Question 509686
(1) {{{ 3c + d = 2g }}}
(2) {{{ c + 2d + 3g = 25 }}}
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(1) {{{ 3c + d - 2g = 0 }}}
(2) {{{ c + 2d + 3g = 25 }}}
Add the equations:
(3) {{{ 4c + 3d + g = 25 }}}
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There are 3 unknowns and only 2 equations, so it isn't 
directly solvable, but trial and error might work
Suppose
g = 6
d = 1
c = 4
Almost
(2) {{{ 24 = 25 }}}
(3) {{{ 25 = 25 }}}
Suppose
g = 5
d = 4
c = 2
I think this works!
(1) {{{ 3c + d = 2g }}}
(1) {{{ 3*2 + 4 = 2*5 }}}
(1) {{{ 6 + 4 = 10 }}}
(1) {{{ 10 = 10 }}}
OK
(2) {{{ c + 2d + 3g = 25 }}}
(2) {{{ 2 + 2*4 + 3*5 = 25 }}}
(2) {{{ 2 + 8 + 15 = 25 }}}
(2) {{{ 25 = 25 }}}
OK
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