91Ó°ÊÓ

Equations in algebra are like puzzles. They help us find unknown values by setting up relationships between different quantities. In investment problems like this, equations can be used to represent how money is distributed among different interest rates.

To solve the problem, we need to set up equations based on the scenario given to us. We use three variables:

• \(X\) for the amount invested at 10%
• \(Y\) for the amount at 12%
• \(Z\) for the amount at 15%

The first equation comes from the total amount of money invested: \(X + Y + Z = 17000\).

The second equation is about the income generated from these investments. Each part of the money earns interest according to its rate, resulting in \(0.10 \times X + 0.12 \times Y + 0.15 \times Z = 2110\).

Finally, we have a relationship between the amounts: \(Y = X + Z - 1000\). This equation links \(Y\) to \(X\) and \(Z\) directly, providing another layer of information. By solving these equations together, we can uncover the unique amounts invested at each rate.

Interest rates represent how much profit an investment generates over time. In this problem, the percentages tell us how money grows over one year.

Each part of the investment grows differently:

• 10% means for every dollar invested, you earn an extra \(0.10.
• 12% results in an extra \)0.12 per dollar.
• 15% gives the highest return of \(0.15 for every dollar.

Interest calculations are straightforward. Multiply the initial investment by its respective rate to find the income from that portion. For example, if \(X\) dollars are invested at 10%, the income is \(0.10 \times X\). Similarly, for other rates, it follows the same procedure.

Understanding these percentages is crucial. They determine how each fraction of the total investment contributes to the overall earnings, aligning the totals with the given \)2110.

Solving investment problems involves a clear logical progression through several steps. First, we define variables to represent the unknowns, setting up equations as described before.

Next, we simplify these equations to make the calculations easier. Substituting the equation \(Y = X + Z - 1000\) into others reduces complexity. This substitution reveals new equations like \(X + Z = 9000\), streamlining the solving process.

Solving algebraic problems often requires focusing on one variable. Here, we solve for \(X\) by relating it to \(Z\). Substituting \(Z = 9000 - X\) into another equation allows us to simplify further, finding \(X = \\(5000\).

Finally, knowing \(X\), we calculate \(Z\) from the simplified expression \(X + Z = 9000\), revealing \(Z = \\)4000\). The final step fills in the value of \(Y\), completing the puzzle with \(Y = \$7000\).

Breaking the problem into steps with checks along the way ensures accuracy, making the task manageable and structured.