it is two fractions added together. the fractions are:
2x "over" 2x+2 "plus" 1 "over" x+1
thanx
2007-07-19 17:08:02 · 9 answers · asked by B. 2 in Science & Mathematics ➔ Mathematics
= 2x / (2x + 2) + 1 / (x + 1)
= 2x / (2) (x + 1) + 1 / (x + 1)
= x / (x + 1) + 1 / (x + 1)
= (x + 1) / (x + 1)
= 1
2007-07-21 21:07:26 · answer #1 · answered by Como 7 · 0⤊ 0⤋
1
2007-07-19 17:13:14 · answer #2 · answered by John S2005 3 · 0⤊ 0⤋
2x / (2x+2) + 1 / (x+1)
factor out the 2
= 2x / 2 (x+1) + 1 / (x+1)
= x / (x + 1) + 1/ (x+1)
add the two factors together (we have the same denominator.
= (x+1) / (x+1)
= 1
2007-07-19 17:11:30 · answer #3 · answered by Sam 3 · 0⤊ 0⤋
I assume you mean
2x / (2x + 2) + 1/(x + 1) = x/(x + 1) + 1/(x + 1)
= (x + 1)/(x + 1) = 1
2007-07-19 17:12:43 · answer #4 · answered by Northstar 7 · 0⤊ 0⤋
Simplify the problem by dividing everything in the first fraction by 2. Then you have:
(x/(x+1))+(1/(x+1))
Since they have a common denominator now, you can add the numerators to get:
(x+1)/(x+1) which cancel each other out, leaving you with 1.
2007-07-19 17:11:54 · answer #5 · answered by sarajschryver 2 · 0⤊ 0⤋
2x / (2x+2) + 1 / (x+1)
= 2x / [2(x+1)] + 1 / (x+1), cancel the 2s:
= x / (x+1) + 1 / (x+1)
Since they have the same denominator we can just add the numerators:
= (x + 1) / (x + 1)
= 1, x â -1.
2007-07-19 17:11:22 · answer #6 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
[2x/(2x+2)] + [1/(x+1)]
= [x/ (x+1)] + [1/(x+1)]
because the base is the same add the numerators
to get,
(x+1) / (x+1)
= 1
hope this helps!!! :-)
2007-07-19 17:15:12 · answer #7 · answered by Sindhoor 2 · 0⤊ 0⤋
2x/(2x+2) + 1/(x+1)
= x/(x+1) + 1/(x+1)
= (x+1)/(x+1)
= 1
2007-07-19 17:11:25 · answer #8 · answered by sahsjing 7 · 0⤊ 0⤋
2x/(2x+2) + (1)/(x+1) =2x/(2x+2) + (1*2)/(2(x+1))
= 2x/(2x+2) + 2/(2x+2)
=(2x+2)/(2x+2)
=1
2007-07-19 17:14:16 · answer #9 · answered by eric l 6 · 0⤊ 0⤋