To solve the given equation:

$3\left(\frac{4 + x}{x^2 - 5x}\right) + \frac{1}{x} = 7\left(\frac{8 - x}{5x - x^2}\right),$

let's follow these steps:

1. Rewrite the equation in a more manageable form.

We can rewrite the equation as:

$3 \cdot \frac{4 + x}{x(x - 5)} + \frac{1}{x} = 7 \cdot \frac{8 - x}{x(5 - x)}.$

2. Notice the denominator similarities:

The denominators $x^2 - 5x$ and $5x - x^2$ are negatives of each other:

$x^2 - 5x = x(x - 5), \quad \text{and} \quad 5x - x^2 = x(5 - x) = -x(x - 5).$

Thus, $5x - x^2 = - (x^2 - 5x)$. The equation can now be rewritten as:

$3 \cdot \frac{4 + x}{x(x - 5)} + \frac{1}{x} = 7 \cdot \frac{8 - x}{-x(x - 5)},$

which simplifies further to:

$3 \cdot \frac{4 + x}{x(x - 5)} + \frac{1}{x} = -7 \cdot \frac{8 - x}{x(x - 5)}.$

3. Combine like terms over a common denominator:

The common denominator for all terms is $x(x - 5)$:

$\frac{3(4 + x) + (x - 5)}{x(x - 5)} = -\frac{7(8 - x)}{x(x - 5)}.$

This simplifies to:

$\frac{3(4 + x) + (x - 5)}{x(x - 5)} = -\frac{7(8 - x)}{x(x - 5)}.$

4. Cross multiply and simplify:

Since the denominators are the same, we can equate the numerators:

$3(4 + x) + (x - 5) = -7(8 - x).$

Expand both sides:

$12 + 3x + x - 5 = -56 + 7x.$

Simplify:

$4x + 7 = -56 + 7x.$

5. Solve for $x$:

Combine like terms:

$7 + 56 = 7x - 4x,$

$63 = 3x.$

Divide by 3:

$x = 21.$

Solution

The solution to the equation is $x = 21$.

Would you like more details on any specific step, or do you have any questions?

Related Questions:

1. How do you handle rational equations with multiple fractions?
2. What is the process to find a common denominator for complex rational expressions?
3. How can you determine if a solution to a rational equation is valid or results in an undefined expression?
4. How do you solve equations involving negative signs in the denominators?
5. What methods can be used to check the solution of a rational equation?

Tip:

Always check the final solution in the original equation to ensure there is no division by zero or other undefined operations.