Solve Rational equatition 3-2x/x+3 - 2x+1/x-4=5x-29/x-4

We are given the rational equation:

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

### Step 1: Combine the fractions on the left-hand side
We need to combine the two fractions on the left-hand side. The denominators are \( x + 3 \) and \( x - 4 \). The least common denominator (LCD) is \( (x + 3)(x - 4) \).

To combine the fractions, we will rewrite them with the LCD as the denominator.

\[
\frac{3 - 2x}{x + 3} = \frac{(3 - 2x)(x - 4)}{(x + 3)(x - 4)}
\]

\[
\frac{2x + 1}{x - 4} = \frac{(2x + 1)(x + 3)}{(x + 3)(x - 4)}
\]

Now subtract the second fraction from the first:

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

### Step 2: Expand the numerators
Let's expand the numerators.

For \( (3 - 2x)(x - 4) \):
\[
(3 - 2x)(x - 4) = 3x - 12 - 2x^2 + 8x = -2x^2 + 11x - 12
\]

For \( (2x + 1)(x + 3) \):
\[
(2x + 1)(x + 3) = 2x^2 + 6x + x + 3 = 2x^2 + 7x + 3
\]

### Step 3: Subtract the expanded expressions
Now subtract the second numerator from the first:
\[
(-2x^2 + 11x - 12) - (2x^2 + 7x + 3) = -2x^2 + 11x - 12 - 2x^2 - 7x - 3 = -4x^2 + 4x - 15
\]

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

### Step 4: Eliminate the denominator
We can eliminate the denominator by multiplying both sides of the equation by \( (x + 3)(x - 4) \). This simplifies to:

\[
-4x^2 + 4x - 15 = (5x - 29)(x + 3)
\]

### Step 5: Expand the right-hand side
Now expand \( (5x - 29)(x + 3) \):

\[
(5x - 29)(x + 3) = 5x^2 + 15x - 29x - 87 = 5x^2 - 14x - 87
\]

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

### Step 6: Move all terms to one side
Move all terms to one side of the equation:

\[
-4x^2 + 4x - 15 - 5x^2 + 14x + 87 = 0
\]

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

### Step 7: Divide by -9
Divide the entire equation by -9:

\[
x^2 - 2x - 8 = 0
\]

### Step 8: Solve the quadratic equation
Now solve \( x^2 - 2x - 8 = 0 \) using factoring:

\[
(x - 4)(x + 2) = 0
\]

So, the solutions are:

\[
x = 4 \quad \text{or} \quad x = -2
\]

### Step 9: Check for extraneous solutions
We need to check for extraneous solutions by substituting these values into the original equation.

- For \( x = 4 \): The denominator in the original equation becomes 0, so \( x = 4 \) is not a valid solution.
- For \( x = -2 \): This is a valid solution.

### Final Answer:
The solution is \( \boxed{x = -2} \).

Would you like more details or have any questions about this process? Here are some related questions:

1. How do you check for extraneous solutions in rational equations?
2. What are the steps to combine fractions with different denominators?
3. How can you simplify expressions involving quadratic terms?
4. What methods can be used to solve quadratic equations?
5. How do you identify the least common denominator in rational equations?

**Tip:** Always check if a solution makes the denominator zero, as it would invalidate the solution in rational equations.

Solve Rational equation 3-2x/x+3 - 2x+1/x-4=5x-29/x-4

Explore a step-by-step solution to the rational equation 3-2x/x+3 - 2x+1/x-4 = 5x-29/x-4. This problem covers algebraic concepts such as rational equations, combining fractions, and solving quadratic equations. Suitable for students in Grades 9-12.

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