Problem 26 - Algebra 1: Concepts and Skills

Question

AEROBICS CLASSES A fitness club offers an aerobics class in the morning and in the evening. Assuming that the number of people in each class can be represented by a linear function, use the information in the table below to predict when the number of people in each class will be the same. $$ \begin{array}{|c|c|c|} \hline \text { Class } & {\text { Current }} & {\text { Increase (people }} \\\ \underline{\phantom{xxx}} & {\text { attendance }} & {\text { per month) }} \\ \hline \text { Morning } & {40} & {2} \\ \hline \text { Evening } & {22} & {8} \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The attendance of both the morning and evening aerobics classes will be the same in 3 months.

Step 1: Formulate the Linear Equations

From the table, it can be noted that the morning class starts with 40 people and increases by 2 per month, let's represent this by the equation: \( M = 40 + 2x \). And the evening class starts with 22 people and increases by 8 per month, represented by: \( E = 22 + 8x \), where x is time in months.

Step 2: Solve for x

To predict when the number of people attending Morning class \( M \) will equal to the number of people attending Evening class \( E \), we need to solve the equation \( M = E \), which results in \( 40 + 2x = 22 + 8x \). Re-arranging the equation gives \( 6x = 18 \) which results in \( x = 3 \).

Step 3: Interpret the Results

The solution \( x = 3 \) implies that the attendance of both classes will be equal in 3 months.

Key Concepts

Algebraic Equations

Algebraic equations form the foundation of solving problems in algebra. They consist of variables, constants, and operations such as addition, subtraction, multiplication, and division that are set equal to each other. An equation represents a statement of equality and poses a question, implying a search for the value(s) of the variable(s) that make the statement true.

For instance, in our exercise, we encounter linear algebraic equations, which graph as straight lines. The linear equation for the morning aerobics class is given as M = 40 + 2x, where M is the number of attendees and x represents the months passed. The coefficient of x indicates the rate of increase. Understanding these relationships helps us form a picture of how the attendance changes over time.

When constructing an algebraic equation from a word problem, remember these essential steps:

• Identify the variables and constants from the context.
• Translate the given information into mathematical expressions.
• Combine these expressions to form an equation reflecting the real-world scenario.

Thus, equations allow us to model and solve practical problems mathematically.

Systems of Equations

In mathematics, a system of equations refers to a set of two or more equations that have a set of variables in common. The goal when solving a system is to find the value(s) for the variable(s) that satisfy all equations simultaneously.

Our example features a system of linear equations where the morning and evening class attendance can be plotted as two lines on a graph, and the point where they intersect represents the time when both classes have the same attendance. The equations M = 40 + 2x and E = 22 + 8x make up our system, and our task is to find where M equals E.

To solve systems of equations, one can use methods such as:

• Graphical representation, plotting each equation and finding the intersection point.
• Substitution, where you solve one equation for a variable and then substitute that expression into the other equation.
• Elimination, which involves adding or subtracting equations to eliminate one of the variables.

In this exercise, we used the elimination method to find the solution, which is efficient for linear systems especially when the equations are already set up to be combined easily.

Solve for Variables

Solving for variables is the process of finding the values of the unknowns in an equation or system of equations that make the equation true. To do so, one must isolate the variable on one side of the equation through a series of algebraic manipulations, while performing the same operations on both sides to maintain the equality.

In our aerobics class problem, we are asked to find the value of x for which the equations M = 40 + 2x and E = 22 + 8x are equal. By setting these two equations equal to each other, we can manipulate the resulting equation 40 + 2x = 22 + 8x to isolate x.

The steps to solve for x are as follows:

• Get all terms containing x on one side of the equation.
• Combine like terms, if there are any.
• Divide or multiply to solve for the variable.

In our solution, we subtracted 2x from both sides to start isolating x, giving us 6x = 18. Dividing both sides by 6 yielded the simple, yet crucial equation x = 3, meaning the classes will have equal attendance after 3 months. This process shows how algebra is a powerful tool for solving real-world problems.