solve c^2+3ac-8a-4b=0 for a (please check work)

[awesome_m]

they ask me to solve for a:
c^2+3ac-8a-4b=0

this is what i did
c^2+3ac-8a-4b=0
i added 4b to both sides
c^2+3ac-8a=0+4b
i subtracted c^2 to both sides
3ac-8a=0+4b-c^2
i divided 3c to both sides
a-8a=4b-c^2/3c

now what do i do?
thank you!

 

[tkhunny]

Re: is this right? i need someone to check it please?!

they ask me to solve for a:
c^2+3ac-8a-4b=0

this is what i did
c^2+3ac-8a-4b=0
i added 4b to both sides
c^2+3ac-8a=0+4b

No need to track the zero. it would have been fine like this:
c^2 + 3ac - 8a = 4b

i subtracted c^2 to both sides
3ac-8a=0+4b-c^2
i divided 3c to both sides
a-8a=4b-c^2/3c

You back up and remember the Distributive Property of Multiplication over Addition and Parentheses.

1) That c^3/3c is very awkward. Notice how you meant (4b - c^2)/3c. That is a very different animal.
2) Why did your division fail to do anything to the 8a?

Back up to here:

3ac - 8a = 4b - c^2

Factor out the 'a', using the Distributive Property of Multiplication over Addition

a(3c - 8) = 4b - c^2

Remember that?

Divide both sides by '3c - 8'

a = (4b - c^2)/(3c - 8)

 

[awesome_m]

Re: is this right? i need someone to check it please?!

oh ok. i see what i did wrong. thank you very much!!