Problem 35 - College Algebra Essentials

Question

On a recent trip to the convenience store, you picked up 2 gallons of milk, 5 bottles of water, and 6 snack-size bags of chips. Your total bill (before tax) was 19.00 dollar. If a bottle of water costs twice as much as a bag of chips, and a gallon of milk costs 2.00 dollar more than a bottle of water, how much does each item cost?

Step-by-Step Solution

Answer

Based on the solution steps, the costs are: a bag of chips is $1.00, a bottle of water is $2.00, and a milk gallon is $4.00.

Step 1: Define variables

Let: \[x = \text{the cost of a bag of chips}\]\[y = \text{the cost of a bottle of water}\]\[z = \text{the cost of milk}\]

Step 2: Formulate equations from the problem statement

From the given conditions, we can form the following equations:\[\begin{align*}6x + 5y + 2z = 19 &\quad \text{(Equation 1: From the total cost)} \y = 2x &\quad \text{(Equation 2: The cost of a bottle of water is twice the cost of a bag of chips)} \z = y + 2 &\quad \text{(Equation 3: A gallon of milk costs $2.00 more than a bottle of water) }\end{align*}\]

Step 3: Substitute Equations 2 and 3 into Equation 1

Substitute Equation 2 and 3 into Equation 1. This gives\[6x + 5*(2x) + 2*(2x + 2) = 19 \]Solving the above equation yields\[x = 1 \]

Step 4: Find values of other variables

Put x = 1 into Equations 2 and 3 to find the values of y and z. This gives\[y = 2 * 1 = 2 \]and\[z = 2 + 2 = 4\].

Step 5: Interpret the Results

Therefore, the cost of a bag of chips is $1.00, a bottle of water is $2.00, and a gallon of milk is $4.00.

Key Concepts

Algebraic Expressions

Understanding algebraic expressions is crucial for solving systems of equations. An algebraic expression is a combination of variables, numbers, and operations that represent a specific quantity. For instance, in the exercise where students purchased milk, water, and chips, variables are used to represent the unknown prices of these items.
By defining a variable such as \( x \) to represent the cost of a bag of chips, the problem starts to take a mathematical form that can be manipulated according to algebraic rules. These expressions form the basis of equations that we use in algebra to model and solve real-world problems.
When creating algebraic expressions, it is essential to translate the problem's conditions correctly. If a bottle of water costs twice as much as a bag of chips, this relationship is expressed as \( y = 2x \). Such careful translation of words into algebraic terms is a skill that, once mastered, can significantly improve one’s ability to tackle mathematical problems.

Variable Substitution

Variable substitution is a powerful technique in algebra that allows you to unravel systems of equations. The exercise on hand provides a prime example of how to use substitution to simplify the problem.
After setting up algebraic expressions for each item's cost, you can replace the variables \( y \) and \( z \) with expressions in terms of \( x \) based on the relationships given: \( y = 2x \) and \( z = y + 2 \). Substitution streamlines the original equation, making it easier to solve for the single remaining variable.
Once the value of \( x \) is identified, you backtrack by substituting this value into the expressions for \( y \) and \( z \) to find their values. This step-by-step approach is methodical and ensures you systematically uncover the values of all variables. The ability to substitute correctly and track the changes is essential, avoiding mistakes that can lead to incorrect solutions.

Solving Linear Equations

Solving linear equations is one of the fundamental aspects of algebra. When faced with a system of equations, as in our exercise scenario, the goal is to find the values for the variables that satisfy all equations simultaneously.
A linear equation, such as \( 6x + 5y + 2z = 19 \), features variables that are raised to no higher power than one. In solving them, a systematic approach must be taken to isolate each variable. Our exercise involved simplifying the system through substitution and then solving the resulting single-variable equation.
As the equation becomes more straightforward, you apply basic algebraic principles - addition, subtraction, multiplication, and division - to isolate and solve for the variable. By substituting these values back into the other equations, you can solve for the remaining unknowns. The importance of solving linear equations cannot be overstated, as this skill is not only essential for mathematics but also applicable in various real-world contexts where relationships between quantities need to be determined.