https://file.aiquickdraw.com/mathbot/file/17258016374367sbc6kxy.jpg

The equation in the image is:

\[
\frac{x+2}{x-4} + \frac{x-4}{x+2} = \frac{17}{4}, \quad x \neq -2, 4
\]

Let's solve this step by step.

### Step 1: Simplify the left-hand side
The two fractions on the left-hand side have different denominators, so we need a common denominator. The common denominator of \( x-4 \) and \( x+2 \) is \( (x-4)(x+2) \). 

Rewriting each fraction:
\[
\frac{x+2}{x-4} = \frac{(x+2)^2}{(x-4)(x+2)}
\]
\[
\frac{x-4}{x+2} = \frac{(x-4)^2}{(x+2)(x-4)}
\]

So, the left-hand side becomes:
\[
\frac{(x+2)^2 + (x-4)^2}{(x-4)(x+2)}
\]

### Step 2: Simplify the numerator
Expand both squares:
\[
(x+2)^2 = x^2 + 4x + 4
\]
\[
(x-4)^2 = x^2 - 8x + 16
\]

Add these two expressions together:
\[
x^2 + 4x + 4 + x^2 - 8x + 16 = 2x^2 - 4x + 20
\]

So, the left-hand side is now:
\[
\frac{2x^2 - 4x + 20}{(x-4)(x+2)}
\]

### Step 3: Equate both sides
The equation becomes:
\[
\frac{2x^2 - 4x + 20}{(x-4)(x+2)} = \frac{17}{4}
\]

### Step 4: Cross-multiply
Cross-multiply to eliminate the denominators:
\[
4(2x^2 - 4x + 20) = 17(x-4)(x+2)
\]

### Step 5: Expand both sides
Expand the left-hand side:
\[
4(2x^2 - 4x + 20) = 8x^2 - 16x + 80
\]

Expand the right-hand side:
\[
(x-4)(x+2) = x^2 - 4x + 2x - 8 = x^2 - 2x - 8
\]
\[
17(x^2 - 2x - 8) = 17x^2 - 34x - 136
\]

### Step 6: Combine and simplify
The equation is now:
\[
8x^2 - 16x + 80 = 17x^2 - 34x - 136
\]

Move all terms to one side:
\[
0 = 17x^2 - 34x - 136 - 8x^2 + 16x - 80
\]
\[
0 = 9x^2 - 18x - 216
\]

### Step 7: Divide by 9
Simplify by dividing through by 9:
\[
0 = x^2 - 2x - 24
\]

### Step 8: Solve the quadratic equation
Solve the quadratic equation \( x^2 - 2x - 24 = 0 \) using the quadratic formula:
\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-24)}}{2(1)}
\]
\[
x = \frac{2 \pm \sqrt{4 + 96}}{2}
\]
\[
x = \frac{2 \pm \sqrt{100}}{2}
\]
\[
x = \frac{2 \pm 10}{2}
\]
\[
x = \frac{2 + 10}{2} = 6 \quad \text{or} \quad x = \frac{2 - 10}{2} = -4
\]

### Step 9: Check for extraneous solutions
Since \( x \neq 4 \) and \( x \neq -2 \), the solution \( x = -4 \) is valid, but \( x = 6 \) is the final answer.

Thus, the solution is \( \boxed{6} \).

Would you like any further explanation or details?

Here are some related questions you might find interesting:
1. How would the solution change if the right-hand side were a different fraction?
2. What happens if the denominators on the left-hand side were more complex?
3. Can we solve this using a different method, like graphing?
4. How would the solution process change if this were a system of equations?
5. What is the role of checking extraneous solutions in rational equations?

**Tip:** Always check if any potential solutions make the denominators zero to avoid extraneous solutions.

Solve the equation (x + 2) / (x - 4) + (x - 4) / (x + 2) = 17 / 4, x ≠ -2, 4

Learn how to solve a rational equation involving two fractions and a quadratic equation: (x + 2) / (x - 4) + (x - 4) / (x + 2) = 17 / 4. Follow a detailed step-by-step breakdown that includes simplifying, expanding, and solving for x. Suitable for high school students in Grades 10-12.

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