https://file.aiquickdraw.com/mathbot/file/1726546783580xb1k7gz4.jpg

The given equation from the image is:

\[
\frac{7}{2x+1} - \frac{8x}{2x-1} = -4
\]

### Let's solve this step by step:

1. **Find a common denominator**:  
   The common denominator for the two fractions is \((2x+1)(2x-1)\).

   So, rewrite both fractions with this common denominator:
   
   \[
   \frac{7}{2x+1} = \frac{7(2x-1)}{(2x+1)(2x-1)}
   \]
   and
   \[
   \frac{8x}{2x-1} = \frac{8x(2x+1)}{(2x+1)(2x-1)}.
   \]

2. **Combine the two fractions**:  
   Now, combine the fractions over the common denominator:
   
   \[
   \frac{7(2x-1) - 8x(2x+1)}{(2x+1)(2x-1)} = -4.
   \]

3. **Expand the numerators**:
   
   Expanding \(7(2x-1)\):
   \[
   7(2x-1) = 14x - 7.
   \]
   
   Expanding \(8x(2x+1)\):
   \[
   8x(2x+1) = 16x^2 + 8x.
   \]
   
   So the equation becomes:
   \[
   \frac{(14x - 7) - (16x^2 + 8x)}{(2x+1)(2x-1)} = -4.
   \]

4. **Simplify the numerator**:
   
   Simplifying the numerator:
   \[
   14x - 7 - 16x^2 - 8x = -16x^2 + 6x - 7.
   \]

   The equation is now:
   \[
   \frac{-16x^2 + 6x - 7}{(2x+1)(2x-1)} = -4.
   \]

5. **Multiply both sides by \((2x+1)(2x-1)\)**:
   
   To eliminate the denominator, multiply both sides by \((2x+1)(2x-1)\):
   \[
   -16x^2 + 6x - 7 = -4(2x+1)(2x-1).
   \]

6. **Expand the right-hand side**:
   
   Expanding \(-4(2x+1)(2x-1)\) using the difference of squares:
   \[
   (2x+1)(2x-1) = 4x^2 - 1,
   \]
   so:
   \[
   -4(4x^2 - 1) = -16x^2 + 4.
   \]

   The equation becomes:
   \[
   -16x^2 + 6x - 7 = -16x^2 + 4.
   \]

7. **Simplify the equation**:
   
   Cancel out \(-16x^2\) from both sides:
   \[
   6x - 7 = 4.
   \]

8. **Solve for \(x\)**:
   
   Add 7 to both sides:
   \[
   6x = 11.
   \]
   
   Divide by 6:
   \[
   x = \frac{11}{6}.
   \]

Thus, the solution is:
\[
x = \frac{11}{6}.
\]

Would you like more details or have any questions? Here are some related problems:

1. How do you solve similar rational equations involving fractions?
2. What are the steps to find a common denominator in rational expressions?
3. How can you check the solution to a rational equation?
4. What happens when the denominator equals zero in a rational equation?
5. How can you solve an equation where the variable is in both the numerator and the denominator?

**Tip:** When solving rational equations, always check for any extraneous solutions by substituting your solution back into the original equation to ensure it doesn't make any denominator zero.

Solve the equation: \[ \frac{7}{2x+1} - \frac{8x}{2x-1} = -4 \]

Learn how to solve the rational equation \(\frac{7}{2x+1} - \frac{8x}{2x-1} = -4\) with a detailed step-by-step solution. This problem covers key algebraic concepts, including finding a common denominator and simplifying rational expressions. Suitable for students in Grades 10-12.

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