^{2}-1)=(x/x

^{2}-1)

In order for the identity

(1/x+1)+(a/x^{2}-1)=(x/x^{2}-1)

to hold for all x, a must equal what?

Tutors, sign in to answer this question.

Hi Dalia;

(1/x+1)+(a/x^{2}-1)=(x/x^{2}-1)

I think this is...

[1/(x+1)]+[a/(x^{2}-1)]=[x/(x^{2}-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/(x^{2}-1)]+[x/(x^{2}-1)]

Let's consider the fact that...

(x^{2}-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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