3y/2(y+2) + 2y-16/4(y+2) = 3y-2/y+2 I need to solve this? Somehow ^^ I cant find LCD to it

Question

3y/2(y+2) +  2y-16/4(y+2) =  3y-2/y+2   I need to solve this? Somehow ^^ I cant find LCD to it

phi · Answer

first, do you see you can simplify the 2nd fraction?  factor out a 2 from top and bottom and cancel it.

anonymous · Answer

thanks, be right back..

anonymous · Answer

ok done

phi · Answer

If you do, then you see that the 2 fractions on the left have the same denominator.  Also, the 3rd fraction only needs a 2 in the bottom to make its denominator the same as the others.

anonymous · Answer

my denominators are:  (y+2), 4(y+2), and (y+2)

phi · Answer

not following you.  The first
3y/2(y+2)     has 2(y+2)
the 2nd 
2y-16/4(y+2)  =  2(y-8)/(4(y+2))= (y-8)/(2(y+2)) has 2(y+2)
the 3rd
3y-2/y+2   has (y+2)

phi · Answer

so multiply the 3rd fraction by 2/2  (2 times the top, 2 times the bottom)

anonymous · Answer

do you want original problem? the above was factored.

anonymous · Answer

3y/2y+4   +   2y-16/4y+8   =  3y-2/y+2

phi · Answer

Yes that is the same as what you posted.  do you see you can simplify the 2nd fraction by canceling a 2 from the top and bottom?

anonymous · Answer

I canceled 2 from left side and top

anonymous · Answer

I know, i did that

phi · Answer

$$\frac{2y-16}{4y+8}= \frac{\cancel{2}(y-8)}{\cancel{2}2(y+2)}$$

phi · Answer

Just fixed it.  the denominator is 2(y+2)

so your problem is now
$$\frac{3y}{2(y+2)} +  \frac{y-8}{2(y+2)} =  \frac{3y-2}{y+2 }$$

phi · Answer

Follow?

now multiply the 3rd fraction by 2/2

anonymous · Answer

why the 4? and not the 2 outside parenthesis? or can you do both?

phi · Answer

you can re-write (4y+8)  as  2(2y+4)   or   4(y+2)    or  2*2*(y+2)
when you distribute each one, they all turn back into 4y+8

anonymous · Answer

i got 4(y+2)

phi · Answer

for what?

anonymous · Answer

when i rewrote 4y+8

phi · Answer

OK, I with you.  now up top, factor out a 2 from 2y-16

anonymous · Answer

I factored everything, its the canceling thats being difficult

phi · Answer

you know that if you have a fraction like
$$ \frac{4}{8}  =\frac{2\cdot 2}{2\cdot 4} = \frac{2}{2} \cdot \frac{2}{4}=\frac{2}{4}$$
because 2/2 = 1 so we can ignore it.

Canceling means find a 2 (for example) in both the  top  and bottom and getting rid of it because it is 2/2 = 1

anonymous · Answer

I know but I did that, when you do, can you only cancel one thing, once?

phi · Answer

Let's look at the 2nd fraction
$$ \frac{2y-16}{4y+8}= \frac{2(y-8)}{4(y+2)} $$

phi · Answer

divide the top by 2  and the bottom by 2  (that is the same as canceling 2)

anonymous · Answer

so only two left? then put 2 on other side of =?

phi · Answer

Slow and steady...
we now have
$$\frac{3y}{2(y+2)} +  \frac{y-8}{2(y+2)} =  \frac{3y-2}{y+2 }$$

Do you agree?

anonymous · Answer

yes

phi · Answer

now multiply the 3rd equation by 2/2 to get a common denominator:
$$\frac{3y}{2(y+2)} +  \frac{y-8}{2(y+2)} =  \frac{2}{2}\frac{(3y-2)}{(y+2) }$$

anonymous · Answer

6y-4 now at top?

phi · Answer

on the right hand side.
now combine the left hand side

anonymous · Answer

4y-8=6y-4

phi · Answer

now solve for y:  subtract 4y from both sides
add +4 to both sides.

anonymous · Answer

=-2

phi · Answer

so y= -2 is the solution.  BUT  (this is awful!)  if we put y=-2 into the original equation, the denominator turns into 0.  We cannot divide by zero, so this particular equation has no solution.

anonymous · Answer

darn lol well thanks.  Im sorry for all the trouble.

phi · Answer

I assume they wanted you to find that this equation had no solution.  There was only one way to find out, which you just did...  I hope you learned a little... got to sign off.

anonymous · Answer

aww, well bye :)