Problem 23 - Algebra and Trigonometry

Question

Things did not go quite as planned. You invested \(\$ 8000\), part of it in stock that paid \(12 \%\) annual interest. However, the rest of the money suffered a \(5 \%\) loss. If the total annual income from both investments was \(\$ 620,\) how much was invested at each rate?

Step-by-Step Solution

Answer

To solve this problem, we need to use algebra to solve the system of equations described in the problem. Implementing these steps should lead to the solution, which are the amounts invested at each rate.

Step 1: Define the Variables

Let's denote the money that was invested with a return rate of \(12\%\) as \(x\) and the sum invested with a \(5\%\) loss as \(y\).

Step 2: Outline the Problem in Equations

From the problem, we can write out the following equations based on the total money invested and total return: \[x + y = 8000 \] - since the total sum of money is $8000. For the return, \[0.12x - 0.05y = 620 \] - since we know that the total income (taking into account losses as negative) totals to $620. The positive \(0.12x\) represents the earnings from the investment that had a 12% yield, while the negative \(0.05y\) denotes a 5% loss from the second investment.

Step 3: Solve the Equations

Now we have a system of two equations which we can solve. We can use substitution or elimination method. First, let's manipulate the first equation by isolating for y, so we get: \[y = 8000 - x\]. Substitute this value of y into the second equation to get: \[0.12x - 0.05(8000 - x) = 620\]. Solve this for x. Then, use the found x-value into the equation for y to solve for y.

Step 4: Verify the Solution

Once we have values for x and y, it is advisable to substitute them into both original equations to ensure that they satisfy the conditions of the problem.