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```        The value of x satisfying (x+1)/(x-1) + (x-2)/(x+2)+(x-3)/(x+3)+(x+4)/(x-4)=4 are

(a)(-5+3451/2)/10
(b) (5+3401/2)/8
(c)(-5+3251/2)/16
(d)none```
8 years ago

17 Points
```										Hi
The given equation is
(x+1)/(x-1) + (x-2)/(x+2)+(x-3)/(x+3)+(x+4)/(x-4)=4
Write (x+1) as (x-1)+2. So the first term becomes
1+ 2/(x-1).
Similarly, the second term becomes 1 + 2/(x-2), and so on.
Do this to all four terms on the LHS. Now the equation becomes
1+ 2/(x-1) + 1+ 2/(x-2) + 1+ 2/(x-3) + 1+ 2/(x-4) = 4
or, 4 + 2/(x-1) + 2/(x-1) + 2/(x-1) + 2/(x-1) = 4
or, 2 ( 1/(x-1) + 1/(x-2) + 1/(x-3) + 1/(x-4)) = 0
or, 1/(x-1) + 1/(x-4) + 1/(x-3) + 1/(x-2) = 0
or,
[(x-4)+(x-1)]/[x2-5x+4] + [(x-3)+(x-2)]/[x2-5x+6] = 0
or,
(2x-5)/[x2-5x+4] = -(2x-5)/[x2-5x+6]
Therefore Either 2x-5 = 0 or 1/[x2-5x+4] = -1/[x2-5x+6]
x = 5/2 or [x2-5x+6]=-[x2-5x+4]
x=5/2 or 2x2-10x+10=0

x=5/2 or x2-5x+5=0
x=5/2 or x = (5+root(5))/2
These are the only solutions. I think there is some typing error in the options but I am sure these 3 are the only solution. I have verified it using MATLAB software also.

```
8 years ago
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