Problem 53 - Introductory Algebra for College Students

Question

You plan to start taking an aerobics class. Nonmembers pay \(\$ 4\) per class. Members pay a \(\$ 10\) monthly fee plus an additional \(\$ 2\) per class. The monthly cost, \(y,\) of taking \(x\) classes can be modeled by the linear system $$\left\\{\begin{array}{l}y=4 x \\ y=2 x+10\end{array}\right.$$ a. Use graphing to solve the system. b. Interpret the coordinates of the solution in practical terms.

Step-by-Step Solution

Answer

The solution can be interpreted as the number of classes where both members and nonmembers pay the same amount. The number of classes (\(x\) coordinate) represents the point at which it becomes more economical to be a member rather than a nonmember, and the cost (\(y\) coordinate) represents the amount both members and nonmembers would pay for that number of classes.

Step 1: Plot the Equations

Graph the two equations \(y=4x\) and \(y=2x+10\), including the entire range of possible values (since there's no maximum mentioned in the exercise). Both equations are in slope-intercept form \(y=mx+b\), which means each can be plotted by marking the y-intercept \(b\) and using the slope \(m\) to find the next point.

Step 2: Find the Intersection Point

The intersection point of two lines represented by the equations is the solution of the system of equations. In the graph, look for the point where the two lines cross. This is the solution (\(x, y\)) of the system of equations.

Step 3: Interpret the Solution

Given that \(x\) represents the number of classes and \(y\) the cost, the solution can be interpreted as the number of classes where both nonmembers and members pay the same amount. The \(x\) coordinate (number of classes) is the point at which it becomes more economical to be a member rather than a nonmember, and the \(y\) coordinate (cost) is the amount both members and nonmembers would pay for that number of classes.

Key Concepts

Graphing Linear Equations

Understanding how to graph linear equations is a fundamental skill in algebra. A linear equation is an equation between two variables that produces a straight line when graphed. These equations typically take the form of \( y = mx + b \), known as the slope-intercept form, where \( m \) represents the slope and \( b \) the y-intercept.

For example, in the given exercise, we have two linear equations: \( y=4x \) and \( y=2x+10 \). To graph these, you would start by plotting the y-intercept \( b \) on the y-axis. For the first equation, the y-intercept is 0 (since it is not explicitly stated), and for the second equation, it is 10. After marking the y-intercepts, you use the slope \( m \) to determine the direction and steepness of the line. In this context, \( m=4 \) for the first equation, meaning that for every unit increase in \( x \), \( y \) increases by 4 units. Similarly, for the second equation, \( m=2 \), indicating a gentler slope compared to the first equation.

Slope-Intercept Form

The slope-intercept form of a line, \( y = mx + b \), is instrumental in quick graphing of linear equations. The 'slope', \( m \), signifies the rate at which \( y \) changes with \( x \) - essentially the angle of the line in relation to the axes. The 'intercept', \( b \), is the point at which the line crosses the y-axis, where \( x=0 \).

This format is particularly useful for visual comparison of different linear equations to understand their rate of change and initial value at a glance. In our exercise, the differing slopes of the equations for nonmembers \( (m=4) \) versus members \( (m=2) \), and the member intercept of \( 10 \) explain visually why and when membership becomes cost-effective.

System of Equations Solution

A 'system of equations' consists of multiple equations that you solve at once, usually to find where they intersect—or the values of \( x \) and \( y \) that satisfy all equations in the system. The point of intersection represents the solution to the system. In a graphical approach, as seen in this exercise, this is the point where the lines representing the equations cross.

In practical applications, like the aerobics classes membership problem, finding the intersection point tells you the exact scenario in which the options offered are equivalent—in this case, the number of classes that equates the cost for members and nonmembers. Solving systems of equations can be done by graphing, substitution, elimination, or using matrices, each with its own use cases and advantages.

Algebraic Problem Solving

Algebraic problem solving typically involves identifying the correct equations based on a given scenario, manipulating these equations, and interpreting their solutions within the context of the problem. In our exercise, we have a real-world situation converted into a system of linear equations. Solving this problem algebraically, you would set the two equations equal to each other (since they both equal \( y \)) and find the value of \( x \) that makes them true.

Once the solution is found, whether graphically or algebraically, it's critical to interpret it in terms of the original problem. Here, we interpret the solution to ascertain at what point a membership becomes advantageous for an individual planning to take aerobics class. This blend of equation solving and contextual application highlights the practical value of algebra in everyday decision making.