anonymous
  • anonymous
Find the solution. I'm having a lot of trouble with this particular problem, and I know what the answer is, I'm just not sure how to get there. (3/(x+1)) - (1/2) = (1/(3x + 3))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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dape
  • dape
A hint is that 3x+3=3(x+1). Use this and multiply both sides by (x+1) and I'm sure you'll get it ;)
anonymous
  • anonymous
OK! \[\frac{3}{x+1}-\frac{1}{2}=\frac{1}{3x+3}\] \[\frac{6-(x+1)}{2(x+1)}=\frac{1}{3x+3}\] \[\frac{5-x}{2x+2}=\frac{1}{3x+3}\] \[(5-x)(3x+3)=(2x+2)\] \[-3(x+1)(x-5)=2(x+1)\] \[-3(x-5)=2\] \[(15-3x)=2\] \[15-2=3x\] \[13=3x\] \[x=\frac{13}{3}\] Correct?
anonymous
  • anonymous
That's correct, I realize what I did wrong when I tried working this problem out before. Silly mistakes. Thank you so much! c:

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anonymous
  • anonymous
Anytime :-)
dape
  • dape
It is correct, but I think there is a more elegant way: \[\frac{3}{x+1}-\frac{1}{2}=\frac{1}{3x+3}=\frac{1}{3(x+1)} \\ \frac{3}{x+1}(x+1)-\frac{1}{2}(x+1)=\frac{1}{3(x+1)}(x+1) \\ 3-\frac{x}{2}-\frac{1}{2}=\frac{1}{3} \Leftrightarrow x=\frac{13}{3}\]
phi
  • phi
It is helpful to notice that 3x+3 = 3(x+1) then a good step is to multiply both sides (and all terms) by x+1 at the beginning \[ \frac{3}{x+1}-\frac{1}{2}=\frac{1}{3(x+1)} \\ 3 -\frac{1}{2}(x+1)= \frac{1}{3}\] we could multiply both sides by 6 to get rid of the fractions \[ 18 - 3(x+1) = 2 \\ -3x-3= -16 \\ -3x= -13\\x= \frac{13}{3}\]
anonymous
  • anonymous
...how elegant ;) haha "all roads lead to rome"
dape
  • dape
But some roads are faster ;)
anonymous
  • anonymous
Oooh touché. Hahah

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