how do you answer -(9y-8z/3y) + (5y+8z/3y)?

-(9y - 8z)/(3y) + (5y + 8z)/(3y)
= (-9y + 8z)/(3y) + (5y + 8z)/(3y)
= (-9y + 8z + 5y + 8z)/(3y)
= (-9y + 5y + 8z + 8z)/(3y)
= (-4y + 16z)/(3y)
= -4(y - 4z)/(3y)

Presentation is suspect due to lack of brackets.

Will take a GUESS that you MIGHT mean :-

- [ (9y - 8z) / (3y) ] + ( 5y + 8z ) / (3y)

[ (8z - 9y) / (3y) ] + ( 5y + 8z ) / (3y)

8z - 9y + 5y + 8z
------------------------------
3y

16 z - 4y
---------------
3y

PS
You MUST use brackets when presenting a question.

- (9y-8z/3y)+(5y+8z/3y)=-9y+8z/3y+5y+8z/3y=-4y+16z/3y=(16z-12y)/3y

you do it one step at a time paying attention to order of operations

-(9y-8z/3y) + (5y+8z/3y) The problem
-9y + 8z/3y + 5y + 8z/3y Distribute the - (technically it is the same as times -1)
-9y + 5y +(8z/3y) + (8z/3y) group the terms added () to show the division as a fraction
-4y + 16z/3y sum the terms (the strange part of this is adding the 2 fractions)

This can be rewritten as:

(-1)(9y-8z/3y) + (5y+8z/3y)

Distribute the -1 first, and remove the parentheses:

-9y + 8z/3y + 5y + 8z/3y

Combine like terms:

-4y + 16z / 3y

Get a common denominator (3y)

-12y^2/3y + 16z/3y

Combine to get the final answer:

(-12y^2 + 16z) / 3y