Problem 34 - Beginning and Intermediate Algebra

Question

Write a system of equations and solve. Diego inherited \(\$ 20,000\) and puts some of it into an account earning \(4 \%\) simple interest and the rest in an account earning \(7 \%\) simple interest. He earns a total of \(\$ 1130\) in interest after a year. How much did he deposit into each account?

Step-by-Step Solution

Answer

The system of equations for the given problem is: \[ \begin{cases} x + y = 20{,}000 \\ 0.04x + 0.07y = 1{,}130 \end{cases} \] Solving the system, we find that Diego deposited \(\$9,000\) into the account earning \(4\%\) simple interest and \(\$11,000\) into the account earning \(7\%\) simple interest.

Step 1: Define the variables

Let's denote the amount Diego deposits in the first account (4% interest) as x and the amount he deposits in the second account (7% interest) as y.

Step 2: Write the system of equations

I. Total amount: Since Diego has $20,000 to invest, we can write the equation as: \(x + y = 20{,}000\) II. Total interest: The total interest earned is $1,130 after a year, and can be represented as: \(0.04x + 0.07y = 1{,}130\) Now, we have a system of two equations with two unknowns: \[ \begin{cases} x + y = 20{,}000 \\ 0.04x + 0.07y = 1{,}130 \end{cases} \]

Step 3: Solve the system of equations

We can use the substitution method or elimination method to solve this system. Let's use the substitution method in this case. I. Solve the first equation for x: \(x = 20{,}000 - y\) II. Substitute this into the second equation: \(0.04(20{,}000 - y) + 0.07y = 1{,}130\) III. Simplify and solve for y: \[ \begin{aligned} 800 - 0.04y + 0.07y &= 1{,}130 \\ 0.03y &= 330 \\ y &= 11{,}000 \end{aligned} \] IV. Substitute the value of y back into the equation for x: \(x = 20{,}000 - 11{,}000 = 9{,}000\)

Step 4: Conclusion

Diego deposited \(\$9,000\) into the account earning \(4\%\) simple interest and \(\$11,000\) into the account earning \(7\%\) simple interest.

Key Concepts

Simple Interest

Simple interest is a way to calculate interest on an amount of money that is deposited or borrowed. It is straightforward because the interest is calculated only on the principal amount. The key formula for simple interest is given by:\[ I = P \cdot r \cdot t \]where:

• \( I \) is the interest earned or paid,
• \( P \) is the principal amount,
• \( r \) is the annual interest rate (in decimal), and
• \( t \) is the time the money is invested or borrowed for, in years.

In Diego's case, the interest earned on his money is spread across two accounts with different rates. The goal is to ensure the total interest, which is \$1,130, matches the combined interest from each account.

Substitution Method

The substitution method is a key technique for solving systems of equations. It involves solving one of the equations for one variable and then substituting that expression into the other equation. This reduces the system to a single equation with one unknown, which is much simpler to solve.
Here's how it works with Diego's problem:

• First, express one variable in terms of the other using one of the equations. Here, solve \(x + y = 20{,}000\) for \(x\), giving \(x = 20{,}000 - y\).
• Next, substitute \(x = 20{,}000 - y\) into the second equation \(0.04x + 0.07y = 1{,}130\). This creates an equation with just \(y\), simplifying the process.
• Finally, solve for \(y\) and then substitute back to find \(x\).

This method is especially useful when one of the variables is easy to isolate.

Elimination Method

The elimination method involves combining the equations in a system to "eliminate" one of the variables. By adding or subtracting the equations, one variable is removed, making the system easier to solve.
For instance, consider Diego's system:

• Multiply the first equation \(x + y = 20{,}000\) by a number that allows one variable to cancel when added or subtracted from the second equation \(0.04x + 0.07y = 1{,}130\).
• Adjust both equations such that adding (or subtracting) them removes one variable, paving the way to solve for the other.
• This method is beneficial when equations are already somewhat aligned for elimination.

It complements the substitution method and can be more efficient depending on the given equations.

Real-Life Application

Understanding real-life applications of algebra, like solving systems of equations, is essential because it reflects how math is used beyond the classroom. Diego's scenario is a common situation where people need to manage finances and investments.
When dealing with different interest rates, balancing the total deposit to meet a specific goal (like a total interest earned) is a practical application of these mathematical techniques. Whether it's handling personal finances, business investments, or even budgeting in engineering projects, systems of equations help create a plan to achieve desired financial outcomes.
By applying the methods of solving systems of equations, such as substitution or elimination, individuals can make informed decisions based on mathematical modeling. This turns complex financial decisions into manageable calculations.