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Can anyone give this a whirl and see what you come up with. :D
Im trying to solve for y.
2pn - (y/3) = 3n + ay + p
6pn -y = 9n + 3ay + 3p
-y = 9n + 3ay + 3p - 6pn
-3ay -y = 9n + 3p -6pn
Now here is where im not sure what to do.
How do i factor the left hand side? y(-1 - 3a) or -(y + 3ay)?
Also do i have to factor the right hand side? 3(3n + p + -2pn)
y = (3(n(2p-3)-p))/(3a+1)
Sorry for the delay. I had to take a second to look at the format your answer was in. :lol: Do you know where i might have went wrong or was i on the right track?
No, you were perfectly fine
AS you are finding for y, you should factor the left-hand-side with the y outside the brakets, as you'll have to divide by whatever is in the brackets later. Also, you don't have to factorise the right-hand-side unless asked to 'simplify'. Otherwise its entirely optional, but usually looks nicer if you do
answer (with p, not n, factored, so the p's don't feel left out:
y=3 ( p (1-2n) + 3n) / - (3a+1)
...
y = 3 ( p (2n-1) - 3n) / (3a + 1)
you will notice here, and with DiGiTaIErRoR, that the top line was negated to remove the negative sign of the denominator, also a good thing to do.
2pn - (y/3) = 3n + ay + p
6pn -y = 9n + 3ay + 3p
-y = 9n + 3ay + 3p - 6pn
-3ay -y = 9n + 3p -6pn
y(3a+1) = 3(2pn-3n-p)
finally
y = 3(2pn-3n-p)/(3a +1)
Thanks for the help guys. I really appreciate it. :D