Solve each problem using a system of equations. A company sells recordable CDs for \(0.80 each and play-only CDs for \)0.60 each. The company receives \$76.00 for an order of 100 CDs. However, the customer neglected to specify how many of each type to send. Determine the number of each type of CD that should be sent.

Verified step by step guidance

1. Define variables to represent the unknown quantities: let \(x\) be the number of recordable CDs and \(y\) be the number of play-only CDs.
2. Write an equation representing the total number of CDs ordered: since the total is 100, we have \(x + y = 100\).
3. Write an equation representing the total cost of the CDs: recordable CDs cost \(0.80\) each and play-only CDs cost \(0.60\) each, so the total cost equation is \(0.80x + 0.60y = 76.00\).
4. Use the system of equations formed by \(x + y = 100\) and \(0.80x + 0.60y = 76.00\) to solve for \(x\) and \(y\). You can use substitution or elimination methods.
5. After solving the system, interpret the values of \(x\) and \(y\) as the number of recordable and play-only CDs, respectively.

Key Concepts

Systems of Linear Equations

A system of linear equations consists of two or more linear equations with the same variables. Solving the system means finding values for the variables that satisfy all equations simultaneously. In this problem, the quantities of two types of CDs are the variables, and the total number and total cost provide two equations.

Setting Up Equations from Word Problems

Translating a word problem into mathematical equations involves identifying variables and expressing relationships described in the problem. Here, the total number of CDs and the total cost are used to form two equations representing the sum of quantities and the sum of costs, respectively.

Solving Systems by Substitution or Elimination

Once the system is set up, methods like substitution or elimination are used to find the values of variables. Substitution involves solving one equation for a variable and plugging it into the other, while elimination involves adding or subtracting equations to eliminate a variable, simplifying the solution process.