# [✓] Solve a system of two rather complicated power equations?

GROUPS:
 HiI am not a Mathematica user, I am just looking for a math package that can solve my problem.I need to solve a system of two rather complicated power equations, like those in the image below. Here x and z are the unknown variables, all other constants are known. Can Mathematica do that?But actually I need even more. I will have a set of values for the known constants I and D (A and a are fixed). So I will need not just one solution but a set of solutions for each pair of (I, D). So, if I feed the set of (I,D) pairs into Mathematica, can it produce a set of solutions for me, not just one solution?
4 months ago
6 Replies
 Gianluca Gorni 2 Votes For this particular set of equation it can: Solve[{i == A z (z^2 + (a + x)^2)^(-1/2), D == A (x + a) (z^2 + (a + x)^2)^(-3/2)}, {x, z}] 
4 months ago
 Thank you for the reply. Looks quite simple. But is it possible to automatically produce a set of solutions for a set of (I, D) pairs? This set is going to be quite long (several hundred pairs) so it does not look practical to run the above solution for each pair separately.
4 months ago
 Gianluca's solution outputs Then of course you can use your (I,d) and the a, A constants to fill in the gaps. So I would say it's easy to implement this with Mathematica.
 Gianluca Gorni 1 Vote Yes, it's easy once you clarify your input and output format. You can repackage the solution into a function, for example sol[{i_, d_}] = Solve[{i == A z (z^2 + (a + x)^2)^(-1/2), d == A (x + a) (z^2 + (a + x)^2)^(-3/2)}, {x, z}] sol[{1, 2}] 
 Jason Biggs 2 Votes I would recommend not using the letter D for a symbol in your equation. D is a built in symbol, which is why it shows up as black in a notebook without you having defined it.It doesn't give an error in Gianluca's code (because it has DownValues but no OwnValues), but it is still a good practice to avoid name conflicts.