X y 5 2x 3y 10 July 8, 2019 16 32 64 128 X y 5 and 2x 3y 10 substitution method Xy =- 5 2x 3y 10 X 5 y 3 z 2 và 2x 3y 100 Elimination method class 10 x y 5 2x 3y 4 X 10 y 5 3y 2z và 2x 3y 4z 330 X 3y 10 2x y 5 solve by substitution method X+y 2 5+z xy