It works. So grateful for your quick response.

Thank you

Dear Gianluca Gorni, Many thanks for your help. Now, I'm trying to apply the same manner that you explained above but in different problem. I don't get expected answer from Mathematica.

Here is the code:

In[1]:= Quit[]

In[1]:= ClearAll;

EE = Solve[[Sigma]y == D[p x y^3 - 2 p x^3 y, y], [Sigma]y]

In[3]:= Solve[[Tau]xy == D[-1.5 p x^2 y^2 + c1 y^2 + .5 p x^4 + c3, x], [Tau]xy]

In[2]:= Solve[[Tau]xy == D[-1.5 p x^2 y^2 + c1 y^2 + .5 p x^4 + c3, x], [Tau]xy] /. EE

Out[2]= {{[Tau]xy -> 2. p x^3 - 3. p x y^2}} /. EE

As you see in the Out[2], I want the answer to be in terms of [segma]y.

Can you please tell me the right code in this case? I attached the file too.

Thank you

Attachments:

Test Replace All.nb

You can solve two equations together:

u = 2 x^2;
sls = Solve[{\[Epsilon]x == D[u, {x, 1}], \[Epsilon]y == 
    D[v, {y, 1}]}, {\[Epsilon]x, \[Epsilon]y}]
Solve[\[Sigma]x == 
   2 G \[Epsilon]x + \[Lambda] (\[Epsilon]x + \[Epsilon]y), \
\[Sigma]x] /. sls
Solve[\[Gamma]xy == D[u, {y, 1}] + D[v, {x, 1}], \[Gamma]xy] /. sls
Solve[\[Sigma]y == 
   2 G \[Epsilon]y + \[Lambda] (\[Epsilon]x + \[Epsilon]y), \
\[Sigma]y] /. sls
Solve[\[Tau]xy == G \[Gamma]xy, \[Tau]xy] /. sls

Gianluca Gorni, This is what i was looking for. Thank you very much

Hi Rohit Namjoshi,

Thanks for your reply. I have segma is function of epsilon, epsilon is function of neu and neu is function of x. i want to find segma in terms of x.

When I SOLVE epsilon i get it in terms of x (which is OK). Then i SOLVE Segma and get it in terms of epsilon ( i want it in terms of x)

is there a simple way to do so?

u = 2 x^2; Solve[[Epsilon]x == D[u, {x, 1}], [Epsilon]x] Solve[[Sigma]x == 2 G [Epsilon]x + [Lambda] ([Epsilon]x + [Epsilon]y), [Sigma]x]

i also attached the file

Many thanks for your help

Attachments:

CH. 3(query).nb