anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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phi
  • phi
first, do you see you can simplify the 2nd fraction? factor out a 2 from top and bottom and cancel it.
anonymous
  • anonymous
thanks, be right back..
anonymous
  • anonymous
ok done

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phi
  • phi
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
  • anonymous
my denominators are: (y+2), 4(y+2), and (y+2)
phi
  • phi
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
  • phi
so multiply the 3rd fraction by 2/2 (2 times the top, 2 times the bottom)
anonymous
  • anonymous
do you want original problem? the above was factored.
anonymous
  • anonymous
3y/2y+4 + 2y-16/4y+8 = 3y-2/y+2
phi
  • phi
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
  • anonymous
I canceled 2 from left side and top
anonymous
  • anonymous
I know, i did that
phi
  • phi
\[\frac{2y-16}{4y+8}= \frac{\cancel{2}(y-8)}{\cancel{2}2(y+2)}\]
phi
  • phi
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
  • phi
Follow? now multiply the 3rd fraction by 2/2
anonymous
  • anonymous
why the 4? and not the 2 outside parenthesis? or can you do both?
phi
  • phi
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
  • anonymous
i got 4(y+2)
phi
  • phi
for what?
anonymous
  • anonymous
when i rewrote 4y+8
phi
  • phi
OK, I with you. now up top, factor out a 2 from 2y-16
anonymous
  • anonymous
I factored everything, its the canceling thats being difficult
phi
  • phi
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
  • anonymous
I know but I did that, when you do, can you only cancel one thing, once?
phi
  • phi
Let's look at the 2nd fraction \[ \frac{2y-16}{4y+8}= \frac{2(y-8)}{4(y+2)} \]
phi
  • phi
divide the top by 2 and the bottom by 2 (that is the same as canceling 2)
anonymous
  • anonymous
so only two left? then put 2 on other side of =?
phi
  • phi
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
  • anonymous
yes
phi
  • phi
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
  • anonymous
6y-4 now at top?
phi
  • phi
on the right hand side. now combine the left hand side
anonymous
  • anonymous
4y-8=6y-4
phi
  • phi
now solve for y: subtract 4y from both sides add +4 to both sides.
anonymous
  • anonymous
=-2
phi
  • phi
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
  • anonymous
darn lol well thanks. Im sorry for all the trouble.
phi
  • phi
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
  • anonymous
aww, well bye :)

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