*To*: mathgroup@smc.vnet.net*Subject*: [mg10331] Re: Simple algebra problem*From*: Allan Hayes <hay@haystack.demon.co.uk>*Date*: Thu, 8 Jan 1998 23:40:54 -0500*References*: <68q6q4$1v8@smc.vnet.net>

Eddy Wibowo wrote: > Suppose that I have an equation of: > Sqrt[x+y]+ 2x + 3y == 0 > > Now, what is the Mathematica-way to find: > Sqrt[x+y] = ??? > y = ??? Eddy: These are separate rearrangements In[1]:= Solve[ (Sqrt[x+y]+ 2x + 3y == 0)/.Sqrt[x+y]->s,s]/.s->Sqrt[x+y] Out[1]= {{Sqrt[x + y] -> -2*x - 3*y}} In[2]:= Solve[ (Sqrt[x+y]+ 2x + 3y == 0),y] Out[2]= {{y -> (1 - 12*x - Sqrt[1 + 12*x])/18}, {y -> (1 - 12*x + Sqrt[1 + 12*x])/18}} -- Allan Hayes Training and Consulting Leicester, UK hay@haystack.demon.co.uk http://www.haystack.demon.co.uk voice: +44 (0)116 271 4198 fax: +44 (0)116 271 8642