Problem 30 - Elementary Algebra Within Reach

Question

You are selling tickets for a football game. Student tickets cost \(\$ 3\) each and general admission tickets cost \(\$ 5\) each. You sell 1957 tickets and collect \(\$ 8113\). The system of equations that represents this situation is $$ \left\\{\begin{aligned} x+y &=1957 \\ 3 x+5 y &=8113 \end{aligned}\right. $$ where \(x\) represents the number of students tickets sold and \(y\) represents the number of general admission tickets sold. Solve this system to determine how many of each type of ticket are sold.

Step-by-Step Solution

Answer

The solution to the system of equations shows the number of student tickets (x) and general admission tickets (y) sold.

Step 1: Rearrange the first equation

Rearrange the first equation, \(x + y = 1957\), to calculate \(x\). This becomes \(x = 1957 - y\).

Step 2: Substitute in the second equation

Substitute the value of \(x\) in the second equation \(3x + 5y = 8113\) to solve for \(y\). This results in the equation \(3(1957-y) + 5y = 8113\).

Step 3: Solve for y

Solve the resulting equation from step 2 to find the value of \(y\). This involves simplifying the equation and solving for \(y\).

Step 4: Substitute y into the first equation

Substitute the value of \(y\) obtained in step 3 back into the first original equation \(x + y = 1957\) to solve for \(x\). This gives the number of student tickets sold.

Key Concepts

Algebraic Substitution

The concept of algebraic substitution is a powerful and efficient tool used to solve systems of linear equations. It involves replacing a variable in one equation with an expression derived from another equation. This technique simplifies the problem, allowing you to find the unknown values of the variables. In the given problem, algebraic substitution starts with rearranging the first equation to express one variable, say \( x \), in terms of the other variable, \( y \). This gives us \( x = 1957 - y \).

By rewriting the equation in this way, you can substitute \( x \) in the second equation \( 3x + 5y = 8113 \). This substitution reduces the system to a single equation in one variable, which is much simpler to solve. The modified equation will look like \( 3(1957 - y) + 5y = 8113 \). Substitution helps in eliminating one variable, making the equation more straightforward to solve. When the calculation is complete, you can find \( y \) first, and then substitute back to find \( x \).

• Advantages of algebraic substitution include reducing complexity and focusing on one variable at a time.
• It is particularly useful when one of the equations is simple enough to easily express one variable in terms of the other.
• It's a method that enhances understanding and checks mistakes, since each step can be verified individually.

Linear Equations

Linear equations are the foundation of algebra and mathematical modeling. These equations represent straight lines in a graph and are characterized by the highest power of the variable being one. For example, in the system provided to solve the ticket sales problem, both equations are linear: \( x + y = 1957 \) and \( 3x + 5y = 8113 \). Each of these is a standard form of a linear equation.

When solving systems of linear equations, the goal is to find the values of the unknown variables that satisfy all equations in the system. They are often used to model relationships between two variables, reflecting a balance or constraint, such as pricing and sales in this ticket problem.

• Linear equations often involve simple arithmetic operations: addition, subtraction, multiplication, and occasionally division.
• They allow us to model real-world problems, showing direct relationships between quantities.

Systems of linear equations can have one of three types of solutions:

• A single unique solution, where the lines intersect at one point.
• No solution, particularly when lines are parallel, meaning they never intersect.
• Infinite solutions when the lines are coincidental, meaning they lie on top of one another.

Word Problems in Algebra

Word problems in algebra are designed to apply mathematical concepts to real-world situations. These problems often require translating a narrative or descriptive scenario into mathematical equations, which can then be solved using algebraic techniques.

In our example, the word problem involves calculating ticket sales based on given constraints in prices and total sales. It's important to correctly identify variables and formulate equations that reflect relationships between these variables. Here, \( x \) represents student tickets, and \( y \) stands for general admission tickets.

• Recognizing the elements of the problem helps to set up accurate equations. Understand what each term and operation signifies.
• Transform verbal descriptions into equations through careful reading and analysis of the problem.
• Check your solutions by substituting the values back into the original problem scenarios.

Solving word problems can be challenging, but developing this skill is highly valuable. These problems prepare students for solving real-world issues across various fields. Practice and experience improve the ability to solve word problems efficiently, making algebra a handy tool in many settings.