Problem 21 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. John goes into a donut shop and buys ten plain donuts and five cream-filled donuts for \(\$ 3.70 .\) Jane goes into the same shop and buys five plain donuts and ten cream-filled donuts for \(\$ 4.10 .\) What is the cost of a plain donut?

Step-by-Step Solution

Answer

The cost of a plain donut is $0.22.

Step 1: Define Variables

Let the cost of one plain donut be denoted by \( p \) and the cost of one cream-filled donut be denoted by \( c \).

Step 2: Set Up the Equations

From the given information, John buys ten plain donuts and five cream-filled donuts for \( 3.70 \) dollars. Therefore, we can write the first equation as: \[ 10p + 5c = 3.70 \]

Step 3: Second Equation

Jane buys five plain donuts and ten cream-filled donuts for \( 4.10 \) dollars. This gives us the second equation: \[ 5p + 10c = 4.10 \]

Step 4: Simplify the Equations

Simplify the second equation by dividing all terms by 5: \[ p + 2c = 0.82 \]

Step 5: Solve the System of Equations

We now have the system of equations: \[ \begin{cases} 10p + 5c = 3.70 \ p + 2c = 0.82 \end{cases} \] Next, multiply the second equation by 5 to align the coefficients of \( p \): \[ 5(p + 2c) = 5(0.82) \ 5p + 10c = 4.10 \]

Step 6: Eliminate Variable

Subtract the first equation from the modified second equation: \[ (5p + 10c) - (10p + 5c) = 4.10 - 3.70 \ -5p + 5c = 0.40 \ -5p = 0.40 - 5c \] Substitute \( c \) from second equation: \[ p + 2c = 0.82 \ p = 0.82 - 2c \] Substitute this back into the first equation to solve for \( c \).

Step 7: Solve for Both Variables

Substitute the simplified expression for \( p \) back into the first equation: \[ 10(0.82 - 2c) + 5c = 3.70 \ 8.20 - 20c + 5c = 3.70 \ 8.20 - 15c = 3.70 \ -15c = 3.70 - 8.20 \ -15c = -4.50 \ c = 0.30 \] Now, substitute \( c = 0.30 \) back into \( p = 0.82 - 2c \) to solve for \( p \): \[ p = 0.82 - 2(0.30) \ p = 0.82 - 0.60 \ p = 0.22 \]

Key Concepts

Algebraic Equations

In this problem, we are using algebraic equations to solve for the costs of plain and cream-filled donuts. Algebraic equations are mathematical statements that use letters to represent unknown values, known as variables, along with numbers and operational symbols like +, -, *, and /.

Here, we represent the cost of a plain donut as \( p \) and the cost of a cream-filled donut as \( c \). The problem involves setting up and solving these equations:

• John bought 10 plain donuts and 5 cream-filled donuts for \(3.70: \( 10p + 5c = 3.70 \)
• Jane bought 5 plain donuts and 10 cream-filled donuts for \)4.10: \( 5p + 10c = 4.10 \)

By correctly setting up these equations, we can move forward to solve them.

Variable Substitution

Variable substitution is a powerful method in algebra for solving systems of equations. The idea is to express one variable in terms of the other and then substitute that expression into another equation to find the unknowns.

In this example, after simplifying **Jane's** equation by dividing it by 5, we get:
\( p + 2c = 0.82 \).

To align coefficients, we multiply this by 5 to get: \( 5p + 10c = 4.10 \).

Next, we align and subtract this new equation from John's equation to eliminate **p**. This simplifies our system of equations, allowing us to substitute back and solve for both variables.

Linear Equations

Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable. Linear equations can be solved graphically or algebraically.

In our problem, both John's and Jane's equations are linear:

• John's equation: \( 10p + 5c = 3.70 \)
• Jane's equation: \( 5p + 10c = 4.10 \)

These linear equations form a system of linear equations. Solving such systems often involves aligning coefficients, substitution, or elimination (just like in our example) to find the values of the variables.

Simplification

Simplification is the process of making an equation easier to solve. This might involve reducing fractions, combining like terms, or performing operations to eliminate one of the variables.

In this problem, we simplified Jane's equation by dividing all terms by 5: \( 5p + 10c = 4.10 \), which becomes \( p + 2c = 0.82 \). This step is crucial and makes it easier to eliminate one variable.

We then use subtraction to simplify further and solve for a single variable, making the math more straightforward and reducing the potential for mistakes. Simplifying at each step keeps the equations manageable and easier to handle.