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