SOLUTION: Could you please help me with this problem Solve for x {{{10 over x^2-2x + 4 over x}}} = 5 over x-2

Question 29996: Could you please help me with this problem
Solve for x
= 5 over x-2
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
10 over x^2-2x + 4 over x = 5 over x-2
10/(x^2-2x) + 4/x = 5/(x-2) ----(1)
10/x(x-2) + 4/x = 5/(x-2)
Multiplying by x(x-2) through out
10+4(x-2)=5x ----(2)
10+4x-8 = 5x
(10-8) = 5x-4x (grouping like terms,changing sign while changing side)
2 = x
That is x = 2
Verification:Putting x= 2 in (1) we arrive at division by zero which is not defined as 0 does not have multiplicative inverse
The value x = 2 holds in the version (2) of the problem where there is no division by the factor (x-2)
Note:Why 0 does not have multiplicative inverse?
Any real number a NOT ZERO is said to have multiplicative iverse b NOT ZERO only when (b)X(a) = 1 = (a)X(b)
We call this b the multiplicative inverse of a and write b as b = (1/a)
Why should a (or for that matter) be NOT ZERO?
Supposing a= 0
and if it should have multiplicative inverse then we should find that something called inverse of zer0 so that
(0)X(something)=1 =(that same something)X(0)
But then we know that (0) X (anything ) is always zero and NOT 1
Hence 0 does not have multiplicative inverse.
That is the multiplicative inverse of 0
namely (1/0)does not exist
That is division by 0 is not defined.
Got it!

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