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Let's break down the math problems one by one:

---

### Problem #28:
Solve the system of equations:

\[
1.3x - 0.2y = 12
\]
\[
0.4x + 17y = 89
\]

**Step 1: Solve for $ x $ or $ y $**
First, multiply the first equation by 5 to eliminate the decimal:

\[
(5)(1.3x - 0.2y) = 5(12)
\]
\[
6.5x - y = 60
\]
So, the system becomes:
\[
6.5x - y = 60 \quad \text{(1)}
\]
\[
0.4x + 17y = 89 \quad \text{(2)}
\]

**Step 2: Solve the system using substitution or elimination.**
We can solve for $ y $ from equation (1):
\[
y = 6.5x - 60
\]

Substitute this expression for $ y $ in equation (2):
\[
0.4x + 17(6.5x - 60) = 89
\]
Expand and solve for $ x $:
\[
0.4x + 110.5x - 1020 = 89
\]
\[
110.9x - 1020 = 89
\]
\[
110.9x = 1109
\]
\[
x = \frac{1109}{110.9} = 10
\]

**Step 3: Solve for $ y $:**
Substitute $ x = 10 $ into the equation $ y = 6.5x - 60 $:
\[
y = 6.5(10) - 60 = 65 - 60 = 5
\]

**Solution:**
\[
x = 10, \quad y = 5
\]

---

### Problem #18: Loan Problem

Let the two loans be:
- Loan 1 at 2.8% interest = $ x $
- Loan 2 at 4.5% interest = $ 31,000 - x $

The total interest earned after one year is $1024.40. Using the simple interest formula $ I = P \times r \times t $ where $ t = 1 $ year, we can write the equation:

\[
0.028x + 0.045(31,000 - x) = 1024.40
\]

**Step 1: Expand the equation:**
\[
0.028x + 0.045(31,000) - 0.045x = 1024.40
\]
\[
0.028x + 1395 - 0.045x = 1024.40
\]

**Step 2: Combine like terms:**
\[
-0.017x + 1395 = 1024.40
\]
\[
-0.017x = 1024.40 - 1395
\]
\[
-0.017x = -370.60
\]
\[
x = \frac{-370.60}{-0.017} = 21,800
\]

**Step 3: Calculate the other loan:**
\[
Loan \ 2 = 31,000 - 21,800 = 9,200
\]

**Solution:**
- Loan 1: $21,800 at 2.8%
- Loan 2: $9,200 at 4.5%

---

### Problem #76:
Simplify:

\[
(10xy - 4x^2y^2 - 3y^3) - (-9x^2y^2 + 4y^3 - 7xy)
\]

**Step 1: Distribute the negative sign:**
\[
10xy - 4x^2y^2 - 3y^3 + 9x^2y^2 - 4y^3 + 7xy
\]

**Step 2: Combine like terms:**
- $ 10xy + 7xy = 17xy $
- $ -4x^2y^2 + 9x^2y^2 = 5x^2y^2 $
- $ -3y^3 - 4y^3 = -7y^3 $

**Solution:**
\[
17xy + 5x^2y^2 - 7y^3
\]

---

### Problem #80:
Multiply:

\[
(2x - y)(2x + y)(4x^2 - y^2)
\]

**Step 1: Multiply the first two factors using the difference of squares:**
\[
(2x - y)(2x + y) = 4x^2 - y^2
\]

**Step 2: Now multiply the result with the third factor:**
\[
(4x^2 - y^2)(4x^2 - y^2) = (4x^2 - y^2)^2
\]

**Solution:**
\[
(4x^2 - y^2)^2 = 16x^4 - 8x^2y^2 + y^4
\]

---

Let me know if you'd like more details or explanations on any of these problems! Here are some related questions you could explore:

1. How can we verify the solution of a system of linear equations graphically?
2. What other methods could be used to solve the loan interest problem?
3. Why is the difference of squares useful in simplifying certain types of polynomials?
4. How would the loan problem change if compounded interest were involved instead of simple interest?
5. What are the real-life applications of systems of linear equations?

**Tip:** When solving a system of equations, it's often useful to double-check your solution by plugging the values of the variables back into the original equations.

Solve the following problems:
- Problem #28: Solve the system of equations: 1.3x - 0.2y = 12, 0.4x + 17y = 89
- Problem #18: Cole's two student loans totaled $31,000. One loan was at 2.8% interest and the other at 4.5%. After one year, the total interest was $1024.40. Find the amount of each loan.
- Problem #76: Subtract: (10xy - 4x^2y^2 - 3y^3) - (-9x^2y^2 + 4y^3 - 7xy)
- Problem #80: Multiply: (2x - y)(2x + y)(4x^2 - y^2)

Explore solutions for various algebraic problems including solving systems of equations, calculating simple interest on student loans, and simplifying polynomial expressions. This page provides detailed step-by-step solutions for high school algebra students in grades 9-11, covering concepts like linear equations, interest calculations, and polynomial operations.

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