ok..sorry.. another problem i cant solve but its a diff. type..?

A

ADIBUGATTI-THELIVERPOOLFAN

New Member
my textbook says<br />
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perform each occasion<br />
<br />
(2y^2 + 9y + 7) / (y+1)<br />
<br />
(here's a snapshot of my paper with problem writtn on it)<br />
<br />
http//tinypic.com/view.php?pic=20l14eq&s=3<br />
<br />
thank you all once again!<br />
 
answer is 2y+7

you added 9y and 2y, that is the mistake
actually you have to subtract 9y and 2y
 
I see your mistake! When you multiplied your 2y by y+1 you got 2y^2+2y which is correct. You placed this answer underneath the 2y^2+9y which is also correct. You then subtracted 2y^2+2y from 2y^2+9y which is the correct idea, but you forgot there are implied paranthesis in that step.

look at this idea from a different perspective. Suppose you need to perform (2y^2+9y)-(2y^2+2y). Because of the paranthesis you would DISTRIBUTE the - sign and get 2y^2+9y-2y^2-2y giving 7y. But because of your sign error, you got 11y and then got the problem wrong. If you had done it correctly, 7y+7 would have been the next group, and you will be able to get a nice round answer.

Always put paranthesis around those groups to avoid sign errors!
 
(2y^2 + 9y + 7) ...when you factor it, it becomes (2y + 7) (y + 1). since it is all over (y +1) the two (y + 1)'s cancel out and you are left with (2y + 7).

if you want to find y.. then 2y=-7. y= -7/2
 
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