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Re: extracting lhs or rhs of equations

• To: mathgroup at smc.vnet.net
• Subject: [mg16870] Re: [mg16853] extracting lhs or rhs of equations
• From: "Kevin J. McCann" <kevinmccann at Home.com>
• Date: Mon, 5 Apr 1999 02:24:15 -0400
• References: <199904020235.VAA11442@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com

Two ways. I made the equation a little more complicated just to make sure
it works with more stuff on the LHS.

Method 1 (easier to see)

eq1 = (Sin[x] + Cos[x])^2 + 4/(2 + 3*x) == x;
 
eq1[[1]]
4/(2 + 3*x) + (Cos[x] + Sin[x])^2

eq1[[2]]
x

Method 2 (more in the spirit of Mathematica and pattern matching)

eq1
4/(2 + 3*x) + (Cos[x] + Sin[x])^2 == x

eq1/.(x_ == y__)->x
4/(2 + 3*x) + (Cos[x] + Sin[x])^2

eq1/.(x_ == y_)->y
x

Kevin


----- Original Message ----- 
From: David P. Johnson <johnson at ae.msstate.edu>
To: mathgroup at smc.vnet.net
Subject: [mg16870] [mg16853] extracting lhs or rhs of equations


> Let's say I have an expression like:
> 
>   In[1]:= eq1= Sin[x] == x;
> 
> Is there a way to get just the left-hand side or right-hand side of the
> equation? Something like:
> 
>   In[2]:= LHS[eq1]
>   Out[2]:= Sin[x]
> 
> TIA.
> 
> -- 
> David
> ->(Signature continues here)
> 
> 

• References:
  • extracting lhs or rhs of equations
    • From: "David P. Johnson" <johnson@ae.msstate.edu>