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I have two simple coupled equations

eqn1 = 0 == -x + P - G x (y + 1);
eqn2 = 0 == -y + G x (y + 1);

If I use Solve[eqn1 && eqn2, y] I get no result. But I can force Mathematica to give me the result by solving eqn1 for x and then substituting that into eqn2 and solving for y

Simplify[Solve[eqn2, y] /. Solve[eqn1, x]]

which gives the result y -> (G P)/(1 + G (1 - P + y)).

Why does the single pass of Solve not work; and can I make it work?

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    $\begingroup$ I think I've seen a few more problems when solving for n equations with m unknowns and m!=n $\endgroup$
    – Bill
    Commented Feb 12, 2021 at 12:44
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    $\begingroup$ Try Solve[{0 == -x + P - G x (y + 1), 0 == -y + G x (y + 1)}, y, x] $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Feb 12, 2021 at 12:45
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    $\begingroup$ @J.M. - If you use a List bracket around the eliminate variable, you can avoid a Solve::bdomv warning. That is, Solve[{0 == -x + P - G x (y + 1), 0 == -y + G x (y + 1)}, y, {x}] $\endgroup$
    – Bob Hanlon
    Commented Feb 12, 2021 at 17:23
  • $\begingroup$ @Bob, you're right; old habits... $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Feb 12, 2021 at 17:25

2 Answers 2

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In general, this system has no solution for y. That's to be expected: usually for a system of two equations you need to solve for two variables. Solve[eqn1 && eqn2, {x, y}] yields two solutions.

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In 12.2

eqn1 = 0 == -x + P - G x (y + 1);eqn2 = 0 == -y + G x (y + 1);
Reduce[eqn1 && eqn2, y]

(x == 0 && P == 0 && y == 0) || (x (-1 - P + x) != 0 && G == -((P - x)/(x (-1 - P + x))) && y == P - x)

You may solve G == -((P - x)/(x (-1 - P + x))) in x and then express y in terms of P and G. I leave it on your own.

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