Question

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))

Answer 1

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 ;)

Answer 2

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?

Answer 3

That's correct, I realize what I did wrong when I tried working this problem out before. Silly mistakes. Thank you so much! c:

Answer 4

Anytime :-)

Answer 5

It is correct, but I think there is a more elegant way:
\[\frac{3}{x+1}-\frac{1}{2}=\frac{1}{3x+3}=\frac{1}{3(x+1)} \\
\frac{3}{x+1}(x+1)-\frac{1}{2}(x+1)=\frac{1}{3(x+1)}(x+1) \\
3-\frac{x}{2}-\frac{1}{2}=\frac{1}{3} \Leftrightarrow x=\frac{13}{3}\]

Answer 6

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
\[ 18 - 3(x+1) = 2 \\ -3x-3= -16 \\ -3x= -13\\x= \frac{13}{3}\]

Answer 7

...how elegant ;) haha "all roads lead to rome"

Answer 8

But some roads are faster ;)

Answer 9

Oooh touché. Hahah