Problem 323 - Elementary Algebra

Question

Tickets for an American Baseball League game for 3 adults and 3 children cost less than \(\$ 75,\) while tickets for 2 adults and 4 children cost less than \(\$ 62\). (a) Write a system of inequalities to model this problem. (b) Graph the system. (c) Could the tickets cost \(\$ 20\) for adults and \(\$ 8\) for children? (a) Could the tickets cost \(\$ 15\) for adults and \(\$ 5\) for children?

Step-by-Step Solution

Answer

The system of inequalities is \(3a + 3c < 75\) and \(2a + 4c < 62\). Tickets cannot cost \(\$20\) for adults and \(\$8\) for children, but they can cost \(\$15\) for adults and \(\$5\) for children.

Step 1: Understanding the Problem

We need to establish inequalities based on the given cost constraints for tickets purchased by adults and children.

Step 2: Define Variables

Let the cost of an adult ticket be represented by \(a\) dollars, and the cost of a child's ticket be represented by \(c\) dollars.

Step 3: Formulate Inequalities

We have two scenarios:1. Tickets for 3 adults and 3 children costing less than \(75\) dollars: \(3a + 3c < 75\).2. Tickets for 2 adults and 4 children costing less than \(62\) dollars: \(2a + 4c < 62\).

Step 4: System of Inequalities

Combining the inequalities from the given scenarios, we have the system of inequalities:1. \(3a + 3c < 75\).2. \(2a + 4c < 62\).

Step 5: Graph the System

To graph the system, first convert the inequalities to equations to find boundary lines:1. \(3a + 3c = 75\) or \(a + c = 25\).2. \(2a + 4c = 62\) or \(a + 2c = 31\).Draw these lines on a graph and shade the region below each line to represent the areas where the inequalities hold.

Step 6: Test Ticket Prices: \(a = 20, c = 8\)

Substitute \(a = 20\) and \(c = 8\) into both inequalities:1. \(3(20) + 3(8) = 60 + 24 = 84\) (not less than 75)2. \(2(20) + 4(8) = 40 + 32 = 72\) (not less than 62)Hence, the tickets cannot cost \(\$20\) for adults and \(\$8\) for children.

Step 7: Test Ticket Prices: \(a = 15, c = 5\)

Substitute \(a = 15\) and \(c = 5\) into both inequalities:1. \(3(15) + 3(5) = 45 + 15 = 60\) (less than 75)2. \(2(15) + 4(5) = 30 + 20 = 50\) (less than 62)Hence, the tickets can cost \(\$15\) for adults and \(\$5\) for children.

Key Concepts

Graphing Inequalities

When graphing inequalities, we're essentially finding locations where conditions are met. First, we convert the inequalities into equations to establish boundary lines. For instance, if the inequality is \(3a + 3c < 75\), we first graph \(3a + 3c = 75\). To simplify, this equation is equivalent to \(a + c = 25\).
To draw this, we find intercepts. If \(a = 0\), then \(c = 25\), and if \(c = 0\), then \(a = 25\). Plotting these, we get a line downward sloping from \(a = 25\) on the horizontal axis to \(c = 25\) on the vertical axis.
Next we shade the region below this line because we are looking for points where \(3a + 3c\) is less than 75. Repeat for the second inequality \(2a + 4c = 62\), simplified to \(a + 2c = 31\). Intercepts here are found similarly. Shading the appropriate region shows areas where both conditions are valid.

Algebraic Modeling

Algebraic modeling translates real-world problems into mathematical expressions. In our exercise, we define variables to represent the cost of tickets: \(a\) for an adult ticket, \(c\) for a child's ticket. By framing our problem this way, we can use math to explore solutions.
We form inequalities to express constraints. For 3 adults and 3 children costing less than \$75: \(3a + 3c < 75\). This means the total cost for this group cannot exceed \$75.
The second inequality, \(2a + 4c < 62\), follows similar logic. These inequalities build a system we can solve or graph. With algebraic models, we're looking to see if given values (e.g., \$15 for adults, \$5 for children) satisfy our conditions. If successful, the solution is valid both algebraically and in real-life scenarios.

Cost Constraints

Cost constraints limit how much we can spend. In our baseball ticket problem, we aim to find combinations of adult and child ticket prices that meet specified limits.
Let's check if tickets costing \$20 for adults and \$8 for children fit our inequalities. Substituting, \(3(20) + 3(8) = 84\), exceeding \$75; thus, not feasible. Similarly, for \(2(15) + 4(5) = 50\), the prices work, keeping both totals within limits.
Cost constraints force us to balance spending within certain boundaries. By applying algebra, we quickly test different prices against these constraints. This ensures the validity of our choices while maintaining budget limits.