Re: Can someone help me?

In article 
<51543823-1e25-4f11-b911-e5753f749645@[EMAIL PROTECTED]
>,
 revdorisj@[EMAIL PROTECTED]
 wrote:

> One of the students in my church is struggling with his algebra
> homework.  Can someone show me the steps to solving these two
> problems:
> 
> A.  Graph the inverse of y = (2/3) to the x power.

Build a table of x-y values for y = (2/3)^x.
Swap the x and y columns.
Plot the new points from the new table
Connect the points with a smooth curve.
> 
> B.  Simplify (sorry this one is hard to lay out):
> 
> 1/a squared - 1/b squared in the numerator
> 
> a to the -2 + 2ab to the -1 + b to the -2 in the denominator

 In newsgroup notation, this might be expressed as 

(1/a^2 - 1/b^2) / ( a^(-2) + (2*a*b)^(-1) + b^(-2) )

Though I suspect you might mean

(1/a^2 - 1/b^2) / ( a^(-2) + 2*(a*b)^(-1) + b^(-2) )

Then simplify one piece at time:
 
(1/a^2 - 1/b^2) = (b^2 - a^2)/(a*b)^2

and, as you described it, 

( a^(-2) + (2*a*b)^(-1) + b^(-2) ) = (b^2 + a*b/2 + a^2)/ (a*b)^2

Then the (a*b)^2 bits cancel out leaving
 (b^2 - a^2) / (b^2 + a*b/2 + a^2)



With the alternate form, I mentioned above, one gets

(1/a^2 - 1/b^2) / ( a^(-2) + 2*(a*b)^(-1) + b^(-2) )
=
(b^2 - a^2) / b^2 - 2*a*b* + a^2)

But b^2 - a^2 = (b-a)(b+a) 
and b^2 - 2*a*b + b^2 = (b-a)(b-a)

so the whole thing reduces to (b+a) / (b - a)