12d - 5d =14 - 7
7d = 7
d = 1
2007-11-08 19:12:53
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answer #1
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answered by Como 7
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12d+7=5d+14
12d-5d=-7+14
7d=7
7d/7= 7/7
d=1
2007-11-08 07:28:10
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answer #2
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answered by Anonymous
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12d + 7 = 5d + 14
subtract 7 from each side
12d = 5d + 7
now subtract 5d from each side
7d = 7
divide each side by 7
d = 1
2007-11-08 07:22:33
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answer #3
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answered by Melissa 3
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d=1
2007-11-08 07:24:58
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answer #4
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answered by K F 2
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1) rearrange it first so d's are on one side and numbers are on the other (HINT: when moving an item from one side of the equation to another it changes form + to - and vica versa)
12d+7=5d+14
12d-5d+7=14
12d-5d=14-7
2) simplify
12d-5d=14-7
7d=7
3) solve: divide both sides by 7)
7d=7
d=1
the value of d is 1
2007-11-08 07:32:43
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answer #5
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answered by Crispy (Chris P) 2
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d=1
To calculate:
12d+7 -7=5d+14-7 which gives
12d = 5d+7
12d -5d = 5d +7 -5d which gives
7d = 7 therefore
d=1
2007-11-08 07:22:30
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answer #6
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answered by Simon S 2
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12d+7=5d+14
-5d -5d
7d+7=14
-7 -7
7d=7
d=1
2007-11-08 07:36:27
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answer #7
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answered by Anonymous
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Ok, this one's easy.
12d+7=5d+14
Subtracting 5d from each side,
7d+7 = 14
Now subtracting 7 from each side,
7d = 7
Dividing by 7, we get
d=1
And there's your answer, d=1.
Do please feel free to drop me a line if you'd like any assistance in generating any other logical number puzzles; this one was a bit too simple methinks.
2007-11-10 05:23:17
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answer #8
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answered by general_ego 3
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ok first you want to get the ds on one side and the number on one side
so lets move the 5d to the left. in order to cancel that out you subtract 5d on both sides so it gets canceled from the right and you have
7d+7=14
2) get the number on the other side and you subtract 7 because it is an addition and to cancel out you have to subtract and you get
7d=7
3) get d alone you divide by 7 on both sides because the opposite of multiplication is division.
and you get
d=1
and that your answer
you can check by plugging it in
12(1) +7= 5(1) +14
and 19=19 so correct!
hope it helps and good luck
2007-11-08 07:27:18
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answer #9
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answered by tobedoc 3
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