Equation with fractions

  • Thread starter Mkorr
  • Start date
  • #1
48
0

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.
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
10
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
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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!
 

Related Threads on Equation with fractions

Replies
3
Views
343
Replies
4
Views
1K
Replies
6
Views
1K
  • Last Post
Replies
13
Views
2K
Replies
4
Views
8K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
12K
  • Last Post
Replies
8
Views
14K
Top