Problem 39 - Algebra A Combined Function

Question

Kathi and Robert Hawn had a pottery stand at the annual Skippack Craft Fair. They sold some of their pottery at the original price of \(\$ 9.50\) each, but later decreased the price of each by \(\$ 2\). If they sold all 90 pieces and took in \(\$ 721,\) find how many they sold at the original price and how many they sold at the reduced price.

Step-by-Step Solution

Answer

They sold 23 pieces at the original price and 67 pieces at the reduced price.

Step 1: Define Variables

Let's define two variables to help us solve the problem. Let \( x \) be the number of pieces sold at the original price of \( \\(9.50 \), and \( y \) be the number of pieces sold at the reduced price of \( \\)7.50 \) (since the price was reduced by \( \$2.00 \)).

Step 2: Set Up Equations

We have two conditions given in the problem: \( x + y = 90 \) because they sold a total of 90 pieces. The second condition arises from the total income: \( 9.5x + 7.5y = 721 \).

Step 3: Solve the System of Equations

We have the system of equations: \[ \begin{align*}x + y &= 90 \9.5x + 7.5y &= 721\end{align*} \] We will solve this system using substitution or elimination. Here, we will use substitution. Solving the first equation for \( y \), we get \( y = 90 - x \).

Step 4: Substitute and Solve for x

Substitute \( y = 90 - x \) into the second equation: \[ 9.5x + 7.5(90 - x) = 721 \] Simplify and solve for \( x \):\[ 9.5x + 675 - 7.5x = 721 \]\[ 2x + 675 = 721 \]\[ 2x = 46 \]\[ x = 23 \]

Step 5: Solve for y

Substitute \( x = 23 \) back into \( y = 90 - x \) to find \( y \):\[ y = 90 - 23 = 67 \]

Step 6: Verify the Solution

Verify by substituting \( x = 23 \) and \( y = 67 \) back into both equations: For total pieces: \( 23 + 67 = 90 \), which is correct. For total income: \( 9.5 \times 23 + 7.5 \times 67 = 218.5 + 502.5 = 721 \), which is also correct.

Key Concepts

System of Equations

When tackling problems like the one with Kathi and Robert's pottery sales, we often use a system of equations. A system of equations consists of two or more equations. These equations share the same set of variables.
Solutions to these systems are the values that satisfy all equations simultaneously. In our pottery sales example, we have two equations:

• The first equation deals with the total number of pieces sold: \( x + y = 90 \).
• The second equation involves the total income: \( 9.5x + 7.5y = 721 \).

Both equations involve the variables \( x \) and \( y \). Solving the system tells us how many pieces were sold at each price. The goal is to find the values of the variables that make both equations true at the same time.

Substitution Method

The substitution method is a common strategy to solve systems of equations. This method involves solving one of the equations for one variable in terms of the other variable. Then, you substitute this expression into the second equation. This reduces the system's number of equations and makes it easier to solve.

In our exercise, we have:

• The first equation: \( x + y = 90 \)

We solve it for \( y \) to get: \( y = 90 - x \).
Substituting \( y = 90 - x \) into the second equation gives:

• \( 9.5x + 7.5(90 - x) = 721 \)

Thus, the problem now only involves the variable \( x \), which simplifies the calculation. Once \( x \) is found, we use its value to determine \( y \). This method is particularly useful when one equation is easy to solve for a single variable.

Linear Equations

Linear equations are equations of the first degree, meaning they involve only the powers of one in the variables. They can be easily recognized because they graph as straight lines. In the context of solving a system of equations, linear equations are handy because they simplify many calculations.

In our problem, the equations \( x + y = 90 \) and \( 9.5x + 7.5y = 721 \) are both linear. This means that:

• Their graphs will intersect in a single point, assuming they are not parallel.
• Solving the equations will give us the exact coordinates of this intersection, which represent the solution.

Understanding and recognizing linear equations is crucial for quickly analyzing and solving algebraic problems like algebraic word problems, where relationships between quantities are often linear.

Algebraic Word Problems

Algebraic word problems translate real-world scenarios into mathematical expressions. These problems typically involve identifying the relationships between quantities and forming equations that depict these relationships.

In many cases, like our pottery stand problem, the challenge is to:

• Define variables that represent the unknown values, like \( x \) for the number of pieces sold at the original price.
• Interpret the problem's conditions into a system of equations, such as total pieces and total income.

Once translated, the math skills like solving systems of equations, help find the values of these variables. Turning word problems into equations can often seem challenging, but with practice, it becomes an enlightening way to see how algebra models real-world situations.