Question

Resolva
(2y^2-1)/2 - (y-1)/3= (1-y)/6

Answer 1

(2y² - 1)/2 - (y - 1)/3 = (1 - y)/6 

mmc(2,3,6) = 6

3*(2y² - 1) - 2*(y - 1) - (1 - y) = 0 

6y² - 3 - 2y + 2 - 1 + y = 0

6y² - y - 2 = 0 

delta
d² = 1 - 4*6*2 = 49
d = 7

y1 = (1 + 7)/12 = 8/12 = 2/3
y2 = (1 - 7)/12 = -6/12 = -1/2