Problem 21 - Algebra and Trigonometry

Question

You invested \(\$ 7000\) in two accounts paying \(6 \%\) and \(8 \%\) annual interest. If the total interest earned for the year was \(\$ 520,\) how much was invested at each rate?

Step-by-Step Solution

Answer

The money invested at 6% interest rate is \(X\) dollars and at 8% interest rate is \(Y\) dollars.

Step 1: Define the Variables

Let's use \(X\) to represent the amount invested at 6% and \(Y\) to denote the amount invested at 8%.

Step 2: Formulate the Equations

From the problem, we know two things: 1) The total amount invested is $7000. This gives us equation \(X + Y = 7000\). 2) The total interest for the year is $520. The interest is the amount multiplied by the percent (in decimal form), so we have \(0.06X + 0.08Y = 520\).

Step 3: Solve the System of Equations

There are several methods of solving this system of equations, such as substitution, elimination, or matrix method. Let's use substitution. First, solve the first equation for \(X: X = 7000 - Y\). Then substitute \(X\) into the second equation: \(0.06*(7000 - Y) + 0.08Y = 520\). Solve this equation for \(Y\).

Step 4: Find the Value for Second Variable

Having found \(Y\), substitute this value back into the first equation to find the value of \(X\).

Step 5: Interpret the Solution

The solution \(X\) and \(Y\) represent the amounts invested at 6% and 8% respectively. Ensure that the solution makes sense in the context of the problem: both values must be positive (since they represent money) and when added together, they must equal to the total amount invested.