Problem 41 - Precalculus Student Solutions Manual 5th

Question

Warehouse Club Membership Membership warehouse clubs offer shoppers low prices, along with rewards of cash back on club purchases. If the yearly fee for a warehouse club membership is \(\$ 100\) and the reward rate is \(2 \%\) on club purchases for the year, then the linear equation $$ y=100-0.02 x $$ models the actual yearly cost of the membership \(y,\) in dollars. Here \(x\) represents the yearly amount of club purchases, also in dollars.(a) Determine the actual yearly cost of the membership if club purchases for the year are \(\$ 2400\) (b) What amount of club purchases would reduce the actual yearly cost of the membership to \(\$ 50 ?\) (c) How much would a member have to spend in yearly club purchases to reduce the yearly membership cost to \(\$ 0 ?\)

Step-by-Step Solution

Answer

(a) \)52(b) \(2500(c) \)5000

Step 1: Understand the Equation

The given linear equation is y = 100 - 0.02xwhere y represents the actual yearly cost of the membership in dollars, and x represents the yearly amount of club purchases in dollars.

Step 2: Calculate Actual Yearly Cost for \(2400 Club Purchases

Substitute x = 2400 into the equation y = 100 - 0.02x to find the actual yearly cost.y = 100 - 0.02(2400)Next, compute the value of 0.02 multiplied by 2400:0.02 * 2400 = 48Finally, subtract 48 from 100:y = 100 - 48 = 52Therefore, the actual yearly cost of the membership for \)2400 club purchases is \(52.

Step 3: Determine Purchases to Reduce Yearly Cost to \)50

Set y = 50 and solve for x in the equation y = 100 - 0.02x.50 = 100 - 0.02xSubtract 100 from both sides to get:-50 = -0.02xDivide both sides by -0.02:x = \frac{-50}{-0.02} = 2500Therefore, \(2500 in purchases would reduce the actual yearly cost to \)50.

Step 4: Determine Purchases to Reduce Yearly Cost to \(0

Set y = 0 and solve for x in the equation y = 100 - 0.02x.0 = 100 - 0.02xSubtract 100 from both sides to get:-100 = -0.02xDivide both sides by -0.02:x = \frac{-100}{-0.02} = 5000Therefore, \)5000 in purchases would reduce the actual yearly cost to \(0.

Key Concepts

Linear Equations

Linear equations are mathematical statements that describe a straight-line relationship between two variables. In our exercise, the linear equation is given by:

\( y = 100 - 0.02x \)

Here, the equation represents the actual yearly cost of a warehouse club membership (y) as a function of the amount spent on club purchases (x).

The equation is in the slope-intercept form: \( y = mx + b \), where \(m\) is the slope and \(b\) is the y-intercept. The slope in our equation is \(-0.02\), indicating a decrease in cost for every dollar spent. The y-intercept is 100, which is the initial yearly cost before any purchases are made.

Solving Equations

Solving equations involves finding the value of variables that make the equation true. Let's break down the steps for solving the equations in our problem.

• In Part (a), we substitute \(x = 2400\) into the equation to find the yearly cost.
• In Part (b), we set \(y = 50\) and solve for \(x\) to find the amount of purchases needed to reduce the cost to \(50.
• In Part (c), we set \(y = 0\) and solve for \(x\) to determine how much needs to be spent to bring the cost to \)0.

Using substitution and algebraic manipulation, we can solve these equations:

For \(x = 2400\): \( y = 100 - 0.02 \times 2400 = 52 \)

For \(y = 50\): \( 50 = 100 - 0.02x \) gives \( x = 2500 \)

For \(y = 0\): \( 0 = 100 - 0.02x \) gives \( x = 5000 \)

Real-World Applications

Linear equations have many real-world applications, especially in financial contexts like our warehouse club membership example.

Understanding and solving linear equations can help in:

• Finance and budgeting: Calculating costs, savings, and profits.
• Shopping: Determining the best deals and understanding membership benefits.
• Business: Evaluating pricing models and revenue projections.

For our warehouse club membership, the linear equation helps members see how much they need to spend to lower their yearly membership costs. This type of calculation is practical for making informed financial decisions. By grasping the basics of linear equations, students can apply these concepts to various everyday scenarios, thus enhancing their financial literacy and decision-making skills.