 5(5.34) \rightarrow 10y + 5z = 26.7 \)Subtract from equation 2:\( (3y + 5z) - (10y + 5z) = 14.31 - 26.7 \)\(-7y = -12.39 \rightarrow y = 1.77\)

Step 7: Solve for the Remaining Variable

Use \( y = 1.77 \) and substitute back into \( 2y + z = 5.34 \):\( 2(1.77) + z = 5.34 \rightarrow 3.54 + z = 5.34 \rightarrow z = 1.80 \).

Step 8: Verify the Solution

Check all the original equations with \( x = 1.22 \), \( y = 1.77 \), and \( z = 1.80 \) to ensure they hold true:1. \( 2(1.22) + 2(1.77) + 1.80 = 7.78 \)2. \( 3(1.22) + 4(1.77) + 2(1.80) = 14.34 \)3. \( 4(1.22) + 3(1.77) + 5(1.80) = 19.19 \)All equations are satisfied.

Key Concepts

Elimination Method

The elimination method is a widely used technique for solving systems of linear equations. It involves manipulating the equations to eliminate one of the variables, making it easier to solve for the remaining variables. In this exercise, we start by writing down the three given equations based on the items and their costs. Our goal is to reduce the system step by step, eventually finding all variables: the costs of catsup, peanut butter, and pickles.

To eliminate a variable, we strategically scale the equations and subtract one from another. For instance, to eliminate the variable \( z \), we multiplied the first equation by 2 and subtracted it from the second equation. This quickly isolated the variable \( x \), allowing us to solve for its value. A similar strategy was used later in the exercise to eliminate another variable and solve for \( y \) and \( z \).

• Multiply the entire equation to help match coefficients.
• Subtract or add adjusted equations to eliminate terms.
• Repeat as necessary to solve the system.

By gradually reducing the system, the elimination method allows us to efficiently find solutions to complex algebraic problems like the one given here.

Linear Equations

Linear equations form the backbone of systems of equations, especially in the exercise we were looking at, which involved calculating costs for different grocery items. Such equations describe relationships that are straight lines when plotted on a graph, characterized by having variables raised only to the power of one.

For our exercise, each equation represents a linear combination of the costs for catsup, peanut butter, and pickles. In the simplest form, a linear equation looks like \( ax + by + cz = d \), where all terms are linear. In our case:

• \(2x + 2y + z = 7.78\)
• \(3x + 4y + 2z = 14.34\)
• \(4x + 3y + 5z = 19.19\)

To solve the problem, we rely on these linear structures, knowing that changes in one variable affect another predictably. This characteristic makes linear equations central to solving many algebraic problems efficiently.

Algebraic Problem Solving

Algebraic problem solving involves identifying unknowns and setting up equations that model real-world situations. In this exercise, we're tasked with finding out the individual costs of catsup, peanut butter, and pickles. Each type of item represents a variable, leading to the setup of a system of equations.

When solving algebraically, we follow these steps:

• Define variables for unknowns. Here, \( x \), \( y \), and \( z \) stand for the costs of the products.
• Relate these variables using given information to form a system of equations. In our task, we created three equations from the costs of item bundles.
• Simplify the system using techniques like substitution or elimination to solve for the unknowns. This means systematically reducing the system until each variable is isolated and solved.

Problem-solving using algebra revolves around logic and mathematical operations. It's about transforming a stated problem into a series of manageable mathematical statements and gradually clarifying until the unknowns are fully understood. In this way, we can take complex scenarios and draw out clear answers with precision.