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Re: broadcasting of Equal ? (newbie question)

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You will probably get a lot of responses to this, but basically you want 
the following:

Solve[{x+y-1==0,2x+y-2==0},{x,y}]

Sometimes it is more convenient to do it the way you indicate, and here 
is how:

Solve[Thread[{x + y - 1, 2 x + y - 2} == {0, 0}], {x, y}]

Pull the Thread part of the above out and try it by itself, and you will 
see that it gives you what you want. A more general thing to try would 
be to replace {0,0} with {a,b} and see what Thread does to that.

Kevin


On 4/14/2011 4:59 AM, Alan wrote:
> I noticed that e.g.
> Solve[{x + y - 1, 2 x + y - 2} == {0, 0}, {x, y}]
> can be written as
> Solve[{x + y - 1, 2 x + y - 2} == 0, {x, y}]
> but I cannot find the rule that allows this.
> (E.g., I do not find it in the help for Equal
> or the help for Solve.)
>
> Can you point me to it?
>
> Thanks,
> Alan
>

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Thread

broadcasting of Equal ? (newbie question) Alan <alan.isaac@gmail.com> - 2011-04-14 08:59 +0000
  Re: broadcasting of Equal ? (newbie question) "Kevin J. McCann" <Kevin.McCann@umbc.edu> - 2011-04-15 07:56 +0000
  Re: broadcasting of Equal ? (newbie question) Alan <alan.isaac@gmail.com> - 2011-04-15 07:57 +0000

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