We are given the equation:

\[
\frac{3 - 2x}{x + 3} - \frac{2x + 1}{x - 4} = \frac{5x - 29}{x - 4}
\]

### Step 1: Get a common denominator on the left-hand side.
The left-hand side has two fractions with different denominators, \(x + 3\) and \(x - 4\). To combine these fractions, we need a common denominator. The common denominator would be \((x + 3)(x - 4)\).

So, we rewrite each term:
\[
\frac{3 - 2x}{x + 3} = \frac{(3 - 2x)(x - 4)}{(x + 3)(x - 4)}
\]
\[
\frac{2x + 1}{x - 4} = \frac{(2x + 1)(x + 3)}{(x + 3)(x - 4)}
\]

### Step 2: Expand the numerators.
For \(\frac{(3 - 2x)(x - 4)}{(x + 3)(x - 4)}\):
\[
(3 - 2x)(x - 4) = 3x - 12 - 2x^2 + 8x = -2x^2 + 11x - 12
\]

For \(\frac{(2x + 1)(x + 3)}{(x + 3)(x - 4)}\):
\[
(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3
\]

Now, the equation becomes:
\[
\frac{-2x^2 + 11x - 12}{(x + 3)(x - 4)} - \frac{2x^2 + 7x + 3}{(x + 3)(x - 4)} = \frac{5x - 29}{x - 4}
\]

### Step 3: Combine the fractions on the left-hand side.
Since both fractions have the same denominator \((x + 3)(x - 4)\), we can subtract the numerators:

\[
\frac{(-2x^2 + 11x - 12) - (2x^2 + 7x + 3)}{(x + 3)(x - 4)}
\]

Simplify the numerator:
\[
-2x^2 + 11x - 12 - 2x^2 - 7x - 3 = -4x^2 + 4x - 15
\]

So the equation now becomes:
\[
\frac{-4x^2 + 4x - 15}{(x + 3)(x - 4)} = \frac{5x - 29}{x - 4}
\]

### Step 4: Eliminate the denominators.
We can multiply both sides of the equation by \((x + 3)(x - 4)\) to eliminate the denominators:
\[
-4x^2 + 4x - 15 = (5x - 29)(x + 3)
\]

### Step 5: Expand the right-hand side.
Expand \((5x - 29)(x + 3)\):
\[
(5x - 29)(x + 3) = 5x^2 + 15x - 29x - 87 = 5x^2 - 14x - 87
\]

Now the equation is:
\[
-4x^2 + 4x - 15 = 5x^2 - 14x - 87
\]

### Step 6: Bring all terms to one side.
Move all terms to the left-hand side:
\[
-4x^2 + 4x - 15 - 5x^2 + 14x + 87 = 0
\]

Simplify:
\[
-9x^2 + 18x + 72 = 0
\]

### Step 7: Divide through by -9.
Divide the entire equation by -9 to simplify:
\[
x^2 - 2x - 8 = 0
\]

### Step 8: Solve the quadratic equation.
Factor the quadratic equation:
\[
x^2 - 2x - 8 = (x - 4)(x + 2) = 0
\]

Thus, \(x = 4\) or \(x = -2\).

### Step 9: Check for extraneous solutions.
We must check for extraneous solutions by substituting \(x = 4\) and \(x = -2\) into the original equation.

- If \(x = 4\), the original equation has a denominator of zero because of the term \(x - 4\), which makes this an invalid solution.
- If \(x = -2\), the equation is valid.

### Final Answer:
The solution is:
\[
x = -2
\]

Would you like further details or have any questions?

Here are 5 related questions to consider:
1. How can we solve a quadratic equation using the quadratic formula?
2. What are extraneous solutions and why do they occur?
3. How do we find a common denominator when combining fractions?
4. What are some methods to factor quadratic equations?
5. How do we expand binomials when simplifying equations?

**Tip:** Always check for extraneous solutions, especially when dealing with rational equations, to avoid including invalid solutions.

3 - 2x/x + 3 - 2x + 1/x - 4 = 5x - 29/x - 4

Learn how to solve a complex rational equation involving quadratic terms: 3 - 2x/(x+3) - (2x+1)/(x-4) = (5x-29)/(x-4). This step-by-step guide covers algebraic methods, such as finding a common denominator and solving a quadratic equation. Suitable for students in Grades 9-11.

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