Problem 63 - A Graphical Approach to Precalculus with Limits

Question

The table shows the selling prices for three representative homes. Price \(P\) is given in thousands of dollars, age \(A\) in years, and home size \(S\) in thousands of square feet. These data may be modeled by the equation \(P=a+b A+c S\). $$\begin{array}{c|c|c} \text { Price (P) } & \text { Age (A) } & \text { Size (S) } \\ \hline 190 & 20 & 2 \\ 320 & 5 & 3 \\ 50 & 40 & 1 \end{array}$$ (a) Write a system of linear equations whose solution gives \(a, b,\) and \(c\) (b) Solve this system of linear equations. (c) Predict the price of a home that is 10 years old and has 2500 square feet.

Step-by-Step Solution

Answer

The predicted price is 160 thousand dollars.

Step 1: Write Equations for Each Home

From the given equation form \(P = a + bA + cS\), we can write the following equations for the provided data: - For the first home: \(190 = a + 20b + 2c\)- For the second home: \(320 = a + 5b + 3c\)- For the third home: \(50 = a + 40b + 1c\).

Step 2: Set Up the System of Equations

The equations obtained are:1. \(190 = a + 20b + 2c\)2. \(320 = a + 5b + 3c\)3. \(50 = a + 40b + c\).

Step 3: Solve the System of Equations

Using methods such as substitution or elimination, solve the system:- Subtract equation 2 from equation 1: \( (190 - 320) = (a - a) + (20b - 5b) + (2c - 3c) \), giving \(-130 = 15b - c\).- Subtract equation 3 from equation 1: \( (190 - 50) = (a - a) + (20b - 40b) + (2c - c) \), giving \(140 = -20b + c\).- Add these results to eliminate \(c\): \(-130 + 140 = 15b - c - 20b + c\), yielding \(10 = -5b\) or \(b = -2\).- Substitute \(b = -2\) into \(-130 = 15b - c\): \(-130 = 15(-2) - c\) => \(-130 = -30 - c\) => \(c = -100\).- Substitute \(b = -2\), \(c = -100\) in equation 1: \(190 = a + 20(-2) + 2(-100)\), giving \(190 = a - 40 - 200\) => \(a = 430\).

Step 4: Predict the Price of the New Home

Substitute \(a = 430\), \(b = -2\), \(c = -100\) into the equation for the price prediction \(P = a + bA + cS\) for a home with \(A = 10\) and \(S = 2.5\):\[P = 430 - 2(10) - 100(2.5)\] Simplify: \[P = 430 - 20 - 250 = 160\] Thus, the predicted price of the new home is 160 thousand dollars.

Key Concepts

Price Prediction

Understanding price prediction involves creating a general model that estimates the cost of a product or, in this case, a home. This estimation is based on known factors that affect price. In real estate, key influences might include age and size of the home.
In the exercise, the formula used for price prediction is given by the linear equation \( P = a + bA + cS \), where \( P \) represents the price, \( A \) is the age, and \( S \) is the size. The variables \( a \), \( b \), and \( c \) are coefficients that determine how much each factor contributes to the price.

By deriving these coefficients from past data, we can predict future prices for homes with specific attributes. This is particularly useful for prospective buyers or sellers who need a general idea of a home's market value before making decisions.

Linear Modeling

Linear modeling is a crucial mathematical concept used for establishing relationships between different variables. It involves creating equations that depict these relationships in a straight line when graphed.
The equation \( P = a + bA + cS \) is an example of a linear model, where \( a \), \( b \), and \( c \) are called coefficients. These coefficients are constants that provide the weight each variable (age, size) holds in influencing the home's price.

One of the main benefits of linear modeling is its simplicity and efficiency in handling real-world data. It helps in forming a clear picture of how changes in one or more variables affect the outcome, in this case, the price of the home.

Substitution Method

The substitution method is a technique to solve systems of equations where you replace one variable with another equivalent expression.
Let's consider the resulting system from the given equations:

• \( 190 = a + 20b + 2c \)
• \( 320 = a + 5b + 3c \)
• \( 50 = a + 40b + c \)

Choose one equation to solve for one variable, then substitute that expression into the others. For example, if we solve the second equation for \( a \), then replace \( a \) in the other equations, eventually continuing until all variables are solved.

This might seem more complex initially, but the substitution method is particularly useful when equations are simple enough or the coefficients are suited for direct substitution. It systematically reduces the amount of unknowns, making the system easier to solve step by step.

Elimination Method

The elimination method involves adding or subtracting equations to eliminate one of the variables, thus simplifying the system.
This exercise presents a scenario where we have:

• \( 190 = a + 20b + 2c \)
• \( 320 = a + 5b + 3c \)
• \( 50 = a + 40b + c \)

By subtracting these equations from each other aptly, we are able to eliminate variable \( a \) first, simplifying our system significantly.

Then, through further elimination, we can determine values for \( b \) and \( c \), which are found as \( b = -2 \) and \( c = -100 \). This allows us to easily solve for \( a \) using any of the original equations, resulting in \( a = 430 \).
Elimination is favored for its direct and straightforward approach, especially when the system’s coefficients allow for simple arithmetic to cancel out a variable.