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I see the math problems in your image. Let's go through and solve them step by step, given that \( x = -12 \) and \( y = 3 \).

### 1. \( 10 \div \frac{(-1 \cdot x + 3y)}{8} \)
Substitute \( x = -12 \) and \( y = 3 \):

\[
10 \div \frac{-1(-12) + 3(3)}{8}
\]

Simplify:
\[
= 10 \div \frac{12 + 9}{8} = 10 \div \frac{21}{8} = 10 \times \frac{8}{21} = \frac{80}{21}
\]

The result is \( \frac{80}{21} \).

---

### 2. \( (3x + 2y)(x - 3y) \)
Substitute \( x = -12 \) and \( y = 3 \):

\[
(3(-12) + 2(3))((-12) - 3(3))
\]
\[
= (-36 + 6)(-12 - 9) = (-30)(-21) = 630
\]

The result is \( 630 \).

---

### 3. \( \frac{2}{3} \times (x - 6y) \)
Substitute \( x = -12 \) and \( y = 3 \):

\[
\frac{2}{3} \times (-12 - 6(3)) = \frac{2}{3} \times (-12 - 18) = \frac{2}{3} \times (-30)
\]
\[
= -20
\]

The result is \( -20 \).

---

### 4. \( x^2 + y^2 \)
Substitute \( x = -12 \) and \( y = 3 \):

\[
(-12)^2 + (3)^2 = 144 + 9 = 153
\]

The result is \( 153 \).

---

### 5. \( -x^2 + (-2y)^2 \)
Substitute \( x = -12 \) and \( y = 3 \):

\[
-(-12)^2 + (-2(3))^2 = -144 + (-6)^2 = -144 + 36 = -108
\]

The result is \( -108 \).

---

Would you like further details on any of these steps?

Here are some related questions you can explore:
1. How do you simplify complex fractions in algebraic expressions?
2. How does the distributive property work in the multiplication of binomials?
3. What is the general formula for calculating the difference of squares?
4. How do negative exponents affect the result in equations?
5. Can you explain how to expand and simplify polynomials?

**Tip:** Always substitute values carefully and remember to check signs when dealing with negatives!

Given x = -12 and y = 3, solve the following math expressions.

This page offers a step-by-step guide to solving algebraic expressions by substituting x = -12 and y = 3. Learn how to apply substitution, simplify binomials, and work with exponents. Suitable for Grades 7-9.

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