## samyflashy Group Title Solve for y. 2 years ago 2 years ago

1. samyflashy Group Title

|dw:1341889966376:dw|

2. zepp Group Title

Cross-multiply.

3. zepp Group Title

$(y-1)(a+b)=(y+1)(a-b)\\(y-1)a+(y-1)b=(y+1)a-(y+1)b\\ay-a+by-b=ay+a-by+1b\\ay-a+by-b-ay-a+by-1b=0\\-2ay-2a+2y-2b=0$

4. zepp Group Title

And it's impossible to solve for y, as there's 3 variable and only 1 equation.

5. samyflashy Group Title

but they gave these 3 choices : A. |dw:1341890553510:dw|

6. Kainui Group Title

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.

7. samyflashy Group Title

can u show me?

8. Kainui Group Title

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!

9. samyflashy Group Title

i dont get it

10. Kainui Group Title

11. samyflashy Group Title

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

12. Kainui Group Title

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.

13. PaxPolaris Group Title

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

14. PaxPolaris Group Title

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$

15. Carniel Group Title

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 ^_^

16. ariushomer Group Title

[(-2a*(a+b))/(a-b-1) =y\]