|
Question 187695This question is from textbook Introductory and Intermediate Algebra
: Hi i'm trying to work out on the practice Test on my txtbook & i'm not sure with my answer can you please help me...
48)Denominators are additive inverses.Add pr subtract as indicated.Simplify the result if possible.
(y-7)/(y^2-16) + (7-y)/(16-y^2)
I did the cross multiplication
(16y-y^3-112+7y^2) / (7y^2-y3-112+16y)
(-y^3 +7y^2 +16y-112)/(-y^3 +7y^2 +16y-112)
what will be the answer is it 1 or 0?
Did I do the right thing? or there is another solution?
Looking forward for your reply.
Thanks
KAttie
This question is from textbook Introductory and Intermediate Algebra
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
OK. LET'S SEE. I'LL DO IT AND THEN WE'LL COMPARE.
(y-7)/(y^2-16) + (7-y)/(16-y^2)
Cross multiplying, we get:
(y-7)(16-y^2)=-y^3+7y^2+16y-112
(7-y)(y^2-16)=-y^3+7y^2+16y-112
Now for the denominator:
(y^2-16)(16-y^2)=-y^4+32y^2-256
I THINK THIS IS WHERE YOU MAY HAVE GOTTEN OFF TRACK
Now, putting the numerator and denominator together, we have:
{Notice, I'm factoring a -2 out of the numerator and a -1 out of the denominator}
-2(y^3-7y^2-16y+112)/-1(y^4-32y+256) and this equals
2((y-7)(y^2-16))/(y^2-16)^2 cancel out the (y^2-16) and we get
2(y-7)/(y^2-16)------------------I think this is the answer
Look at the second term; if we factor out -1 from the numerator and denominator, what do we get?
(y-7)/(y^2-16)+(-1)(y-7)/(-1)(y^2-16)) Whoaaaa!! the -1's cancel and we now have:
y-7)/(y^2-16)+(y-7)/(y^2-16)) =2(y-7)/(y^2-16)
Maybe an easier way???
Hope this helps, Kattie---ptaylor
|
|
|
| |
