https://file.aiquickdraw.com/mathbot/file/1725848532785m794slyi.jpg

I've displayed the image. I'll now extract the mathematical problems and provide solutions step by step.

Let's analyze the equations:

1. \( \frac{x}{7} = \frac{7}{9} \)
2. \( 2x - 15 = \frac{-5}{2} \)
3. \( \frac{2}{x-3} + \frac{5}{x+3} = 0 \)
4. \( \frac{x}{x-1} - \frac{1}{x+1} = 0 \)
5. \( \frac{2}{x-2} + \frac{x+2}{x-3} = \frac{20}{x^2 - 5x + 6} \)

Now, I'll go through each equation step by step.

### 1. \( \frac{x}{7} = \frac{7}{9} \)

Multiply both sides by 7 to isolate \( x \):

\[
x = 7 \times \frac{7}{9} = \frac{49}{9}
\]

### 2. \( 2x - 15 = \frac{-5}{2} \)

Add 15 to both sides:

\[
2x = \frac{-5}{2} + 15 = \frac{-5 + 30}{2} = \frac{25}{2}
\]

Now, divide by 2:

\[
x = \frac{25}{2} \times \frac{1}{2} = \frac{25}{4}
\]

### 3. \( \frac{2}{x-3} + \frac{5}{x+3} = 0 \)

Find a common denominator and combine:

\[
\frac{2(x+3) + 5(x-3)}{(x-3)(x+3)} = 0
\]

Simplify the numerator:

\[
2(x+3) + 5(x-3) = 2x + 6 + 5x - 15 = 7x - 9
\]

So, the equation becomes:

\[
\frac{7x - 9}{(x-3)(x+3)} = 0
\]

For this to be zero, the numerator must be zero:

\[
7x - 9 = 0
\]

Solve for \( x \):

\[
x = \frac{9}{7}
\]

### 4. \( \frac{x}{x-1} - \frac{1}{x+1} = 0 \)

Find a common denominator:

\[
\frac{x(x+1) - (x-1)}{(x-1)(x+1)} = 0
\]

Simplify the numerator:

\[
x(x+1) - (x-1) = x^2 + x - x + 1 = x^2 + 1
\]

Thus, the equation becomes:

\[
\frac{x^2 + 1}{(x-1)(x+1)} = 0
\]

Since the numerator \( x^2 + 1 \) is never zero for real numbers, there is no real solution.

### 5. \( \frac{2}{x-2} + \frac{x+2}{x-3} = \frac{20}{x^2 - 5x + 6} \)

Factor the denominator on the right-hand side:

\[
x^2 - 5x + 6 = (x-2)(x-3)
\]

Now, rewrite the equation:

\[
\frac{2}{x-2} + \frac{x+2}{x-3} = \frac{20}{(x-2)(x-3)}
\]

Multiply both sides by \( (x-2)(x-3) \):

\[
2(x-3) + (x+2)(x-2) = 20
\]

Expand both sides:

\[
2x - 6 + (x^2 - 4) = 20
\]

Combine terms:

\[
x^2 + 2x - 10 = 20
\]

Subtract 20 from both sides:

\[
x^2 + 2x - 30 = 0
\]

Solve this quadratic using the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-30)}}{2(1)}
\]

\[
x = \frac{-2 \pm \sqrt{4 + 120}}{2} = \frac{-2 \pm \sqrt{124}}{2} = \frac{-2 \pm 2\sqrt{31}}{2}
\]

\[
x = -1 \pm \sqrt{31}
\]

So, the solutions are:

\[
x = -1 + \sqrt{31}, \quad x = -1 - \sqrt{31}
\]

Would you like more details on any of these steps? Here are 5 questions to extend the understanding:

1. How do we deal with rational equations that involve zero denominators?
2. Can you explain how the quadratic formula works in detail?
3. What happens when there is no real solution to a rational equation?
4. Why is it important to factor complex denominators in rational equations?
5. Can we always use cross-multiplication in rational equations?

**Tip:** Always check for extraneous solutions by substituting the found values back into the original equation.

Explore step-by-step solutions to rational equations, including how to solve algebraic problems like quadratic equations using key formulas and theorems. This guide is ideal for students in Grades 8-10 and provides detailed explanations of the quadratic formula, cross-multiplication, and factoring techniques.

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