Problem 13 - Elementary Algebra

Question

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question. Laurie has \(\$ 46,000\) invested in stocks and bonds. The amount invested in stocks is \(\$ 8,000\) less than three times the amount invested in bonds. How much does Laurie have invested in bonds?

Step-by-Step Solution

Answer

Laurie has \(\$13,500\) invested in bonds.

Step 1: Define Variables

Let the amount invested in bonds be denoted by the variable \(b\). The amount invested in stocks can be denoted by \(s\).

Step 2: Set Up the Equations

Laurie has a total of \(\$46,000\) invested in stocks and bonds.Thus, \[ b + s = 46000 \] Furthermore, the amount invested in stocks is \(\$8,000\) less than three times the amount invested in bonds.Thus, \[ s = 3b - 8000 \]

Step 3: Substitute and Solve

Substitute the expression for \(s\) from the second equation into the first equation: \[ b + (3b - 8000) = 46000 \] Combine like terms: \[ 4b - 8000 = 46000 \] Solve for \(b\): \[ 4b = 54000 \] \[ b = 13500 \] Thus, Laurie has \(\$13,500\) invested in bonds.

Step 4: Verification

To verify, calculate the amount invested in stocks: \[ s = 3(13500) - 8000 = 40500 - 8000 = 32500 \] Sum the investments in stocks and bonds: \[ 13500 + 32500 = 46000 \] The total is indeed \(\$46,000\), confirming our solution.

Step 5: Write a Complete Sentence

Laurie has \(\$13,500\) invested in bonds.

Key Concepts

defining variables

Word problems in algebra can be tricky to solve, but the first crucial step is to define your variables clearly. Variables represent the unknowns in the problem. In this case, we need to determine two unknowns: the amount Laurie has invested in bonds, and the amount she has invested in stocks.
Let’s choose a variable for each unknown. Let's use the variable \(b\) to represent the amount invested in bonds. For the stocks, we can use \(s\) to denote the amount invested in stocks. By doing this, we simplify the problem and create a foundation for establishing equations.
Clearly defining variables keeps our work organized and makes it easier to set up and solve the equations.

setting up equations

Next, we must translate the word problem into mathematical equations. Let's use the information given:
- Laurie has a total of \$46,000\ invested in stocks and bonds. So we can write: \[ b + s = 46000 \]
- The amount invested in stocks is \$8,000\ less than three times the amount invested in bonds. Hence: \[ s = 3b - 8000 \]
By setting up these equations, we create a mathematical model of the word problem that we can solve using algebraic techniques.

substitution method

To solve our equations, we can use the substitution method. This involves substituting one variable’s expression from one equation into another equation, allowing us to solve for one variable at a time.
In our case, let's substitute the expression for \(s\) from the second equation into the first equation:
\[ b + (3b - 8000) = 46000 \]
This results in a single equation with one variable, making it much simpler to solve. This technique helps in breaking down complex problems, making them more manageable.

solving linear equations

Once we have substituted one equation into another, we need to solve the resulting linear equation. For our problem, the substitution gives us:
\[ b + 3b - 8000 = 46000 \]
Combine like terms:
\[ 4b - 8000 = 46000 \]
Add \$8000\ to both sides:
\[ 4b = 54000 \]
Divide by \$4\:
\[ b = 13500 \]
This tells us that Laurie has \$13,500\ invested in bonds. Solving linear equations step-by-step helps ensure accuracy and completeness.

verification of solution

After finding a solution, it is important to verify it to make sure it’s correct. Using the value we found for \(b\), we can calculate the amount invested in stocks: \[ s = 3(13500) - 8000 = 40500 - 8000 = 32500 \]
Now, verify the total investment:
\[ 13500 + 32500 = 46000 \]
The total matches the information given in the problem, confirming that our solution is correct. Always verify your solutions to ensure they meet all the problem’s conditions.