Problem 30 - Algebra and Trigonometry Real Mathematics, Real People

Question

You are given the yearly interest earned from a total of \$18,000 invested in two funds paying the given rates of simple interest. Write and solve a system of equations to find the amount invested at each rate. Yearly Interest \$ 840 Rate 1 \(6 \%\) Rate 2 \(3 \%\)

Step-by-Step Solution

Answer

The amount invested at 6% is \$10,000 and at 3% is \$8,000.

Step 1: Formulate the equations

Let \(x\) denote the amount invested at 6% and \(y\) the amount at 3%. Since the total amount invested is $18,000, the first equation would be:\(x + y = 18000\)The total interest from both investments is $840. The amount of interest is obtained by multiplying the money invested by the interest rate. Hence, the second equation is:\(0.06x + 0.03y = 840\)This gives a system of two linear equations:\[\begin{align*}x + y &= 18000 \0.06x + 0.03y &= 840\end{align*}\]

Step 2: Solve the system of equations

First, multiply the second equation by 100 to eliminate the decimals:\[\begin{align*}x + y &= 18000 \6x + 3y &= 84000\end{align*}\]Then, isolate \(x\) in the first equation: \(x = 18000 - y\)Substitute \(x\) in the second equation: \(6(18000 - y) + 3y = 84000\)Simplify and solve for \(y\): \(108000 - 6y + 3y = 84000\)\(3y = 24000\)\(y = 8000\)Substitute \(y = 8000\) in the first equation to solve for \(x\):\(x = 18000 - 8000\)\(x = 10000\)

Step 3: Verify the solution

Check the solution by substituting \(x = 10000\) and \(y = 8000\) into the original equations:\[\begin{align*}x + y &= 10000 + 8000 = 18000 \0.06x + 0.03y &= 0.06*10000 + 0.03*8000 = 600 + 240 = 840\end{align*}\]The solutions satisfy both equations, so the amount invested at 6% is \$10,000 and at 3% is \$8,000.

Key Concepts

Simple Interest

Simple interest is a way to calculate the interest you earn or pay on an investment or loan. You can think of it as steadily earning a fixed amount over time, rather than increasing over time like compound interest. The key formula for simple interest is \( I = P \times r \times t \), where:

• \( I \) is the interest amount
• \( P \) is the principal amount (initial investment)
• \( r \) is the rate of interest per year in decimal
• \( t \) is the time period in years

In the problem, simple interest helps to determine how much each part of the investment contributes to the total interest earned.
This method is straightforward, making it easier to use in problems like this where calculating interest on investments is required.

Linear Equations

Linear equations are equations of the first order. They make a straight line when graphed, and usually have one or more variables with no exponents. In this system of linear equations:

• We have two equations to work with
• The equations are related and describe a relationship in the context of the problem

The given equations are:

• \( x + y = 18000 \)
• \( 0.06x + 0.03y = 840 \)

These equations allow us to find exact values for the variables \( x \) and \( y \), which represent the amounts invested at two different interest rates.
Linear equations are crucial in solving many real-world problems, especially those involving planning and budget, like investments.

Investment Problems

Investment problems often deal with the allocation and distribution of funds to achieve certain financial goals. In this case, you need to decide how much to invest at different interest rates to get a specific total return. Here's how you approach the problem:

• Identify total amounts and interest rates
• Set up equations based on these amounts and rates
• Solve the equations to find out how much should be invested in each option

These types of problems are common in finance and personal financial planning.
They teach valuable lessons on balancing funds to achieve better returns and emphasize the importance of understanding interest rates in investment decisions.