Solve for x: a/(ax-1) + b/(bx-1) = a+b

Steps to Solve

1. Find a common denominator and combine the fractions on the left-hand side The common denominator is $(ax-1)(bx-1)$. Multiply the first term by $\frac{bx-1}{bx-1}$ and the second term by $\frac{ax-1}{ax-1}$: $$ \frac{a(bx-1)}{(ax-1)(bx-1)} + \frac{b(ax-1)}{(ax-1)(bx-1)} = a+b $$

Combine the fractions: $$ \frac{a(bx-1) + b(ax-1)}{(ax-1)(bx-1)} = a+b $$

2. Simplify the numerator on the left-hand side Expand the terms in the numerator: $$ \frac{abx - a + abx - b}{(ax-1)(bx-1)} = a+b $$ Combine like terms: $$ \frac{2abx - a - b}{(ax-1)(bx-1)} = a+b $$

3. Multiply both sides by the denominator $(ax-1)(bx-1)$ $$ 2abx - a - b = (a+b)(ax-1)(bx-1) $$

4. Expand the right-hand side Expand $(ax-1)(bx-1)$ first: $$ (ax-1)(bx-1) = abx^2 - ax - bx + 1 $$ Now multiply $(a+b)$ by the expanded form: $$ (a+b)(abx^2 - ax - bx + 1) = a^2bx^2 - a^2x - abx + a + ab^2x^2 - abx - b^2x + b $$ Simplify the right-hand side: $$ a^2bx^2 + ab^2x^2 - a^2x - 2abx - b^2x + a + b $$

5. Rewrite the equation Now we have: $$ 2abx - a - b = a^2bx^2 + ab^2x^2 - a^2x - 2abx - b^2x + a + b $$

6. Move all terms to one side of the equation $$ 0 = a^2bx^2 + ab^2x^2 - a^2x - 4abx - b^2x + 2a + 2b $$

7. Factor out common terms Factor out $x^2$, $x$ terms: $$ 0 = ab(a+b)x^2 - (a^2 + 4ab + b^2)x + 2(a+b) $$

8. Solve for x by using the quadratic formula The quadratic formula is $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, where $A = ab(a+b)$, $B = -(a^2 + 4ab + b^2)$, and $C = 2(a+b)$.

$$ x = \frac{(a^2 + 4ab + b^2) \pm \sqrt{(a^2 + 4ab + b^2)^2 - 4ab(a+b)(2(a+b))}}{2ab(a+b)} $$ $$ x = \frac{(a^2 + 4ab + b^2) \pm \sqrt{(a^2 + 4ab + b^2)^2 - 8ab(a+b)^2}}{2ab(a+b)} $$ This looks complicated.

9. Try a different approach: Rearrange the original equation Original Equation: $\frac{a}{ax-1} + \frac{b}{bx-1} = a+b$ Rewrite: $\frac{a}{ax-1} - a + \frac{b}{bx-1} - b = 0$ $$ \frac{a - a(ax-1)}{ax-1} + \frac{b - b(bx-1)}{bx-1} = 0 $$ $$ \frac{a - a^2x + a}{ax-1} + \frac{b - b^2x + b}{bx-1} = 0 $$ $$ \frac{2a - a^2x}{ax-1} + \frac{2b - b^2x}{bx-1} = 0 $$ $$ \frac{a(2-ax)}{ax-1} + \frac{b(2-bx)}{bx-1} = 0 $$ $$ \frac{a(2-ax)}{ax-1} = - \frac{b(2-bx)}{bx-1} $$ Cross-multiply: $$ a(2-ax)(bx-1) = -b(2-bx)(ax-1) $$ $$ a(2bx - 2 - abx^2 + ax) = -b(2ax - 2 - abx^2 + bx) $$ $$ 2abx - 2a - a^2bx^2 + a^2x = -2abx + 2b + ab^2x^2 - b^2x $$ $$ - a^2bx^2 - ab^2x^2 + a^2x + b^2x + 4abx - 2a - 2b = 0 $$ $$ -ab(a+b)x^2 + (a^2 + b^2 + 4ab)x - 2(a+b) = 0 $$ $$ ab(a+b)x^2 - (a+b)^2 x + 2(a+b) = 0 $$ This must be a mistake somewhere. Let's go back to: $$ \frac{a(2-ax)}{ax-1} + \frac{b(2-bx)}{bx-1} = 0 $$ $$ a(2-ax)(bx-1) + b(2-bx)(ax-1) = 0 $$ $$ a(2bx-2-abx^2+ax) + b(2ax-2-abx^2+bx) = 0 $$ $$ 2abx - 2a - a^2bx^2 + a^2 x + 2abx - 2b - ab^2x^2 + b^2 x = 0 $$ $$ -ab(a+b)x^2 + (a^2 + b^2 + 4ab)x - 2(a+b) = 0 $$ $$ ab(a+b)x^2 - (a+b)^2 x + 2(a+b) = 0 $$ Divide by a+b $$ abx^2 - (a+b)x + 2 = 0 $$ Using quadratic formula $$ x = \frac{(a+b) \pm \sqrt{(a+b)^2 - 8ab}}{2ab} $$ $$ x = \frac{(a+b) \pm \sqrt{a^2 + 2ab + b^2 - 8ab}}{2ab} $$ $$ x = \frac{(a+b) \pm \sqrt{a^2 - 6ab + b^2}}{2ab} $$ Still too complicated

10. Try a simple value for x If $x = \frac{2}{a+b}$: LHS: $$ \frac{a}{a(\frac{2}{a+b}) - 1} + \frac{b}{b(\frac{2}{a+b}) - 1} = \frac{a}{\frac{2a}{a+b} - 1} + \frac{b}{\frac{2b}{a+b} - 1} = \frac{a}{\frac{2a - a - b}{a+b}} + \frac{b}{\frac{2b - a - b}{a+b}} = \frac{a(a+b)}{a-b} + \frac{b(a+b)}{b-a} $$ $$ \frac{a(a+b)}{a-b} - \frac{b(a+b)}{a-b} = \frac{a^2 + ab - ab - b^2}{a-b} = \frac{a^2 - b^2}{a-b} = \frac{(a-b)(a+b)}{a-b} = a+b $$

So $x = \frac{2}{a+b}$ is the answer.