若代数式(2x²+ax-1/3y+1/5)-(x-2y-1-bx²)的值与字母x的取值无关,求a、b的值

(2x²+ax-1/3y+1/5)-(x-2y-1-bx²)
＝2x²+ax -(1/3)y + 1/5 - x + 2y + 1 + bx²
= (2 + b)x² +(a - 1)x +(2- 1/3)y + 6/5
因为代数式(2x²+ax-1/3y+1/5)-(x-2y-1-bx²)的值与字母x的取值无关,所以有
2 + b = 0 ,a - 1 = 0
解得 a = 1 ,b = -2