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## Homework Statement

Solve the following equations:

[tex]\frac{1}{x-1} -\frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}[/tex]

## Homework Equations

See above.

## The Attempt at a Solution

Rearrangement gives

[tex]\frac{1}{x-1} -\frac{1}{x-2} - \frac{1}{x-3} + \frac{1}{x-4} = 0[/tex]

Conversion to same denominator gives

[tex]\frac{(x-2)(x-3)(x-4)}{(x-1)(x-2)(x-3)(x-4)} -\frac{(x-1)(x-3)(x-4)}{(x-1)(x-2)(x-3)(x-4)} - \frac{(x-1)(x-2)(x-4)}{(x-1)(x-2)(x-3)(x-4)} + \frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Writing as one fraction gives

[tex]\frac{(x-2)(x-3)(x-4) -(x-1)(x-3)(x-4) - (x-1)(x-2)(x-4) + (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Partially multiplying binomials gives

[tex]\frac{(x^{2} - 5x + 6)(x-4) -(x^{2}-4x +3)(x-4) - (x^{2} - 3x +2)(x-4) + (x^{2} - 3x + 2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Complete multiplication (just looking at the denominator for simplification) gives

[tex]\frac{x^3 - 4x^2 - 5x^2 + 20x + 6x - 24 - (x^3 - 8x^2 + 16x - 12) - (x^3 - 7x^2 - 12x + 2x - 8) + x^3 - 6x + 9x + 2x -6}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Removal of the parenthesis gives

[tex]\frac{x^3 - 4x^2 - 5x^2 + 20x + 6x - 24 - x^3 + 8x^2 - 16x + 12 - x^3 + 7x^2 + 12x - 2x + 8 + x^3 - 6x^2 + 9x + 2x -6}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Adding together everything gives

[tex]\frac{31x - 10}{(x-1)(x-2)(x-3)(x-4)} = 0[/tex]

Rewriting

[tex]\frac{1}{(x-1)(x-2)(x-3)(x-4)} \cdot (31x - 10) = 0[/tex]

The left factor equals to zero lacks a solution so

31x - 10 = 0

x = 10/31

but this is clearly wrong as checking does not produce the equality when checked.