First, recognize that 3/9 is 3/(3 x 3) = 1/3 So,
1/3 + 1/4
Now, we need to get the denominators of the fractions the same so we can add them. What number is divisible by both 3 and 4? Well, 12 will do (it is the smallest), but any number divisible by both 3 and 4 would do. Let's use 12. We multiply each fraction by 12/12 (that's just 1, and multiplying anything by 1 gives us that thing back again). So,
(1/3)(12/12) + (1/4)(12/12) =
12/(3 x 12) + 12/(4 x 12)
But that is just (dividing 12 by 3 on the left and 12 by 4 on the right)
4/12 + 3/12 = 7/12
To show you that any number divisible by 3 and 4 will do, let's pick 48 (UGH!). We do the same thing:
(1/3)(48/48) + (1/4)(48/48) = 48/(3 x 48) + 48/(4 x 48) =
16/48 + 12/48 = 28/48 = (4 x 7)/(4 x 12) = 7/12
HTH
Charles
2007-10-09 15:21:26
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answer #1
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answered by Charles 6
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3/9 + 1/4 = 1/3 + 1/4 = (4 + 3)/12 = 7/12
2007-10-09 22:14:23
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answer #2
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answered by Ivan B 5
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First make them equal denominator
= 3(4) / 9(4) + 1(9) / 4(9)
= 12 / 36 + 9 / 36
= 21 / 36
= 7 / 12
2007-10-09 22:14:31
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answer #3
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answered by Rayan Ghazi Ahmed 4
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12 / 36 + 9 / 36 = 21 / 36 = 7 / 12
2007-10-10 14:34:23
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answer #4
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answered by Como 7
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First get the Least Common Denominator (LCD), which, in this case is 36.
Then multiply 36 by each fraction. Use cancellation:
12/36 + 9/36 = 21/36
To reduce to its simplest term, divide both the numerator and the denominator by their Greatest Common Factor (GCF), which, is 3. The answer would be:
7/12
2007-10-09 22:12:50
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answer #5
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answered by edith p 3
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3/9 + 1/4
=1/3 + 1/4
=4/12 + 3/12
= 7/12.
Simple!
2007-10-09 22:12:31
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answer #6
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answered by student 2
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So, first of all make them have equivalent denominators, to add them. Let's have the denominator 36.
3/9=12/36 and 1/4=9/36.
So, adding numerators, 12+9=21/36.
Reduce and that's, 7/12
2007-10-09 22:12:23
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answer #7
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answered by smile ♥ 4
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7/12
2007-10-09 22:10:17
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answer #8
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answered by john 4
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12/36 + 9/36 = 21/36
Reduce that and it's
7/12
2007-10-09 22:09:28
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answer #9
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answered by Ms. Exxclusive 5
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21/36
2007-10-09 22:10:39
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answer #10
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answered by ElectroJewelz 5
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