Re: solving algebraic equations

On Thu, 21 Feb 2008 09:03:23 -0600, OP
<facetious_nickname@[EMAIL PROTECTED]
> wrote:


>Wow, talk about rusty!  I thought there was some kind of crazy 
>voodoo going on there. Thanks for the push in the right direction.
>
>Once I plugged "B = b - Aa" into the second equation it took me 
>about 5 minutes of staring at it to recognize that
>
>Ac - Aa = d - b
>
>is equal to

"is equivalent to"

>A(c - a) = d - b
>
>is equal to

"is equivalent to"

>A = (d - b)/(c - a)

 .. assuming c not equal a.

>Nice to know they haven't repealed the distributive property since 
>the last time I used my brain. Of course, I'm still probably doing 
>it wrong, but oh well! :-)

While you are bru****ng up, you might as well relearn correct
terminology. Good luck.

--Lynn