Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
1998
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 1998

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Adding equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg14402] Re: [mg14342] Adding equations
  • From: BobHanlon at aol.com
  • Date: Sun, 18 Oct 1998 15:10:07 -0400
  • Sender: owner-wri-mathgroup at wolfram.com

In a message dated 10/15/98 5:13:03 AM, lwalker701 at earthlink.net wrote:

>I've been trying to add two equations as follows:
>
>In[4]:=
>    a==b;
>In[5]:=
>    c==d;
>In[6]:=
>    %4+%5
>Out[6]=
>    (a==b)+(c==d)
>
>The result I was looking for is (a+c)==(b+d). The following works but it
>seems clumsy. In[7]:=
>    %4[[1]]+%5[[1]]==%4[[2]]+%5[[2]] Out[7]=
>    a+c==b+d
>Is there a way to elegantly add (sub, mult, div) two equations?  Also is
>there an elegant way to add (sub, mult, div) an expression to both
>sides of an equation?

One approach:

eqnAdd[(a_) == (b_), (c_) == (d_)] := a + c == b + d;  eqnAdd[a_, (c_)
== (d_)] := a + c == a + d;  eqnAdd[a_, x_List] := (eqnAdd[a, #1] & )
/@ x;  eqnMult[a_, (b_) == (c_)] := a*b == a*c;  eqnMult[a_, x_List] :=
(eqnMult[a, #1] & ) /@ x;  eqnSqrt[(a_) == (b_)] := 
   {Sqrt[a] == Sqrt[b], Sqrt[a] == -Sqrt[b]}; 

a*x^2 + b*x + c == 0; 

eqnMult[4*a, %]

4*a*(c + b*x + a*x^2) == 0

eqnAdd[b^2 - 4*a*c, %]

b^2 - 4*a*c + 4*a*(c + b*x + a*x^2) == b^2 - 4*a*c

ExpandAll[%]

b^2 + 4*a*b*x + 4*a^2*x^2 == b^2 - 4*a*c

Factor /@ %

(b + 2*a*x)^2 == b^2 - 4*a*c

eqnSqrt[%]

{Sqrt[(b + 2*a*x)^2] == Sqrt[b^2 - 4*a*c], 
  Sqrt[(b + 2*a*x)^2] == -Sqrt[b^2 - 4*a*c]}

PowerExpand[%]

{b + 2*a*x == Sqrt[b^2 - 4*a*c], 
  b + 2*a*x == -Sqrt[b^2 - 4*a*c]}

eqnAdd[-b, %]

{2*a*x == -b + Sqrt[b^2 - 4*a*c], 
  2*a*x == -b - Sqrt[b^2 - 4*a*c]}

eqnMult[1/(2*a), %]

{x == (-b + Sqrt[b^2 - 4*a*c])/(2*a), 
  x == (-b - Sqrt[b^2 - 4*a*c])/(2*a)}

Bob Hanlon


  • Prev by Date: Re: Two programming challenges
  • Next by Date: Re: Adding equations
  • Previous by thread: RE: Adding equations
  • Next by thread: Re: Adding equations