Solving Equations with Fractions

[Mkorr]

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.

 

[eumyang]

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]

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

 

[HallsofIvy]

By the way- when you combine all the fractions, you will eventually set it equal to 0, then multiply both sides by the common denominator, eliminating it. It is simpler just to get it equal to 0, then multiply both sides of the equation by that common denominator immediately. That is, go from
[tex]\frac{1}{x-1} -\frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}[/tex]
to
[tex]\frac{1}{x-1} -\frac{1}{x-2}- \frac{1}{x-3}+ \frac{1}{x-4}= 0[/tex]
to
[tex](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[/tex]

It at least saves writing all of those fractions!