Community Tip - Learn all about PTC Community Badges. Engage with PTC and see how many you can earn! X
According to some web tutorials, I have tried to solve the equations system you can find in the attached figure.
I want to solve the system in order to find variables x and e. All the other variables are defined (you can see defined values also for e and x, they are guess values according to tutorials).
I am not able to find out what is my mistake, if I click on "sol" I see the error "This variable is undefined".
You find also the Mathcad file attached.
Can someone help me?
Sincerely
Stefano Grandi
Solved! Go to Solution.
Two issues here. One solvable, one not.
Firstly, to use the Find(x) function to solve a system, you must place a Given command before your two equation constraints.
Secondly, before solving anything, always plot your data. If we rearrange the first equation, putting variable e on the left side (it is separable) and call this e1(x), we can get an idea of how this equation behaves. Rearranging the second equation in terms of e and calling this e2(x) gives a look at how both equations look and where they may intersect.
They do not intersect. There is no solution.
Hi,
The "Given" keyword needs to be placed after the guess values and before the equations of the solve block.
With the two equations as they stand Fm.c=5 and T=2000 you can set x, e to get 5 and 2000 but not with the same values.
Please check your equations.
Solve block works Fm.c=5 and T=2 to show you it can work if equations are solvable.
Cheers
Terry
Thank you very much @terryhendicott and @JeffH1!
I solved my issue and I will keep in mind to plot curves before.
Regards
Two issues here. One solvable, one not.
Firstly, to use the Find(x) function to solve a system, you must place a Given command before your two equation constraints.
Secondly, before solving anything, always plot your data. If we rearrange the first equation, putting variable e on the left side (it is separable) and call this e1(x), we can get an idea of how this equation behaves. Rearranging the second equation in terms of e and calling this e2(x) gives a look at how both equations look and where they may intersect.
They do not intersect. There is no solution.