anonymous
  • anonymous
solve the equation. 1/ 3x + 9 - 2/ x + 3 = 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ 1 }{ 3x + 9 } - \frac{ 2 }{ x + 3 } = 2\]
Jhannybean
  • Jhannybean
oh okay.
anonymous
  • anonymous
?

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More answers

Luigi0210
  • Luigi0210
you got this?
anonymous
  • anonymous
No. I don't know why she said oh okay.
bahrom7893
  • bahrom7893
I GOT THIS YALL!
Jhannybean
  • Jhannybean
No lol. im guessing we're going to multiply the fractions with the denominators of both... can't find an easy common denominator :(
Luigi0210
  • Luigi0210
Alright,YEA BAHROM!
Luigi0210
  • Luigi0210
here I'll start you guys off:|dw:1370400070613:dw|
bahrom7893
  • bahrom7893
http://www.reactiongifs.com/wp-content/uploads/2012/12/you-got-it-dude.gif
anonymous
  • anonymous
Could you try using the equation editor? That's super hard to read
anonymous
  • anonymous
lol @bahrom7893
Luigi0210
  • Luigi0210
Ha, your awesome Bahrom \[\frac{ 1 }{ 3x+9 }*(3x+9)-\frac{ 2 }{ x+3 }*(3x+9)=2(3x+9)\]
anonymous
  • anonymous
Now what?
Luigi0210
  • Luigi0210
multiply it out, what do you get?
anonymous
  • anonymous
I don't know how. so many numbers. so confused.
Luigi0210
  • Luigi0210
|dw:1370400991389:dw|
anonymous
  • anonymous
1- 6 = 2(3x +4)?
Luigi0210
  • Luigi0210
that's a 9 sorry :P
anonymous
  • anonymous
It's okay. So, 1 - 6 = 2(3x + 4). Now what?
Luigi0210
  • Luigi0210
1-6=2(3x+9) Distribute and solve :)
anonymous
  • anonymous
Idk how to distribute. I thought I told you that. lol
Jhannybean
  • Jhannybean
wait... \[\large \frac{ 1 }{ 3x+9 }*(3x+9)-\color{red}{\frac{ 2 }{ x+3 }*(3x+9)}=2(3x+9)\]how did you cancel out the denominator there? The highlighted part.
Luigi0210
  • Luigi0210
Where have you been? D: and factor out a 3: 3(x+3)
Jhannybean
  • Jhannybean
woudn't you still have 3 left over?...
Luigi0210
  • Luigi0210
yup so multiply to the 2
Jhannybean
  • Jhannybean
2/ (x+3) * [(3x+9)/3(x+3)] = 2(3x+9)/3?
Luigi0210
  • Luigi0210
\[\frac{ 2 }{ x+3 }*3x+9=\frac{ 2(3)(x+3) }{ x+3 }\]
Jhannybean
  • Jhannybean
\[\frac{2}{\cancel{x+3}}*\frac{3\cancel{(x+3)}}{3(x+3)}= \frac{6}{3(x+3)}= \frac{2}{x+3}\]
Luigi0210
  • Luigi0210
|dw:1370401905448:dw| Sorry we've been ignoring you @Anonymous120 D:
Jhannybean
  • Jhannybean
oh okay nvm then.
anonymous
  • anonymous
It's okay
Luigi0210
  • Luigi0210
I think I messed up somewhere.. Got a fraction answer :/
anonymous
  • anonymous
Answer choices are a. x=1 b. x= -23/6 c. x= -5/2 d. x= -5
whpalmer4
  • whpalmer4
\[\frac{1}{3x+9}-\frac{2}{x+3} = 2\]Notice that \(3x+9 = 3(x+3)\). Multiply both sides by \(3(x+3)\) to remove fractions: \[\frac{1}{\cancel{3x+9}}*\cancel{(3(x+3))}-\frac{2}{\cancel{x+3}}*(3\cancel{(x+3)}) =2*3(x+3)\]\[1-2*3=2*3(x+3)\]\[-5=6x+18\]\[6x=-23\]\[x=\]
johnweldon1993
  • johnweldon1993
right @Luigi0210 I think you got it right! :)
Luigi0210
  • Luigi0210
Okay, i was wondering if I did something wrong.. thanks for the support
whpalmer4
  • whpalmer4
Yep, you got it right, and given how often textbook problems come out to have nice, round answers, it's understandable that you were suspicious. Personally, I think students would be better off if they had a few more problems with answers like -23/6 and not quite so many with answers like 2 :-)
Luigi0210
  • Luigi0210
Yea, thanks Palmer and John :)
Jhannybean
  • Jhannybean
\[\large \frac{ 1 }{ 3x + 9 } - \frac{ 2 }{ x + 3 } = 2\]\[\large \frac{ 1 }{ 3x+9 }*\frac{{3x+9}}{3x+9}-{\frac{ 2 }{ x+3 }*\frac{3x+9}{3x+9}}\]\[\large \frac{1}{3x+9}-\frac{6}{3x+9}=2\]\[\large \frac{-5}{3x+9}=2\]\[\large -5=2(3x+9)\]\[\large -5=6x+18\]\[\large -23=6x\]
Jhannybean
  • Jhannybean
lol..... John :)
anonymous
  • anonymous
So it's -23/6?
Luigi0210
  • Luigi0210
And thank you too Jhann :P
Luigi0210
  • Luigi0210
that would be correct
anonymous
  • anonymous
Thank you all so much. If I could give everyone medals I would lol
Jhannybean
  • Jhannybean
lol np
Jhannybean
  • Jhannybean
Your problems are so ridiculously long :\
Luigi0210
  • Luigi0210
I'll take metal.. I prefer Ag
Jhannybean
  • Jhannybean
AUUUUUUU
Luigi0210
  • Luigi0210
*Au
Luigi0210
  • Luigi0210
err -_-
anonymous
  • anonymous
@Luigi0210 comment on one of my old questions with no best response and I'll give you a medal
johnweldon1993
  • johnweldon1993
lol nice attempt @Luigi0210 :P
Luigi0210
  • Luigi0210
Haha, you can't blame me for trying :)
Luigi0210
  • Luigi0210
And naw, I'd rather earn them honestly
anonymous
  • anonymous
http://openstudy.com/users/anonymous120#/updates/51ae9e2de4b05b167ed21b6b I woulda given you the orginal medal but somehow you already got 3 lol. And you definetly did earn it! Teaching me is not easy!
Luigi0210
  • Luigi0210
Good day everyone!

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