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Let's go through each of the equations in the image step by step and solve them. Here’s the content as it appears in the image:

**1)** \(-5x + 3 = -3x + 2\)

**2)** \((4x + 6)(3 - 7x) = 0\)

**3)** \((3x + 1)(1 - 6x) = 0\)

**4)** \(5x^2 - 4x = 0\)

**5)** \(x^2 - 9 = 16\)

**6)** \((x + 2)^2 = 9\)

**7)** \(\frac{x+3}{x-3} = \frac{x-3}{x+3}\)

**8)** \(\frac{x+3}{x-3} = \frac{2}{2 - x}\)

---

### Solutions:

#### **1)** \(-5x + 3 = -3x + 2\)
Move all the terms involving \(x\) to one side and constants to the other:
\[
-5x + 3 + 3x = 2
\]
\[
-2x + 3 = 2
\]
Subtract 3 from both sides:
\[
-2x = -1
\]
Divide by \(-2\):
\[
x = \frac{1}{2}
\]

---

#### **2)** \((4x + 6)(3 - 7x) = 0\)
This is a product of two terms equal to zero, so either:
\[
4x + 6 = 0 \quad \text{or} \quad 3 - 7x = 0
\]
Solve each equation:
1. \(4x + 6 = 0\):
   \[
   4x = -6 \quad \Rightarrow \quad x = -\frac{3}{2}
   \]
2. \(3 - 7x = 0\):
   \[
   7x = 3 \quad \Rightarrow \quad x = \frac{3}{7}
   \]

Thus, \(x = -\frac{3}{2}\) or \(x = \frac{3}{7}\).

---

#### **3)** \((3x + 1)(1 - 6x) = 0\)
Again, set each factor to zero:
\[
3x + 1 = 0 \quad \text{or} \quad 1 - 6x = 0
\]
Solve each equation:
1. \(3x + 1 = 0\):
   \[
   3x = -1 \quad \Rightarrow \quad x = -\frac{1}{3}
   \]
2. \(1 - 6x = 0\):
   \[
   6x = 1 \quad \Rightarrow \quad x = \frac{1}{6}
   \]

Thus, \(x = -\frac{1}{3}\) or \(x = \frac{1}{6}\).

---

#### **4)** \(5x^2 - 4x = 0\)
Factor out \(x\):
\[
x(5x - 4) = 0
\]
So, either:
\[
x = 0 \quad \text{or} \quad 5x - 4 = 0
\]
Solve for \(x\):
\[
5x = 4 \quad \Rightarrow \quad x = \frac{4}{5}
\]

Thus, \(x = 0\) or \(x = \frac{4}{5}\).

---

#### **5)** \(x^2 - 9 = 16\)
Add 9 to both sides:
\[
x^2 = 25
\]
Take the square root of both sides:
\[
x = \pm 5
\]

Thus, \(x = 5\) or \(x = -5\).

---

#### **6)** \((x + 2)^2 = 9\)
Take the square root of both sides:
\[
x + 2 = \pm 3
\]
Solve for \(x\):
1. \(x + 2 = 3\) gives \(x = 1\)
2. \(x + 2 = -3\) gives \(x = -5\)

Thus, \(x = 1\) or \(x = -5\).

---

#### **7)** \(\frac{x+3}{x-3} = \frac{x-3}{x+3}\)
Cross-multiply:
\[
(x+3)^2 = (x-3)^2
\]
Expand both sides:
\[
x^2 + 6x + 9 = x^2 - 6x + 9
\]
Simplify:
\[
6x = -6x
\]
This gives:
\[
12x = 0 \quad \Rightarrow \quad x = 0
\]

Thus, \(x = 0\).

---

#### **8)** \(\frac{x+3}{x-3} = \frac{2}{2 - x}\)
Cross-multiply:
\[
(x+3)(2-x) = 2(x-3)
\]
Expand both sides:
\[
2x + 6 - x^2 - 3x = 2x - 6
\]
Simplify:
\[
-x^2 - x + 6 = -6
\]
Move all terms to one side:
\[
x^2 + x - 12 = 0
\]
Solve this quadratic equation using the factorization method:
\[
(x - 3)(x + 4) = 0
\]
Thus, \(x = 3\) or \(x = -4\).

---

Would you like further details or explanations on any of these solutions?

Here are 5 related questions you might find interesting:
1. How can you solve quadratic equations using the quadratic formula?
2. What are the methods to solve polynomial equations?
3. What does it mean when a quadratic has complex solutions?
4. Can systems of nonlinear equations be solved similarly?
5. How does cross-multiplication work for solving rational equations?

**Tip**: Always check for restrictions in rational equations to avoid division by zero!

Solve the following equations: 1) -5x + 3 = -3x + 2 2) (4x + 6)(3 - 7x) = 0 3) (3x + 1)(1 - 6x) = 0 4) 5x^2 - 4x = 0 5) x^2 - 9 = 16 6) (x + 2)^2 = 9 7) (x+3)/(x-3) = (x-3)/(x+3) 8) (x+3)/(x-3) = 2/(2 - x)

This page walks through solving a variety of algebraic equations, including linear, quadratic, and rational equations, using methods like factoring and cross-multiplication. Suitable for high school students in Grades 9-11.

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