3((4 + x)/(x ^ 2 - 5x)) + 1/x = 7((0 - x)/(5x - x ^ 2))

Let's solve the equation:

\[
3\left(\frac{4 + x}{x^2 - 5x}\right) + \frac{1}{x} = 7\left(\frac{0 - x}{5x - x^2}\right)
\]

### Step 1: Simplify the Equation

First, notice that \(x^2 - 5x\) can be factored as:

\[
x^2 - 5x = x(x - 5)
\]

Similarly, \(5x - x^2\) can be written as:

\[
5x - x^2 = -x(x - 5)
\]

So the equation becomes:

\[
3\left(\frac{4 + x}{x(x - 5)}\right) + \frac{1}{x} = 7\left(\frac{-x}{-x(x - 5)}\right)
\]

Simplifying further:

\[
3\left(\frac{4 + x}{x(x - 5)}\right) + \frac{1}{x} = 7\left(\frac{1}{x(x - 5)}\right)
\]

### Step 2: Combine the Fractions

To combine the fractions, multiply each term by \(x(x - 5)\), the least common denominator (LCD):

\[
x(x - 5) \cdot 3\left(\frac{4 + x}{x(x - 5)}\right) + x(x - 5) \cdot \frac{1}{x} = x(x - 5) \cdot 7\left(\frac{1}{x(x - 5)}\right)
\]

This simplifies to:

\[
3(4 + x) + (x - 5) = 7
\]

### Step 3: Expand and Combine Like Terms

Expand the left side:

\[
12 + 3x + x - 5 = 7
\]

Combine like terms:

\[
4x + 7 = 7
\]

### Step 4: Solve for \(x\)

Subtract 7 from both sides:

\[
4x = 0
\]

Divide by 4:

\[
x = 0
\]

### Step 5: Check the Solution

Substitute \(x = 0\) back into the original equation to check for any undefined expressions:

In the original equation, the terms \(\frac{1}{x}\) and \(\frac{1}{x(x - 5)}\) are undefined when \(x = 0\). Thus, \(x = 0\) is not a valid solution.

### Conclusion:
The equation has **no valid solutions** because the only possible solution leads to an undefined expression.

Would you like further details or have any questions?

### Related Questions
1. How do you identify and handle undefined expressions in equations?
2. What are the rules for factoring quadratic expressions?
3. How can we solve rational equations systematically?
4. What is the importance of checking solutions in rational equations?
5. How do extraneous solutions arise in algebra?

### Tip
When solving equations involving fractions or rational expressions, always check if any potential solutions make the denominator zero, as these would invalidate the solution.

Explore how to solve a complex rational equation involving fractions and quadratic expressions step-by-step. This problem covers rational equations, simplifying fractions, and factoring quadratic expressions. Suitable for Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation