Problem 24 - Algebra and Trigonometry

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 14% interest is \$6000 and the amount invested at a loss of 6% is also \$6000.

Step 1: Define the Variables

Let \(x\) represent the amount invested at an interest rate of 14% and \(y\) represent the amount invested at a loss of 6%. Using these variables, it can be expressed that the total investment is \(x+y = 12000\).

Step 2: Formulate the second equation

The total annual income, which is the sum of the income from the first investment and the loss from the second investment, is given by \(0.14x - 0.06y = 680\). This equation reflects the total profit made. Notice the minus in front of \(0.06y\), this is because the second investment led to a loss.

Step 3: Solve the system of linear equations

The system of linear equations is now:\[x + y = 12000\]\[0.14x - 0.06y = 680\]The first step in solving this system is to eliminate one of the variables. Multiply the first equation by 0.14 and subtract the second equation from the result. This gives:\[0.14x + 0.14y - (0.14x - 0.06y) = 0.2y = 1200\]With this, the money invested at a loss of 6% is found by dividing both sides of the equation by 0.2, which gives \(y = 6000\).

Step 4: Solve for the other variable

To find \(x\), which is the money invested at 14%, substitute \(y = 6000\) into the first equation from step 1. This gives \(x = 12000 - 6000 = 6000\). Thus, the money invested at 14% interest is \$6000.