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!