Problem 42 - College Algebra Essentials

Question

Things did not go quite as planned. You invested \(\$ 12,000\), part of it in stock that paid \(14 \%\) annual interest. However, the rest of the money suffered a \(6 \%\) loss. If the total annual income from both investments was \(\$ 680,\) how much was invested at each rate?

Step-by-Step Solution

Answer

The amount invested at a 14% interest rate is $8,000 and the amount that suffered a 6% loss is $4,000.

Step 1 - Define the Variables

First, we need to define the variables to represent the unknowns. Let x be the amount invested at 14% and y be the amount that suffered a 6% loss.

Step 2 - Write, Simplify and Solve the Equations

We have a system of two equations that reflect the problem: \• The total amount invested, which is \$12,000, is the sum of x and y. This gives us equation (1) as x + y = 12,000. \• The interests from both investments amounted to \$680. The interest from investment x is 14% of x and the loss from investment y is 6% of y. Therefore we have equation (2) as 0.14x - 0.06y = 680. \Now, to solve for x and y, we could multiply the first equation by 0.14 and subtract the second equation from the first. We obtain y = 4000. Substitute y into the first equation, we get x = 8000.

Step 3 - Check the Results

To ensure that the solution is correct, we need to substitute x = 8,000 and y = 4,000 back into the second equation to verify whether the left-hand side equals to the right-hand side (680). Doing so, we verify that indeed 0.14*8000 - 0.06*4000 = 680.