2 Added code for simplifying the result edited Jan 18 '18 at 16:56 Sjoerd Smit 6,5951313 silver badges2727 bronze badges You could try Eliminate to eliminate either g22 or g12: Eliminate[{ eq1 == 2 a g12 (b (-1 + b c) g12^2 + b (1 - b c) g11 g22 + c g22^2) + g22 ((-1 + b^2 c^2) g12^2 + c^2 g22 (-b^2 g11 + g22)) + a^2 (g12^2 g22 + b^2 g11 (g12^2 - g11 g22)), g11 a^2 + 2 g12 a c + g22 c^2 == g11 }, g22 ]  If necessary you can then use Solve and Simplify to solve for the temporary variable eq1. : Simplify[eq1 /. First @ Solve[%, eq1]]  You could try Eliminate to eliminate either g22 or g12: Eliminate[{ eq1 == 2 a g12 (b (-1 + b c) g12^2 + b (1 - b c) g11 g22 + c g22^2) + g22 ((-1 + b^2 c^2) g12^2 + c^2 g22 (-b^2 g11 + g22)) + a^2 (g12^2 g22 + b^2 g11 (g12^2 - g11 g22)), g11 a^2 + 2 g12 a c + g22 c^2 == g11 }, g22 ]  If necessary you can then use Solve to solve for the temporary variable eq1. You could try Eliminate to eliminate either g22 or g12: Eliminate[{ eq1 == 2 a g12 (b (-1 + b c) g12^2 + b (1 - b c) g11 g22 + c g22^2) + g22 ((-1 + b^2 c^2) g12^2 + c^2 g22 (-b^2 g11 + g22)) + a^2 (g12^2 g22 + b^2 g11 (g12^2 - g11 g22)), g11 a^2 + 2 g12 a c + g22 c^2 == g11 }, g22 ]  If necessary you can then use Solve and Simplify to solve for the temporary variable eq1: Simplify[eq1 /. First @ Solve[%, eq1]]  1 answered Jan 18 '18 at 16:35 Sjoerd Smit 6,5951313 silver badges2727 bronze badges You could try Eliminate to eliminate either g22 or g12: Eliminate[{ eq1 == 2 a g12 (b (-1 + b c) g12^2 + b (1 - b c) g11 g22 + c g22^2) + g22 ((-1 + b^2 c^2) g12^2 + c^2 g22 (-b^2 g11 + g22)) + a^2 (g12^2 g22 + b^2 g11 (g12^2 - g11 g22)), g11 a^2 + 2 g12 a c + g22 c^2 == g11 }, g22 ]  If necessary you can then use Solve to solve for the temporary variable eq1.