To solve the equation \(\frac{6}{b} - \frac{ac - x^2}{ax} = \frac{7x - bc}{bx}\) for \(x\), let's follow these steps:

1. **Clear the fractions** by finding a common denominator.
2. **Simplify** the equation.
3. **Isolate** \(x\) and solve.

### Step-by-Step Solution:

1. **Clear the fractions**:
   Multiply every term by the common denominator, which is \(bax\).

   \[
   bax \left(\frac{6}{b}\right) - bax \left(\frac{ac - x^2}{ax}\right) = bax \left(\frac{7x - bc}{bx}\right)
   \]

2. Simplify each term:

   \[
   \frac{bax \cdot 6}{b} - \frac{bax \cdot (ac - x^2)}{ax} = \frac{bax \cdot (7x - bc)}{bx}
   \]

   Simplifying the fractions:

   \[
   6ax - b(ac - x^2) = a(7x - bc)
   \]

3. **Distribute** and **simplify**:

   \[
   6ax - bac + bx^2 = 7ax - abc
   \]

4. Combine like terms:

   \[
   6ax - bac + bx^2 = 7ax - abc
   \]

   Move all terms to one side to set the equation to zero:

   \[
   bx^2 + 6ax - bac - 7ax + abc = 0
   \]

   Combine like terms:

   \[
   bx^2 - ax + abc - bac = 0
   \]

   Notice that \(abc - bac = 0\):

   \[
   bx^2 - ax = 0
   \]

5. **Factor** out \(x\):

   \[
   x(bx - a) = 0
   \]

   This gives us two solutions:

   \[
   x = 0 \quad \text{or} \quad bx - a = 0
   \]

6. Solve for \(x\):

   \[
   x = 0 \quad \text{or} \quad bx = a \implies x = \frac{a}{b}
   \]

### Final Answer:

The solutions for \(x\) are:

\[
x = 0 \quad \text{or} \quad x = \frac{a}{b}
\]

Would you like more details or have any questions?

Here are 5 questions you might be interested in:

1. How do you solve quadratic equations using the quadratic formula?
2. Can you explain the concept of polynomial factorization?
3. What is the method for solving systems of linear equations?
4. How do you find the derivative of a function?
5. What are the applications of integrals in real life?

**Tip**: Always check your solutions by substituting them back into the original equation to verify their correctness.

Learn how to solve the equation (6/b) - [(ac - x^2)/ax] = (7x - bc)/bx for x step-by-step. This problem involves algebraic manipulation of fractions and terms, suitable for advanced high school students. Gain insights into clearing fractions and simplifying equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation