Problem 79 - Algebra 2

Question

CONCERTS. The price of two lawn seats and a pavilion seat at an outdoor amphitheater is \(\$ 75 .\) The price of three lawn seats and two pavilion seats is \(\$ 130 .\) How much do lawn and pavilion seats cost?

Step-by-Step Solution

Answer

Lawn seats cost $20 and pavilion seats cost $35.

Step 1: Define Variables

Let's define the variables we will use for solving the problem. Let \( x \) represent the price of a lawn seat, and \( y \) represent the price of a pavilion seat.

Step 2: Set Up Equations

Using the information provided, set up two equations. The first equation comes from the first statement: "The price of two lawn seats and a pavilion seat at an outdoor amphitheater is \(\\( 75\)." This gives us the equation: \( 2x + y = 75 \). The second equation is derived from the second statement: "The price of three lawn seats and two pavilion seats is \(\\) 130\)." This equation is: \( 3x + 2y = 130 \).

Step 3: Solve the System of Equations for One Variable

From the first equation \( 2x + y = 75 \), solve for \( y \) in terms of \( x \): \( y = 75 - 2x \). Substitute this expression into the second equation \( 3x + 2y = 130 \) to solve for \( x \).

Step 4: Substitute to Get One Equation

By substituting \( y = 75 - 2x \) into the second equation, we have: \[ 3x + 2(75 - 2x) = 130 \]. Simplify this to obtain: \[ 3x + 150 - 4x = 130 \].

Step 5: Solve for x

Simplify the equation \( 3x + 150 - 4x = 130 \) which results in the equation \( -x + 150 = 130 \). Solving for \( x \) gives: \[ -x = 130 - 150 \], which is \( -x = -20 \). Thus, \( x = 20 \).

Step 6: Solve for y

Substitute \( x = 20 \) back into the equation \( y = 75 - 2x \): \[ y = 75 - 2(20) \]. Simplifying this gives \( y = 75 - 40 = 35 \). So, \( y = 35 \).

Step 7: Solution Verification

Verify the solution by substituting \( x = 20 \) and \( y = 35 \) back into both of the original equations. For the first equation, \( 2(20) + 35 = 75 \), which holds true. For the second equation, \( 3(20) + 2(35) = 130 \), which also holds true. Thus, the solution is verified.

Key Concepts

Simultaneous Equations

Simultaneous equations involve multiple equations that share common variables, and the goal is to find values for these variables that satisfy all the equations.
In our example, we have two equations: \(2x + y = 75\) and \(3x + 2y = 130\), where \(x\) and \(y\) are the variables representing the prices of lawn and pavilion seats respectively.
Solving these equations simultaneously means finding a pair of values of \(x\) and \(y\) that works for both equations. When you solve such systems, you can get one of the three outcomes:

• A unique solution (one specific \(x\) and \(y\) pair), indicating the lines intersect at exactly one point.
• No solution, which happens when the lines are parallel, meaning they never intersect.
• Infinite solutions, which occurs when the equations are essentially the same line, overlapping completely.

In this case, both equations have a unique solution where \(x = 20\) and \(y = 35\), meaning the price of a lawn seat is \(\\(20\) and a pavilion seat is \(\\)35\).

Substitution Method

The substitution method is an effective approach to solving simultaneous equations. One of the equations is manipulated to express one variable in terms of the other and then substituted into the second equation.
For example, starting with the equation \(2x + y = 75\), we solve for \(y\), which gives us \(y = 75 - 2x\).
This expression for \(y\) is then substituted into the second equation \(3x + 2y = 130\), resulting in a single equation with one unknown: \(3x + 2(75 - 2x) = 130\).
The benefit of substitution is that once you have one variable expressed in terms of the other, you can reduce the complexity of the system.

• It's particularly useful when one of the equations can be easily manipulated to isolate a variable.
• This method often simplifies the solving process, as it focuses on reducing the number of variables one step at a time.

After finding \(x\), we substitute it back into the expression \(y = 75 - 2x\) to find \(y\), completing the solution.

Algebra Problem Solving

Algebra problem solving often involves translating word problems into mathematical equations and solving them systematically.
This process requires careful translation of the verbal description into equations. In this problem, sentences like "the price of two lawn seats and a pavilion seat is \(\$75\)" are converted into the equation \(2x + y = 75\).
It's crucial to define variables clearly as this guides the formulation of accurate equations. Once equations are established as simultaneous equations, methods such as substitution or elimination are used to find the solution.

• Verify your solutions by substituting back into the original equations to ensure they hold true.
• Consistency and checking help confirm that no steps were missed or errors made in calculations.

With practice, algebra problem solving builds a strong foundation for tackling more complex mathematical problems and real-world situations.