3t=4/6-1/6=3/6
t=1/6
2006-12-31 03:10:45
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answer #1
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answered by raj 7
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3t+1/6=2/3
<=> 3t = 2/3 - 1/6 <=> t = 1/6
2006-12-31 11:13:02
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answer #2
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answered by James Chan 4
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3 T + 1/16 = 2/3
3 T = 2/3 - 1/16
3 T = (2/3 - 1/16) divided by 48 common demonator of 3 and 16
3 T = (32 - 3)/48
3 T = 29/48
T = 29/48 x 1/3
T = 29/144
Proof: 3 T - 1/16 = 2/3
3 T = 29/48
1/16 = 3/48
29/48 + 3/48 = 32/48
32/48 reduces to 2/3
Therefore T = 29/144
some of the other answers say that T = 1/16
but they do not prove to be 1/16
ie: 3T + 16 = 2/3
3 T = 3 x 1/16 = 3/16
3 T + 1/16 is supposed to equal 2/3
but 3 T = 3/16 + 1/16 = 1/4 so it is they are the wrong answers
2006-12-31 13:11:06
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answer #3
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answered by David C 2
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3t+1/6=2/3
=> 3t= 2/3-1/6
=> 3t = 3/6
=> t=1/6
2006-12-31 11:39:25
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answer #4
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answered by Riddhi 2
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3t + 1/6 = 2/3
3t = 2/3 - 1/6
3t = 1/2
t = 1/2 divided by 3
t = 1/6
2006-12-31 11:14:15
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answer #5
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answered by summerglow 5
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3t=2/3-1/6 ( Make varible on 1 side).
3t=3/6
t=3/6 divide by 3 (divide by coffeient)
t=1/6
2007-01-01 15:45:57
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answer #6
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answered by lulu 3
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to do this problem turn the 2/3 into 4/6. Now it's easy. obviously 3/6 + 1/6 = 4/6 so it would be 3 times 1/6 + 1/6 = 4/6. so t=1/6.
2006-12-31 11:21:29
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answer #7
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answered by Ring Ring Ring Bananaphone 5
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If ( 3t +1) /6 =2/3
cross multiply,
3(3t+1) =2 *6
3t+1 =4
3t=3
t=1
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If 3t + 1/6 =2/3
3t =2/3 -1/6
=4/6 -1/6
3/6 =1/2
t=1/6
2006-12-31 11:12:29
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answer #8
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answered by amudwar 3
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First: get rid of fractions > take the denominators and multiply them by everything and combine:
6(3)(3t) + 6(3)(1/6) = 6(3)(2/3)
54t + 3 = 12
Second: solve for "t" and keep it on one side by itself > subtract 3 from both sides (opposite side means opposite sign):
54t + 3 -3 = 12 - 3
54t = 9
Third: divide both sides by 54:
54t/54 = 9/54
t = 9/54 simplify into...
t = 1/6
2006-12-31 18:29:57
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answer #9
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answered by ♪♥Annie♥♪ 6
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well, make the two fractions common fractions. then subtract 1/6 from both sides and you should end up with 3t on one side and a fraction on the other. sorry if this answer is vague, but see if you can try to solve this yourself
2006-12-31 11:11:00
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answer #10
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answered by bizzle 1
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