Solve fails to find a solution for trivial linear system

I'm trying to find the value of G in the following set of equations by code:

clear all; clc;
syms R T S u yr y B A G

eq1 = R*u == T*yr - S*y;
eq2 = y == (B/A)*u;
eq3 = y == G*yr;

solve([eq1,eq2,eq3], G)

However, solve fails with an empty solution set, although this system has a trivial answer.

It's easy to see how it can be done, from eq1:

u == (T*yr - S*y)/R

From eq2:

y == (B/A)*(T*yr - S*y)/R

Thus:

y = yr*(B*T)/(A*R + B*S)

Finally, from eq3:

G = (B*T)/(A*R + B*S)

I can't understand why solve fails in this case. I appreciate any help I can get.

Answer 1

[Walter Roberson]

Actually it is just a matter of telling it all three variables to solve for:

solve(eq1, eq2, eq3, y, u, G)

[Daniel]

Wow, that was exactly it! Is there a specific reason why I need two more variables?

I used:

res = solve([eq1,eq2,eq3], [y,u,G])

and it worked perfectly. Thank you, Walter!

[Walter Roberson]

When you do not specify the other variables, the best you could have hoped for was that G was returned as y/yr, from the third equation, as it would have had no reason to know which other variables to eliminate.

Answer 2

[Star Strider]

Sometimes, you have to lead it gently by the hand:

syms R T S u yr y B A G

eq1 = R*u == T*yr - S*y;
eq2 = y == (B/A)*u;
eq3 = y == G*yr;
eq4  = solve(eq2, eq3);
yr = eq4.yr;
eq5 = subs(eq1, yr);
G_solved = solve(eq5, G)

G_solved =
-(B*S*u)/(A*(R*u - T*yr))

[Daniel]

Star, as always, great answer and you're one of the most helpful people here but I'll have to hand this one to Walter Roberson because he hit the nail on the head. Thank you!