1/y+1 - 2/y-1 + 3/y×y-1
3 answers
4/3
assuming the usual carelessness with parentheses,
1/(y+1) - 2/(y-1) + 3/(y(y-1))
using a common denominator of y(y+1)(y-1), we have
[y(y-1) - 2y(y+1) + 3(y+1)]/(y(y^2-1))
(3-y^2)/(y(y^2-1))
1/(y+1) - 2/(y-1) + 3/(y(y-1))
using a common denominator of y(y+1)(y-1), we have
[y(y-1) - 2y(y+1) + 3(y+1)]/(y(y^2-1))
(3-y^2)/(y(y^2-1))
consistent with her carelessness with brackets, I think
her denominator for the last term was probably y^2 - 1
which makes y^2 - 1 the LCD, and a final answer of
-y/(y^2 - 1) or y/(1 - y^2)
her denominator for the last term was probably y^2 - 1
which makes y^2 - 1 the LCD, and a final answer of
-y/(y^2 - 1) or y/(1 - y^2)