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anonymous
 4 years ago
Solve for y.
anonymous
 4 years ago
Solve for y.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1341889966376:dw

zepp
 4 years ago
Best ResponseYou've already chosen the best response.1\[(y1)(a+b)=(y+1)(ab)\\(y1)a+(y1)b=(y+1)a(y+1)b\\aya+byb=ay+aby+1b\\aya+bybaya+by1b=0\\2ay2a+2y2b=0\]

zepp
 4 years ago
Best ResponseYou've already chosen the best response.1And it's impossible to solve for y, as there's 3 variable and only 1 equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but they gave these 3 choices : A. dw:1341890553510:dw

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0Simply subtract all the terms out that don't contain y, and then factor out the yterm and divide both sides by all the nony stuff to isolate it.

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0So here is an example: 2xy+3yx=0 take out all the nony 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 nony stuff! y=x/(2x+3) there you go, that example should help you out!

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0What don't you get? I can help you but I wont' do your homework for you.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Simply subtract all the terms out that don't contain y, and then factor out the yterm and divide both sides by all the nony stuff to isolate it. what u said

Kainui
 4 years ago
Best ResponseYou've already chosen the best response.0So 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
 4 years ago
Best ResponseYou've already chosen the best response.0use the rule if...\[\Large \frac mn = \frac pq\] then ...\[\Large {m+n \over mn}={p+q \over p q}\] ...the inverse is also true

PaxPolaris
 4 years ago
Best ResponseYou've already chosen the best response.0or if you are using zepp's equations ... they are only right til the 2nd step \[(y1)(a+b)=(y+1)(ab) \]\[\implies a(y1)+b(y1)=a(y+1)b(y+1) \]\[\implies \Large \cancel{ay}a+by\cancel b=\cancel{ay}+aby\cancel b\] now: bring all the terms with y on one side of the = sign, and all nony terms to the other\[\implies \Large 2b \cdot y=2a\]\[\Large \implies y= \frac ab\]

Carniel
 4 years ago
Best ResponseYou've already chosen the best response.1This 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
 4 years ago
Best ResponseYou've already chosen the best response.0[(2a*(a+b))/(ab1) =y\]
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