Solve for x: `a/(ax-1)+b/(bx-1)=a+b; x!= 1/a, 1/b`

Answer

To solve the equation \( \frac{a}{ax - 1} + \frac{b}{bx - 1} = a + b \), we will follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \frac{a}{ax - 1} + \frac{b}{bx - 1} = a + b \]

### Step 2: Move one term to the right side
Rearranging gives:
\[ \frac{a}{ax - 1} = a + b - \frac{b}{bx - 1} \]

### Step 3: Find a common denominator
The common denominator for the left side is \( (ax - 1)(bx - 1) \). Rewrite the left-hand side:
\[ \frac{a(bx - 1) + b(ax - 1)}{(ax - 1)(bx - 1)} = a + b \]

### Step 4: Simplify the numerator
Expanding the numerator:
\[ abx - a + abx - b = 2abx - (a + b) \]
So we have:
\[ \frac{2abx - (a + b)}{(ax - 1)(bx - 1)} = a + b \]

### Step 5: Cross-multiply
Cross-multiplying gives:
\[ 2abx - (a + b) = (a + b)(ax - 1)(bx - 1) \]

### Step 6: Expand the right-hand side
Expanding the right-hand side:
\[ (a + b)(abx^2 - (a + b)x + 1) \]

### Step 7: Set the equation to zero
Rearranging gives us:
\[ 0 = (a + b)(abx^2 - (a + b)x + 1) - 2abx + (a + b) \]

### Step 8: Collect like terms
Combine like terms to form a quadratic equation in terms of \( x \):
\[ (2ab - (a + b)(a + b))x^2 + (2(a + b) - (a + b)(a + b))x + (a + b) = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a, b, c \) are the coefficients from the quadratic equation.

### Step 10: Find the solutions
After solving, we find:
\[ x = \frac{a + b}{ab} \quad \text{and} \quad x = \frac{2}{a + b} \]

### Final Answer
The solutions for \( x \) are:
\[ x = \frac{a + b}{ab} \quad \text{and} \quad x = \frac{2}{a + b} \]

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