Question

The average price \(B\) (in dollars) of brand name prescription drugs from 1998 to 2005 can be modeled by \(B=6.928 t-3.45, \quad 8 \leq t \leq 15\) where \(t\) represents the year, with \(t=8\) corresponding to 1998 . Use the model to find the year in which the price of the average brand name drug prescription exceeded \(\$ 75\).

Short answer

After performing the calculations, the result obtained is 't' is approximately 11.32. Adding this to 1990 (since we are considering t=8 corresponds to 1998, thus t=1 corresponds to 1991), we conclude that the average cost of a brand name prescription drug first exceeded $75 in the year 2001.

Step-by-step solution

Identify and Set Up the Equation

The equation to solve is \(B=6.928 t-3.45\), where \(B\) represents the cost of the drug. We want to find when the cost of the drug first exceeded $75, so we need to solve for 't' when \(B = 75\). Thus, the equation we need to solve is \(75 = 6.928t - 3.45\).

Solve the Equation

In order to solve the equation for 't', we first need to isolate 't' by adding 3.45 to both sides of the equation. This gives us: \(75 + 3.45 = 6.928t\). Simplifying the left side, we get \(78.45 = 6.928t \).

Divide Both Sides by 6.928

Now, divide both sides of the equation by 6.928 to solve for 't': \(t = \frac{78.45}{6.928}\).\nThis gives the numerical value of 't'.

Adjust for the Year

The variable 't' represents the year, but it's offset such that t=8 corresponds to 1998. To find the actual year, we add 1990 to the value of 't' from the previous step. This results in the year in which the cost of a brand name prescription drug first exceeded $75.

Key concepts

Modeling with Linear Equations

Linear equations are a staple in college algebra, serving as the foundation to model various real-world problems. In this particular exercise, we're using a linear equation to model the average price of brand name prescription drugs over time. The equation is given as \( B=6.928t-3.45 \), where \( B \) is the price in dollars, and \( t \) is the time in years, with \( t=8 \) corresponding to 1998.
This equation helps us understand trends over time, allowing us to predict when certain thresholds are crossed, such as when the price exceeds a specific amount.
To create a linear model, we often rely on data collected over time. The coefficients in the equation—like 6.928 here—indicate the rate of change, or how much the dependent variable (price) increases with the independent variable (year).
The constant term, -3.45, adjusts the model vertically to fit the observed data better.

Solving Linear Equations

Solving linear equations is a crucial skill in algebra, as it allows us to find unknown variables. In this exercise, we need to find the value of \( t \) when the price \( B \) exceeds $75. This means setting up the equation \( 75 = 6.928t - 3.45 \) and solving for \( t \).
Each step in solving involves operations aimed at isolating the variable of interest. Thus, we first add 3.45 to both sides, leading to \( 78.45 = 6.928t \). Dividing both sides by the coefficient of \( t \), which is 6.928, yields \[ t = \frac{78.45}{6.928} \].
This calculation gives the numerical value for \( t \), which we interpret in the context of the problem.

Applications of Algebra

Algebra is more than just solving equations; it's about applying mathematical theories to analyze and solve real-world issues. In this scenario, the linear equation is used to estimate costs over time, a common practice in economics and business.
Understanding these applications helps in making informed decisions. For instance, businesses can use models like this to forecast future prices, adjust budgets, and strategize pricing. Moreover, policymakers could leverage such models to assess the economic impact of drug pricing on public health.
Thus, algebra provides a means to translate complex data into actionable insights, underscoring its importance in various fields.

Real-World Algebra Problems

Real-world algebra problems often involve modeling change over time, as seen in this exercise with drug pricing. Algebra helps to simplify complex relationships into manageable equations.
Through this approach, we can pose and address questions critical to society, like understanding financial growth or predicting resource needs.
When tackling these problems, it's essential to consider the assumptions and limitations of the model. Knowing the range in which the model is valid (here, \( 8 \leq t \leq 15 \)) ensures that we remain within the bounds of reliable prediction.
Such practice not only enhances our mathematical skills but also deepens our understanding of the world around us.