Problem 75 - College Algebra

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. A store clerk sold 60 pairs of sneakers. The high-tops sold for \(\$ 98.99\) and the low-tops sold for \(\$ 129.99 .\) If the receipts for the two types of sales totaled \(\$ 6,404.40\) , how many of each type of sneaker were sold?

Step-by-Step Solution

Answer

45 high-tops and 15 low-tops were sold.

Step 1: Define the variables

Let's define two variables to represent the unknown quantities. Let \( x \) represent the number of high-tops sold and \( y \) represent the number of low-tops sold.

Step 2: Set up the equations

We know there are a total of 60 pairs of sneakers sold. This gives us our first equation: \[x + y = 60\]We also know the total sales are valued at \( \\(6,404.40 \). The high-tops sold for \( \\)98.99 \) each and the low-tops for \( \$129.99 \) each. This information gives us our second equation:\[98.99x + 129.99y = 6404.40\]

Step 3: Solve the system of equations

Now we have a system of equations:\[1. \ x + y = 60\]\[2. \ 98.99x + 129.99y = 6404.40\]We can solve these equations using substitution or elimination. First, solve the first equation for \( y \):\[y = 60 - x\]Substitute \( y = 60 - x \) into the second equation:\[98.99x + 129.99(60 - x) = 6404.40\]

Step 4: Simplify and solve for x

Distribute 129.99 in the second equation:\[98.99x + 7799.4 - 129.99x = 6404.40\]Combine like terms:\[-31.00x + 7799.4 = 6404.40\]Subtract 7799.4 from both sides to isolate terms with \( x \):\[-31.00x = -1395\]Now, solve for \( x \) by dividing both sides by -31:\[x = \frac{-1395}{-31} = 45\]

Step 5: Find y using the value of x

Substitute \( x = 45 \) back into the equation \( y = 60 - x \):\[y = 60 - 45 = 15\]

Step 6: Interpret the solution

The solution to the system indicates that 45 high-tops and 15 low-tops were sold.

Key Concepts

System of Linear Equations

A system of linear equations involves multiple equations that are solved together to find common solutions. In this exercise, we have two equations connected by two variables: the number of high-top sneakers and low-top sneakers sold.

Such systems are often used to solve real-world problems where different factors relate to each other through certain conditions. In our example:

• The first equation (x + y = 60) represents the total number of sneakers sold.
• The second equation (98.99x + 129.99y = 6404.40) reflects the total sales revenue from those sneakers.

By solving these equations together, we determine how many of each type were sold at the store.

Variables

Variables are symbols that stand for unknown values in mathematical expressions or equations. They may take on different values and are used to express the relationships between quantities.

In our system:

• x represents the number of high-top sneakers sold.
• y represents the number of low-top sneakers sold.

Using variables simplifies the problem-solving process, allowing us to model complex scenarios effectively. By defining these variables clearly, we convert the word problem into a mathematical model, which can then be solved using algebraic techniques.

Substitution Method

The substitution method is a technique used to solve a system of equations by expressing one variable in terms of the other. This method is particularly useful when one equation is easy to solve for one variable.

In this case, we solved the first equation, x + y = 60, for y, which gives us y = 60 - x. We then substituted this expression into the second equation:
98.99x + 129.99(60 - x) = 6404.40
This substitution simplifies the problem to a single equation with one variable, x, making it much easier to solve.

• First, simplify the equation.
• Then solve for x by isolating the variable.

Finally, we use the solution for x to find y, completing the process efficiently.

Elimination Method

The elimination method is another popular technique for solving systems of linear equations. This method works by adding or subtracting equations to eliminate one variable, making it easier to solve for the remaining variable.

In our problem, although we used the substitution method, the elimination method would involve multiplying one or both equations by constants so that adding or subtracting them would cancel out one variable.
Let's consider:

• We could multiply the first equation, x + y = 60, by 98.99 to align it with the coefficients in the second equation.
• This might make it easier to subtract the equations and eliminate x or y.

Choosing between substitution and elimination often depends on the form of the equations and which technique seems more straightforward for the given problem. Understanding both methods gives you greater flexibility in approaching systems of equations.