Question

(1/3y²-4/7y+5)-(2/7y-2/3y²+2)-(1/7y-3+2y²)​

Answer 1

Answer:

-y^2-y+6

Step-by-step explanation:

= (1/3y^2+2/3y^2-2y^2)+(-4/7y-2/7y-1/7y)+(5-2+3)

= (1y^2+2y^2-6y^2/3)+(-4y-2y-1y/7)+(5-2+3)

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= -3/3y^2-7/7y+6

= -y^2-y+6