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The problem you provided asks:

**"Mitch has $11,040 part of which he invests at 5% interest and the rest at 8%. If his income for one year from the two investments was $732, how much did he invest at each rate?"**

Let's define the variables:

- Let $ x $ be the amount Mitch invested at 5%.
- The remaining amount $ 11,040 - x $ is what he invested at 8%.

The total interest earned is $732, so we set up the equation based on the interest earned from both investments:

\[
0.05x + 0.08(11,040 - x) = 732
\]

Now, let's solve the equation.

\[
0.05x + 0.08(11,040 - x) = 732
\]

Expanding the equation:

\[
0.05x + 883.2 - 0.08x = 732
\]

Combining like terms:

\[
-0.03x + 883.2 = 732
\]

Subtract 883.2 from both sides:

\[
-0.03x = 732 - 883.2
\]
\[
-0.03x = -151.2
\]

Now, divide by -0.03:

\[
x = \frac{-151.2}{-0.03} = 5040
\]

So, Mitch invested $5,040 at 5%.

The amount invested at 8% would be:

\[
11,040 - 5040 = 6000
\]

Mitch invested $5,040 at 5% and $6,000 at 8%.

Would you like a detailed breakdown of the steps? Here are 5 related questions:

1. How would the interest change if the rates were different, say 6% and 7%?
2. If Mitch wanted to earn $800 instead, how should he adjust his investments?
3. What would happen if Mitch only had $10,000 to invest?
4. How would the equation change if only part of the money was invested and the rest was saved without interest?
5. What is the impact of compounding interest on such investments over multiple years?

**Tip:** When solving word problems, it's helpful to define the variables clearly and write down the corresponding equation based on the relationships described in the problem.

Explore how to solve the investment interest problem where Mitch invests $11,040 at 5% and 8%, earning $732 in total. Learn the step-by-step process involving linear equations and interest calculations. Suitable for Grades 9-12.

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