anonymous
  • anonymous
Solve for y.
Biology
schrodinger
  • schrodinger
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anonymous
  • anonymous
|dw:1341889966376:dw|
zepp
  • zepp
Cross-multiply.
zepp
  • zepp
\[(y-1)(a+b)=(y+1)(a-b)\\(y-1)a+(y-1)b=(y+1)a-(y+1)b\\ay-a+by-b=ay+a-by+1b\\ay-a+by-b-ay-a+by-1b=0\\-2ay-2a+2y-2b=0\]

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zepp
  • zepp
And it's impossible to solve for y, as there's 3 variable and only 1 equation.
anonymous
  • anonymous
but they gave these 3 choices : A. |dw:1341890553510:dw|
Kainui
  • Kainui
Simply subtract all the terms out that don't contain y, and then factor out the y-term and divide both sides by all the non-y stuff to isolate it.
anonymous
  • anonymous
can u show me?
Kainui
  • Kainui
So here is an example: 2xy+3y-x=0 take out all the non-y terms to try to isolate them by subtracting them from both sides. 2xy+3y=x now factor out the y y(2x+3)=x divide out the non-y stuff! y=x/(2x+3) there you go, that example should help you out!
anonymous
  • anonymous
i dont get it
Kainui
  • Kainui
What don't you get? I can help you but I wont' do your homework for you.
anonymous
  • anonymous
Simply subtract all the terms out that don't contain y, and then factor out the y-term and divide both sides by all the non-y stuff to isolate it. what u said
Kainui
  • Kainui
So basically all you're doing here is trying to get y all by itself, you're solving for y, right? Follow what the guy up there did to get y into an easier form to look like this, −2ay−2a+2y−2b=0 This way you won't have to deal with fractions or anything weird. When you do that, you will notice that some of the terms have a y in them and some don't. So then you can subtract the terms without y in them so they're on the other side. When they're on the other side, you can divide y out. for example if you have y(2x+3) if you multiply that out you get 2xy+3y, so really you're just going backwards to try to extract the y from it.
PaxPolaris
  • PaxPolaris
use the rule if...\[\Large \frac mn = \frac pq\] then ...\[\Large {m+n \over m-n}={p+q \over p -q}\] ...the inverse is also true
PaxPolaris
  • PaxPolaris
or if you are using zepp's equations ... they are only right til the 2nd step \[(y-1)(a+b)=(y+1)(a-b) \]\[\implies a(y-1)+b(y-1)=a(y+1)-b(y+1) \]\[\implies \Large \cancel{ay}-a+by-\cancel b=\cancel{ay}+a-by-\cancel b\] now: bring all the terms with y on one side of the = sign, and all non-y terms to the other\[\implies \Large 2b \cdot y=2a\]\[\Large \implies y= \frac ab\]
Carniel
  • Carniel
This is for Biology group people.Plz go there to have your question answered.d Btw guys your not suppose to answer the question if its in the wrong group. Tell him or her to go to the correct group. Thx ^_^
anonymous
  • anonymous
[(-2a*(a+b))/(a-b-1) =y\]

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