@Kitonum That's what i meant. 

But in my worksheet appears some error with solve. I don't know why.

I ran yours sheet and at the and I see error. For simple equations "solve" works, like 2*x-4=0, but for my not.

Model_Maple2.mw

Probably I not quite clearly described what effect I want to get out of these equations. I know a little English poorly, and Google translate does not always work.
The system of equations I calculate in Matlab Simulink. However, there is a requirement that the equations were assigned the highest derivatives.
Let's say I now have a system of equations as:
ax '+ bx + gy' + x "= 0
ey '+ fy + gx "s" = 0
gz "-ky" = 0

On the other hand I have to get the first equation where one side is the only x "and rządnych other y" can no longer in the equation to be. 

How to solve a system of equations by substitution in Maple if x "is unknown?

But what next i must doo whit this? Two derivatives of the same order still exist in the same equation, and I want to separate them. How?

I put here to do download a file of 3 equations. And as I wrote before I want to appoint the highest derivatives, so that the equation there was no other in the same order

http://andrzejdebowski.wat.edu.pl/Forum.html#about1

Okay, maybe another way, because I can not cope.
How to designate from the system of three equations highest derivatives. The very function of "Isolate" does not settle the matter, because they still are other derivatives of the same order in equation.
What can I do in this case?

I see this:

Maybe another question I ask. Can Solve requires an load some libraries using with(...)? Where to find information about this?

@Carl Love 

Thank you.
Now I will digest it and see if another problem did not appear.

Thank you for the quick reply!

And as you have to modify your record Acer to affected system of three equations with three variables?

@spalinowy 

I found a solution, so I write.
Just use the "expand".

I see that ensued quite interesting discussion.

But now I have a problem again. The solution proposed by Kitonum suits me the best because I completely understand it. However, I noticed that the structure of my equations is too compressed. And these formulas do not read anything from the parenthesis, just move them in full.
Therefore, I have another question ...
How can I force a fully unravel the form, so that there was no factors before parenthesis?

One of my equations below:

Q1 := (diff(theta1(t), t, t))*(J1[XY]+J1[YZ])+9.81*m1*(Vector(1, {(1) = -sin(theta1(t))*x1+cos(theta1(t))*z1}))+(1/2)*(Vector(1, {(1) = (cos(theta[1](t))*x[S1]+sin(theta[1](t))*z[S1]-cos(theta[1](t))*cos(psi[2](t))*x[S2]+cos(theta[1](t))*sin(psi[2](t))*y[S2]-sin(theta[1](t))*z[S2])^2*k[1, x]+(y[S1]-sin(psi[2](t))*x[S2]-cos(psi[2](t))*y[S2])^2*k[1, y]+(-sin(theta[1](t))*x[S1]+cos(theta[1](t))*z[S1]+sin(theta[1](t))*cos(psi[2](t))*x[S2]-sin(theta[1](t))*sin(psi[2](t))*y[S2]-cos(theta[1](t))*z[S2])^2*k[1, z]}))+(1/2)*(Vector(1, {(1) = (-(diff(theta[1](t), t))*sin(theta[1](t))*x[A1]+(diff(theta[1](t), t))*cos(theta[1](t))*z[A1]-(-(diff(theta[1](t), t))*sin(theta[1](t))*cos(psi[2](t))-cos(theta[1](t))*(diff(psi[2](t), t))*sin(psi[2](t)))*x[A2]-((diff(theta[1](t), t))*sin(theta[1](t))*sin(psi[2](t))-cos(theta[1](t))*(diff(psi[2](t), t))*cos(psi[2](t)))*y[A2]-(diff(theta[1](t), t))*cos(theta[1](t))*z[A2])^2*c[1, x]+(-(diff(psi[2](t), t))*cos(psi[2](t))*x[A2]+(diff(psi[2](t), t))*sin(psi[2](t))*y[A2])^2*c[1, y]+(-(diff(theta[1](t), t))*cos(theta[1](t))*x[A1]-(diff(theta[1](t), t))*sin(theta[1](t))*z[A1]-(-(diff(theta[1](t), t))*cos(theta[1](t))*cos(psi[2](t))+sin(theta[1](t))*(diff(psi[2](t), t))*sin(psi[2](t)))*x[A2]-((diff(theta[1](t), t))*cos(theta[1](t))*sin(psi[2](t))+sin(theta[1](t))*(diff(psi[2](t), t))*cos(psi[2](t)))*y[A2]+(diff(theta[1](t), t))*sin(theta[1](t))*z[A2])^2*c[1, z]})) = j[1]

Carl Love 

Sorry, probably too much work...

Kitonum 

Incoherently I am written, because in a hurry.

In my final equations everything is written without a multiplication sign "*". For this reason, I have matrices with zeros.

Carl Love 

I just saw that the same problem is in your solution. So I have to manually rewrite the whole equation to add these operators?

I ask because I have less than equations, but they look like this:

http://pl.tinypic.com/r/t0qiqp/9

This deviation is the first solution. Or check and the second.

This solution almost works for my equations. But as you can see in the equations do not have a multiplication operator. For this reason, he calls me matrix M1, M2 and M3 with all zeros. Can it get around?

Thank you very much for your help. I'll try how it works :)