A bank loaned out ​$12 ,000​, part of it at the rate of 9 % per year and the rest at 19 % per year. If the interest received in one year totaled ​$1500​, how much was loaned at 9 % ?                                      How much of the ​$12 ,000 did the bank loan out at 9 % ?

To solve this problem, let's use a system of equations approach. Let's denote:

- $ x $ as the amount loaned at 9%.
- $ y $ as the amount loaned at 19%.

We are given two pieces of information:

1. The total amount loaned out is $12,000.
   \[
   x + y = 12,000
   \]

2. The total interest received in one year is $1,500.
   \[
   0.09x + 0.19y = 1,500
   \]

### Step 1: Express $ y $ in terms of $ x $ using the first equation.
\[
y = 12,000 - x
\]

### Step 2: Substitute $ y $ into the second equation.
\[
0.09x + 0.19(12,000 - x) = 1,500
\]

### Step 3: Simplify and solve for $ x $.
\[
0.09x + 0.19(12,000) - 0.19x = 1,500
\]
\[
0.09x + 2,280 - 0.19x = 1,500
\]
\[
-0.10x + 2,280 = 1,500
\]
\[
-0.10x = 1,500 - 2,280
\]
\[
-0.10x = -780
\]
\[
x = \frac{-780}{-0.10} = 7,800
\]

### Step 4: Find $ y $.
\[
y = 12,000 - 7,800 = 4,200
\]

### Conclusion
- The bank loaned out **$7,800** at 9%.
- The remaining **$4,200** was loaned at 19%.

Would you like further details or have any questions?

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Here are five related questions to consider:

1. How much interest did the bank earn from the loan at 9%?
2. What is the total interest earned from the loan at 19%?
3. If the bank wanted to increase the total interest to $1,800, how would the amounts at each rate need to change?
4. What would the total interest be if the entire $12,000 was loaned at 9%?
5. What would the total interest be if the entire $12,000 was loaned at 19%?

**Tip:** When dealing with percentages in problems like this, converting the percentage to its decimal form (e.g., 9% to 0.09) makes the calculations much easier.

Learn how to calculate the distribution of a $12,000 loan between 9% and 19% interest rates, and how to compute the total interest earned. Detailed step-by-step solution provided.

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