Problem 71 - College Algebra with Corequisite Support

Question

For the following exercises, use a system of linear equations with two variables and two equations to solve. An investor who dabbles in real estate invested 1.1 million dollars into two land investments. On the first investment, Swan Peak, her return was a \(110 \%\) increase on the money she invested. On the second investment, Riverside Community, she earned \(50 \%\) over what she invested. If she earned \(\$ 1\) million in profits, how much did she invest in each of the land deals?

Step-by-Step Solution

Answer

The investor invested $0.75 million in Swan Peak and $0.35 million in Riverside Community.

Step 1: Define Variables

Let's define two variables to represent the amounts invested in each deal. Let \(x\) be the amount invested in Swan Peak, and \(y\) be the amount invested in Riverside Community.

Step 2: Set Up First Equation

Based on the problem, the total investment was 1.1 million dollars. This can be represented by the equation:\[ x + y = 1.1 \text{ million} \]

Step 3: Set Up Second Equation

The profits from the investments are described in the problem. Swan Peak gives a 110% return, so the profit is \(1.1x\). Riverside Community gives a 50% return, resulting in a profit of \(0.5y\). The total profit is $1 million, hence:\[ 1.1x + 0.5y = 1 \text{ million} \]

Step 4: Solve the System of Equations

We have the system of equations:\[1) \quad x + y = 1.1 \2) \quad 1.1x + 0.5y = 1\]First, solve the first equation for \(y\):\[ y = 1.1 - x \]Substitute this expression for \(y\) into the second equation:\[ 1.1x + 0.5(1.1 - x) = 1 \]Simplify and solve for \(x\):\[ 1.1x + 0.55 - 0.5x = 1 \]\[ 0.6x = 0.45 \]\[ x = 0.75 \]Now substitute \(x = 0.75\) back into the expression for \(y\):\[ y = 1.1 - 0.75 = 0.35 \]

Step 5: Confirm Solution

Now we confirm the solution by checking both conditions.The first condition is \(x + y = 1.1\), and it satisfies since \(0.75 + 0.35 = 1.1\).The second condition for profits is checked as: \[1.1 \times 0.75 + 0.5 \times 0.35 = 0.825 + 0.175 = 1\]Both equations are satisfied, confirming the investment amounts are correct.

Key Concepts

Linear Equations with Two Variables

Understanding linear equations with two variables is fundamental in algebra. These equations typically take the form \( ax + by = c \), where \(x\) and \(y\) are the variables, and \(a\), \(b\), and \(c\) are constants. In real-world scenarios, each variable represents a different quantity that we are trying to find. In this case, the equation \( x + y = 1.1 \text{ million} \) represents the total amount invested in two properties.

To manage these equations effectively, always start by clearly defining your variables. This delineation guides you in setting up correct equations and helps prevent mistakes. With two variables, solutions often require a second equation, allowing you to solve for each variable accurately.

Solving Systems of Equations

Solving systems of equations involves finding the values of the unknowns (variables) that satisfy all equations simultaneously. This process often requires techniques like substitution, elimination, or graphing. In our scenario, we used substitution.

• Start by isolating one variable in one of the equations. For example, from \( x + y = 1.1 \), we rearranged to find \( y = 1.1 - x \).
• Next, substitute this expression for \( y \) in the second equation. This step lets you solve for \( x \) directly.
• Once \( x \) is found, substitute back to find \( y \).

This method is efficient and often simplifies complex equations into more manageable ones, allowing for clearer analysis of the solution's accuracy.

Real-Life Application of Algebra

Algebra isn't restricted to theoretical or abstract problems. It finds robust real-life applications, especially in finance and investment scenarios. Here, we used algebra to tackle a real estate investment question.

Investors frequently encounter situations requiring precise calculations to make informed decisions. By learning and applying algebra to problems like these, you can predict financial outcomes, maximize returns, and minimize risks. Understanding linear equations and systems of equations enables individuals to analyze various investment proposals, helping them in strategizing and planning effectively.

Profit Calculations

Profit calculations are a common application of systems of equations, especially in business and investment situations. In the provided scenario:

• Swan Peak's investment yielded a 110% increase, or \(1.1x\).
• Riverside Community's investment provided a 50% increase, resulting in \(0.5y\).

The total profit was the sum of these gains, equating to \(1\) million dollars. By structuring these profit factors into a second equation (\(1.1x + 0.5y = 1\)), and using it alongside the total investment equation, we calculated the individual investments.

Real-world business often involves similar multi-variable profit scenarios. Getting adept at these calculations means equipping yourself with skills crucial for financial analysis.