Problem 36 - Precalculus Student Solutions Manual 5th

Question

Buying and Selling Land Roger bought two plots of land for a total of \(\$ 120,000\). When he sold the first plot, he made a profit of \(15 \% .\) When he sold the second, he lost \(10 \% .\) His total profit was \(\$ 5500 .\) How much did he pay for each piece of land?

Step-by-Step Solution

Answer

Roger paid \( \$70,000 \) for the first piece of land and \( \$50,000 \) for the second.

Step 1: Set up variables

Let the cost of the first piece of land be denoted by \( x \) and the cost of the second piece of land be denoted by \( y \). According to the problem, \( x + y = 120,000 \). Also, the total profit from selling both plots is \( 5500 \).

Step 2: Express profits in terms of variables

The profit from the first plot is \( 15\% \) of \( x \), which is \( 0.15x \). The loss from the second plot is \( 10\% \) of \( y \), which is \( 0.1y \). The net profit equation therefore is: \( 0.15x - 0.1y = 5500 \).

Step 3: Solve the system of equations

We now have the system of equations: \[ x + y = 120,000 \] and \[ 0.15x - 0.1y = 5500 \ \]. Solve the first equation for \( y \): \[ y = 120,000 - x \]. Substitute this into the second equation: \[ 0.15x - 0.1(120,000 - x) = 5500 \].

Step 4: Simplify and solve for \( x \)

Expand and combine like terms: \[ 0.15x - 12,000 + 0.1x = 5500 \]. Combine like terms: \[ 0.25x - 12,000 = 5500 \]. Add \( 12,000 \) to both sides: \[ 0.25x = 17500 \]. Divide by \( 0.25 \): \[ x = 70,000 \].

Step 5: Solve for \( y \)

Substitute \( x = 70,000 \) back into \[ y = 120,000 - x \]: \[ y = 120,000 - 70,000 = 50,000 \].

Key Concepts

System of Equations

In this exercise, we deal with a system of equations to find out the individual costs of two plots of land. A system of equations is essentially a set of equations with multiple variables. These equations are solved together to find a common solution for the variables involved.

For Roger’s land purchase problem, we establish two equations based on the total cost and the net profit:

• Equation 1: The total cost of the two plots is \(120,000.
• Equation 2: The total profit from selling the plots is \)5,500.

By solving these equations simultaneously, we can find the prices of the individual plots of land.

Profit and Loss Calculation

Let’s break down the profit and loss calculations. When Roger sold the first plot of land, he made a 15% profit. For the second plot, he incurred a 10% loss.

Profit and loss percentages help us determine the net gain or loss from an investment:

• Roger’s profit from the first plot:
  
  < \(Profit = (15/100) \times Cost of first plot \)
  = 0.15 x
• His loss from the second plot:
  
  \(Loss = (10/100) \times Cost of second plot \)
  = 0.1 y

Using these expressions, the net equation for Roger’s overall profit looks like this: 0.15x - 0.1y = 5500.
This equation represents the balance between the profit and loss, giving us critical insight into how investments performed.

Variable Substitution

Variable substitution allows us to solve a system of equations effectively. We start by isolating one variable in one of the equations and substituting this expression into the other equation.

For Roger’s case, we first solve the total cost equation for one variable: x + y = 120,000
y = 120,000 - x

Then, we substitute this expression for y into the profit equation: (0.15 x - 0.1) (120,000 - x) = 5500

This substitution converts the profit equation to a single-variable equation. We can now simplify and solve it step-by-step:

Expanded and simplified: 0.15 x - 12,000 + 0.1 x = 5500
Combining terms gives: 0.25 x - 12,000 = 5500
Adding 12,000: 0.25 x = 17,500
Dividing by 0.25: x = 70,000


Using this value of x, we solve for y: y = 120,000 - 70,000
y = 50,000

Now, we have that Roger paid \(70,000 for the first plot and \)50,000 for the second plot. By systematically applying variable substitution, we navigate through complex problems efficiently.