Question

(1/3y^2-4/7y+11) (1/7y-3+2y) -(2/7y-2/3y^2+2)​

Answer 1

Answer:

Step-by-step explanation:

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

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

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

Add the values of y^2,y and constant value

We get,

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

= -y^2 –y +6