Problem 7 - Understanding Elementary Algebra with Geometry

Question

Solve each of the following verbal problems algebraically. You may use either a one or a two-variable approach. Carmen invests a total of \(\$ 1,700\) in two stocks. One stock pays a yearly dividend of \(7 \%,\) while the other pays \(6 \% .\) If Carmen received \(\$ 110\) in combined dividends from the two stocks, how much did she invest in each?

Step-by-Step Solution

Answer

Carmen invested \$800 in the first stock and \$900 in the second stock.

Step 1: Define Variables

Let \( x \) be the amount invested in the stock that pays a 7% dividend, and \( 1700 - x \) be the amount invested in the stock that pays a 6% dividend.

Step 2: Set Up the Equation

The total dividend received from both stocks is given by combining the dividends:Dividend from the first stock: \( 0.07x \)Dividend from the second stock: \( 0.06(1700 - x) \)These dividends add up to \$110. Therefore, the equation is:\[ 0.07x + 0.06(1700 - x) = 110 \]

Step 3: Simplify the Equation

Distribute the 0.06 in the equation:\[ 0.07x + 0.06 \times 1700 - 0.06x = 110 \]This simplifies to:\[ 0.07x + 102 - 0.06x = 110 \]

Step 4: Combine Like Terms

Combine the \( x \)-terms and constants:\[ 0.01x + 102 = 110 \]

Step 5: Isolate the Variable

Subtract 102 from both sides:\[ 0.01x = 8 \]

Step 6: Solve for x

Divide both sides by 0.01 to solve for \( x \):\[ x = \frac{8}{0.01} = 800 \]

Step 7: Determine the Investment in the Other Stock

Since Carmen invested \( \$1700 - x \) in the second stock:\[ 1700 - 800 = 900 \]

Key Concepts

Investment Problems

Investment problems in algebra involve the distribution and return of money invested in different options or stocks. They commonly require determining how much was invested in each option based on given returns. Here's a summary of how to approach these problems:

• Identify the total investment amount and the individual dividend rates for each option.
• Use variables to represent the unknown amounts invested in each option.
• Set up an equation based on the dividends received from each investment and solve for the variables.

In the given problem, Carmen invested a total of \( \$1700 \) in two stocks. We used the following steps to find out how much she invested in each stock:

• Defined \( x \) as the amount invested in the 7% dividend stock and \( 1700 - x \) for the 6% dividend stock.
• Set up an equation based on the total dividend received, which was \( \$110 \).
• Solved the equation to determine the values of \( x \).

Solving Equations

Solving equations is a fundamental skill in algebra. It involves finding the value of the unknown variable(s) that make the equation true. Here, we solved the investment problem with one equation. Let's break it down:

First, you need to set up a proper equation by equating the dividends. We set up the equation:
\[ 0.07x + 0.06 (1700 - x) = 110 \]
Next, we simplified and solved for \( x \). Distributive property helped us to simplify:
\[ 0.07x + 102 - 0.06x = 110 \]
Combining like terms gave us:
\[ 0.01x + 102 = 110 \]
We isolated the variable by subtracting \( 102 \) from both sides:
\[ 0.01x = 8 \]
Finally, we solved for \( x \) by dividing both sides by \( 0.01 \):
\[ x = \frac{8}{0.01} = 800 \]

This process shows how combining like terms, using basic arithmetic operations, and isolating the variable lead to a solution.

Variables in Algebra

Variables are symbols (often letters) used to represent unknown values. They are essential in algebra because they help generalize arithmetic calculations and create formulas. In this exercise, we used variables to represent the unknown investments.

Key steps to handle variables effectively include:

• Assign a variable to each unknown quantity.
• Express other unknowns in terms of these variables.
• Set up equations using these expressions.

We used:

• \( x \) for the amount invested in the 7% dividend stock.
• \( 1700 - x \) for the amount invested in the 6% dividend stock.

By converting the word problem into an algebraic equation, we could solve for the variable \( x \), which guided us to find the exact amounts invested. This approach leverages the power of algebra to solve real-world problems efficiently.