Can someone help me in my math?

bc+3bd-2ac-6ad
------------------------
2bc+ac+3ad+6bd

...b(c+3d)-2a(c+3d)
=---------------------------
...c(a+2b)+3d(a+2b)

...(b-2a)(c+3d)
=-------------------
...(c+3d)(a+2b)

... b-2a
=------------ answer//
.. a+2b

wait,,,is the left side of the slash (/) all over the right side??

like this: (bc+3bd-2ac-6ad)/(2bc+ac+3ad+6bd)??

then:
[ b(c+3d) - 2a(c+3d) ] / [ c(2b+a) + 3d(a+2b) ] =>(partial factoring)
= [ (b-2a)(c+3d) ] / [ (c+3d)(a+2b) ] => simplifying (factoring completely)
= (b-2a) / (a+2b) => final answer!!

ciao!

(bc + 3bd - 2ac - 6ad)/(2bc + ac + 3ad + 6bd)
= [(bc + 3bd) - (2ac + 6ad)]/[ac + 2bc + 3ad + 6bd]
= [b(c + 3d) - 2a(c + 3d)]/[(ac + 2bc) + (3ad + 6bd)]
= [(c + 3d)(b - 2a)]/[c(a + 2b) + 3d(a + 2b)]
= [(c + 3d)(b - 2a)]/[(a + 2b)(c + 3d)]
= (b - 2a)/(a + 2b)

bc+3bd-2ac-6ad/2bc+ac+3ad+6bd
Just cancel out the same terms and you'll get
1/2bc+1/2bd-2ac-2ad

First, you factorise both numerator and denominator:
b( c+3d ) - 2a( c+3d ) / c ( 2b + a ) + 3d ( a +2b )

Then, you put it together:
( b - 2a ) ( c + 3d ) / ( c + 3d ) ( a +2b )

Then, you simplify the equation, thus removing both ( c + 3d ):
( b - 2a ) / ( a +2b )
This is the answer.

Well first we're going to have to simplify down that equation to look at it better.

bc+3bd-2ac-6ad
--------------------------
2bc+ac+3ad+6bd

From here, you can see that you can take out a few terms, right? Nothing can actually be combined in this problem, as of yet.

b(c+3d)-2a(c+3d)
----------------------------
c(2b+a)+3d(a+2b)

Here is where you can combine like terms. Just think of this example for every time you get stuck on a test on one of these problems: xy-ay=y(x-a), because you still just multiply y to both of the terms and you get the same thing. Here's what it simplifies to:

(b-2a)(c+3d)
--------------------
(c+3d)(b+2a)

From here you can see that the (c+3d) term can cancel each other out, because the same numerator over the same denominator will cancel each other out:

b-2a
--------
b+2a

And no, you can't cancel those terms out because 1) they're unknown variables and 2) they're not like terms.