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Chapter 4: Problem 29

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?

Short Answer

Expert verified
The solution will yield the amount of money invested at a 6% interest rate (x), and the amount invested at an 8% interest rate (y).

Step by step solution

01

Define the Variables

Let \( x \) be the amount invested at \(6 \% \) and \( y \) be the amount invested at \(8 \% \), both in dollars.
02

Formulate the Equations

Now, according to the problem, we can set up the following two equations: \[ x + y = 7000 \] (total amount invested is \$7000) \[0.06x + 0.08y = 520 \] (total interest earned in a year was \$520)
03

Solve the System of Equations

Next, the system of equations is solved by substitution or elimination method. Multiply the first equation by 0.06 and subtract it from the second equation to solve for \( y \). Then, substitute the value of \( y \) from the second equation into the first to solve for \( x \).
04

Calculate the Amount Invested at each rate

Solving the equations using the substitution or elimination method would give the values of \( x \) and \( y \) which are the amounts invested at each rate.

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