# Equation with fractions

## Homework Statement

Solve the following equations:

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

See above.

## The Attempt at a Solution

Rearrangement gives

$$\frac{1}{x-1} -\frac{1}{x-2} - \frac{1}{x-3} + \frac{1}{x-4} = 0$$

Conversion to same denominator gives

$$\frac{(x-2)(x-3)(x-4)}{(x-1)(x-2)(x-3)(x-4)} -\frac{(x-1)(x-3)(x-4)}{(x-1)(x-2)(x-3)(x-4)} - \frac{(x-1)(x-2)(x-4)}{(x-1)(x-2)(x-3)(x-4)} + \frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0$$

Writing as one fraction gives

$$\frac{(x-2)(x-3)(x-4) -(x-1)(x-3)(x-4) - (x-1)(x-2)(x-4) + (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0$$

Partially multiplying binomials gives

$$\frac{(x^{2} - 5x + 6)(x-4) -(x^{2}-4x +3)(x-4) - (x^{2} - 3x +2)(x-4) + (x^{2} - 3x + 2)(x-3)}{(x-1)(x-2)(x-3)(x-4)} = 0$$

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

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

Removal of the parenthesis gives

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

$$\frac{31x - 10}{(x-1)(x-2)(x-3)(x-4)} = 0$$

Rewriting

$$\frac{1}{(x-1)(x-2)(x-3)(x-4)} \cdot (31x - 10) = 0$$

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.

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

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

The 2nd polynomial:
x3 - 8x2 + 16x - 12
should be
x3 - 8x2 + 19x - 12

The 3rd polynomial:
x3 - 7x2 - 12x + 2x - 8
should be
x3 - 7x2 + 12x + 2x - 8 or x3 - 7x2 + 14x - 8

The last polynomial:
x3 - 6x + 9x + 2x -6
should be
x3 - 6x2 + 9x + 2x -6 or x3 - 6x2 + 11x - 6

Last edited:
HallsofIvy
$$\frac{1}{x-1} -\frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}$$
$$\frac{1}{x-1} -\frac{1}{x-2}- \frac{1}{x-3}+ \frac{1}{x-4}= 0$$
$$(x-2)(x-3)(x-4)- (x-1)(x-3)(x-4)- (x-1)(x-2)(x-4)+ (x-1)(x-2)(x-3)= 0$$