We are given the equation:

$\frac{1}{4x - 3} = \frac{5}{4x^2 - 11x + 6} - \frac{x+5}{4x^2 - 11x + 6}$

Step 1: Simplify the right-hand side

The right-hand side has a common denominator, so we can combine the two fractions:

$\frac{5}{4x^2 - 11x + 6} - \frac{x+5}{4x^2 - 11x + 6} = \frac{5 - (x + 5)}{4x^2 - 11x + 6}$

Simplifying the numerator:

$5 - (x + 5) = 5 - x - 5 = -x$

Thus, the right-hand side simplifies to:

$\frac{-x}{4x^2 - 11x + 6}$

Now, the equation becomes:

$\frac{1}{4x - 3} = \frac{-x}{4x^2 - 11x + 6}$

Step 2: Factor the quadratic expression

The denominator $4x^2 - 11x + 6$ can be factored as:

$4x^2 - 11x + 6 = (4x - 3)(x - 2)$

Now, the equation is:

$\frac{1}{4x - 3} = \frac{-x}{(4x - 3)(x - 2)}$

Step 3: Cross-multiply

To eliminate the fractions, we can cross-multiply:

$1 \cdot (4x - 3)(x - 2) = (4x - 3) \cdot (-x)$

Step 4: Cancel common terms

Since $4x - 3$ appears on both sides of the equation, we can cancel it (as long as $4x - 3 \neq 0$, or $x \neq \frac{3}{4}$):

$x - 2 = -x$

Step 5: Solve for $x$

Now, solve the equation:

$x - 2 = -x$

Add $x$ to both sides:

$2x - 2 = 0$

Add 2 to both sides:

$2x = 2$

Divide by 2:

$x = 1$

Step 6: Check for extraneous solutions

We need to check if $x = 1$ is a valid solution by plugging it back into the original equation.

At $x = 1$:

• Left-hand side: $\frac{1}{4(1) - 3} = \frac{1}{1} = 1$
• Right-hand side: $\frac{5}{4(1)^2 - 11(1) + 6} - \frac{1 + 5}{4(1)^2 - 11(1) + 6} = \frac{5}{-1} - \frac{6}{-1} = -5 + 6 = 1$

Both sides are equal, so $x = 1$ is a valid solution.

Final Answer:

The solution is $x = 1$.

Would you like more details or further clarification?

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Here are 5 follow-up questions related to this problem:

1. Can you factor quadratic expressions like $4x^2 - 11x + 6$ easily?
2. Why is cross-multiplying a valid strategy when solving rational equations?
3. What conditions lead to extraneous solutions in rational equations?
4. How does factoring help simplify complex rational expressions?
5. What happens if you forget to check for restrictions on the variables?

Tip: Always check your solutions for rational equations to avoid extraneous solutions caused by dividing by zero.