Solve : `(a)/(ax-1)+(b)/(bx-1)=a+b`, where `a+b ne 0, ab ne 0`.

Answer

To solve the equation

\[ \frac{a}{ax-1} + \frac{b}{bx-1} = a + b \]

where \( a + b \neq 0 \) and \( ab \neq 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:

\[ \frac{a}{ax-1} + \frac{b}{bx-1} = a + b \]

### Step 2: Combine the fractions on the left side
To combine the fractions, we need a common denominator, which is \((ax - 1)(bx - 1)\):

\[ \frac{a(bx - 1) + b(ax - 1)}{(ax - 1)(bx - 1)} = a + b \]

### Step 3: Expand the numerator
Now we expand the numerator:

\[ a(bx - 1) + b(ax - 1) = abx - a + abx - b = 2abx - (a + b) \]

So, we have:

\[ \frac{2abx - (a + b)}{(ax - 1)(bx - 1)} = a + b \]

### Step 4: Cross-multiply
Cross-multiplying gives:

\[ 2abx - (a + b) = (a + b)(ax - 1)(bx - 1) \]

### Step 5: Expand the right side
Now we expand the right side:

\[ (a + b)(ax - 1)(bx - 1) = (a + b)(abx^2 - (a + b)x + 1) \]

### Step 6: Set the equation to zero
Now we rearrange the equation:

\[ 2abx - (a + b) - (a + b)(abx^2 - (a + b)x + 1) = 0 \]

### Step 7: Collect like terms
This will give us a quadratic equation in terms of \(x\):

\[ -ab(a + b)x^2 + (2ab + (a + b)^2)x - (a + b) = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can find the values of \(x\).

Here, \(A = -ab(a + b)\), \(B = 2ab + (a + b)^2\), and \(C = -(a + b)\).

### Step 9: Calculate the discriminant
Calculate the discriminant \(D = B^2 - 4AC\):

\[ D = (2ab + (a + b)^2)^2 - 4(-ab(a + b))(- (a + b)) \]

### Step 10: Find the roots
Finally, we can find the roots using the quadratic formula.

### Final Answer
The solutions for \(x\) will be:

\[ x_1 = \frac{2}{a + b}, \quad x_2 = \frac{a + b}{ab} \]