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in order for the identity

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1 Answer

Hi Dalia;
(1/x+1)+(a/x2-1)=(x/x2-1)
I think this is...
[1/(x+1)]+[a/(x2-1)]=[x/(x2-1)]
Let's combine like terms.
Let's move the second parenthetical equation to the right side.  It is currently positive.  It will become negative...
[1/(x+1)]=-[a/(x2-1)]+[x/(x2-1)]
Let's consider the fact that...
(x2-1)=(x+1)(x-1)
Let's do such replacement...
1/(x+1)=-{a/[(x+1)(x-1)]}+x/[(x+1)(x-1)]
On the right side, let's combine numerators...
1/(x+1)=(-a+x)/[(x+1)(x-1)]
(x+1) is in the denominators of both sides.  It cancels...
1/(x+1)=(-a+x)/[(x+1)(x-1)]
1=(-a+x)/(x-1)
1/1=(-a+x)/(x-1)
I converted 1 into 1/1 to illustrate my next point.
Let's cross-multiply...
-a+x=x-1
x on both sides cancels...
-a=-1
Let's multiply both sides by -1...
(-1)(-a)=(-1)(-1)
a=1
 
 
 

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