Step-by-Step Solution

Solve the equation: $\frac{a}{ax-1} + \frac{b}{bx-1} = a + b \quad \left(x \ne \frac{1}{a}, \; \frac{1}{b}\right)$

Step-by-Step Solution

Step 1: Regroup terms to create a common factor

Transpose $a + b$ to the left by pairing $b$ with the first fraction and $a$ with the second: $\left(\frac{a}{ax - 1} - b\right) + \left(\frac{b}{bx - 1} - a\right) = 0$

Combine each bracket over its denominator: $\frac{a - b(ax - 1)}{ax - 1} + \frac{b - a(bx - 1)}{bx - 1} = 0$

$\frac{a + b - abx}{ax - 1} + \frac{a + b - abx}{bx - 1} = 0$

Step 2: Factor out the common numerator $(a + b - abx)$

$(a + b - abx) \left(\frac{1}{ax - 1} + \frac{1}{bx - 1}\right) = 0$

Step 3: Solve each branch

• Branch 1: $a + b - abx = 0 \implies abx = a + b \implies x = \frac{a + b}{ab}$

• Branch 2: $\frac{1}{ax - 1} + \frac{1}{bx - 1} = 0$ $(bx - 1) + (ax - 1) = 0$ $(a + b)x - 2 = 0 \implies x = \frac{2}{a + b}$

$\boxed{\text{Solution Set: } \left\{\frac{a + b}{ab}, \; \frac{2}{a + b}\right\}}$