X y 2 4 x 3 2y 5 July 8, 2019 16 32 64 128 X y 2 4 x 3 2y 5 by substitution method X y 2 4 x 3 2y 5 by elimination method Solve x y 2 4 x 3 2y 5 Implicit differentiation x 4 x 2y 2 y 3 5 X 1 2 y 4 11 2 and x 3 2 2y 3 17 5 3 x 2 y 12 5 x 2y 4
