Question

Solve for x:

$\frac{1}{x-1} - \frac{4}{x-2} + \frac{4}{x-3} - \frac{1}{x-4} = \frac{1}{30}$

Answer 1

-2, -1, 6, 7

Answer 2

The given equation is $\frac{1}{x-1} - \frac{4}{x-2} + \frac{4}{x-3} - \frac{1}{x-4} = \frac{1}{30}$ We group the terms as follows: $\left( \frac{1}{x-1} - \frac{1}{x-4} \right) - \left( \frac{4}{x-2} - \frac{4}{x-3} \right) = \frac{1}{30}$ Combine the terms in each group: $\frac{(x-4) - (x-1)}{(x-1)(x-4)} - 4 \left( \frac{1}{x-2} - \frac{1}{x-3} \right) = \frac{1}{30}$ $\frac{x-4-x+1}{(x-1)(x-4)} - 4 \left( \frac{(x-3) - (x-2)}{(x-2)(x-3)} \right) = \frac{1}{30}$ $\frac{-3}{x^2 - 5x + 4} - 4 \left( \frac{x-3-x+2}{x^2 - 5x + 6} \right) = \frac{1}{30}$ $\frac{-3}{x^2 - 5x + 4} - 4 \left( \frac{-1}{x^2 - 5x + 6} \right) = \frac{1}{30}$ $\frac{-3}{x^2 - 5x + 4} + \frac{4}{x^2 - 5x + 6} = \frac{1}{30}$ Let $y = x^2 - 5x$. The equation becomes: $\frac{-3}{y+4} + \frac{4}{y+6} = \frac{1}{30}$ Combine the terms on the left side: $\frac{-3(y+6) + 4(y+4)}{(y+4)(y+6)} = \frac{1}{30}$ $\frac{-3y - 18 + 4y + 16}{y^2 + 10y + 24} = \frac{1}{30}$ $\frac{y - 2}{y^2 + 10y + 24} = \frac{1}{30}$ Cross-multiply: $30(y - 2) = y^2 + 10y + 24$ $30y - 60 = y^2 + 10y + 24$ Rearrange into a quadratic equation in $y$: $y^2 + 10y - 30y + 24 + 60 = 0$ $y^2 - 20y + 84 = 0$ Factor the quadratic equation: $(y - 6)(y - 14) = 0$ So, $y = 6$ or $y = 14$. Substitute back $y = x^2 - 5x$.

Case 1: $x^2 - 5x = 6$ $x^2 - 5x - 6 = 0$ $(x - 6)(x + 1) = 0$ So, $x = 6$ or $x = -1$.

Case 2: $x^2 - 5x = 14$ $x^2 - 5x - 14 = 0$ $(x - 7)(x + 2) = 0$ So, $x = 7$ or $x = -2$.

The possible values for $x$ are -2, -1, 6, and 7. We check that none of these values make the original denominators zero. The denominators are $x-1, x-2, x-3, x-4$. None of the values -2, -1, 6, 7 are equal to 1, 2, 3, or 4. Thus, all four solutions are valid.