It's about fractions. Can you help me with these few problems below? Best detalied will get 10 Yahoo points.?

Question

1/3 + 1/4= 1/6 + 4/15=

1/3 + 2/4= 2/3 + 1/4=

1/2 + 1/3= 1/5 + 1/4=

1/5 + 1/3= 2/5 + 1/3=

3/5 - 1/4= 2/3 - 1/5=

Answer 1

I'm not going to do your homework for you.  However, here's a page that shows you how to add and subtract fractions:

http://www.purplemath.com/modules/fracti...

Answer 2

Perhaps these problems were copied incorrectly. The equation given on each line shown above is not true, so you cannot find a solution. I suspect that you really have 10 problems to do, not just 5.

1/3 + 1/4= ______             1/6 + 4/15= ______

1/3 + 2/4= ______             2/3 + 1/4= ______

1/2 + 1/3= ______             1/5 + 1/4= ______

1/5 + 1/3= ______             2/5 + 1/3= ______

3/5 - 1/4=  ______             2/3 - 1/5= ______

Here is how to do the first problem:

When adding 1/3 + 1/4, it is necessary to find a common denominator.  That is a number that both 3 and 4 will go into. In this case, 12 would be a good choice. This is very important if you are trying to add or subtract fractions.

Convert 1/3 into an equivalent fraction with a denominator of 12.  The equation will look like this:  1/3 = x/12.  Solve by cross-multiplying to get 3x = 12, or x = 4.  Thus 4/12 is a fraction equivalent to 1/3.

Next, convert 1/4 into an equivalent fraction with a denominator of 12.  The equation will look like this:  1/4 = x/12.  Solve by cross-multiplying to get 4x = 12, or x = 3.  Thus 3/12 is a fraction equivalent to 1/4.

So, if you are adding 1/3 + 1/4, you can substitute equal fractions to get an equivalent expression.  We will replace 1/3 with 4/12 and we will replace 1/4 with 3/12.  Our new equation looks like this:  4/12 + 3/12 =?  When fractions have the same denominator, you just add the numerators to get the answer.  The equation is saying "What do you get when you add four-twelfths and three twelfths?" The answer is 7/12.

You did not mention the grade you are in.  If you are in grade school, I may have used some words you are not familiar with.  Let's try the second problem with an easier vocabulary.

1/6 + 4/15=_________

Find a number that 6 and 15 both go into. 30, 60, and 90 all work.  We pick the smallest one, 30, to keep the problem easy.  You want to change the fraction 1/6 into another fraction that means the same amount, but has a 30 in the denominator (bottom number of the fraction).  Think of a pizza cut into 6 equal slices and you ate 1, or 1/6 of the pizza. Now think of that same pizza cut into 30 equal pieces.  How many of the 30 pieces would you have to eat to equal the first piece you ate (1/6 of the pizza)?  You might be able to draw this and find the answer.  Or think about how many times 6 will go into 30.  The answer is 5.  So you have to eat 5 of the skinny slices to equal eating 1/6 of the pizza. Thus, 1/6 is the same amount as 5/30.

Do the same thing to change 4/15 into a fraction that has 30 in the denominator. If your first pizza had 15 pieces, imagine that it now has 30.  That is double.  So if you ate 4 out of 15 pieces, that is the same thing as eating 8 out of 30. Thus, the number 4/15 is equal to the number 8/30.

Back to the problem of 1/6 + 4/15=?  Replace 1/6 with 5/30 and replace 4/15 with 8/30.  So, another way to write this problem is 5/30 + 8/30= ?  Once the denominators are the same number, just add up the top numbers.  Since 5 + 8 = 13, we can say that five-thirtieths plus eight-thirtieths equals thirteen-thirtieths, or 13/30.

These steps will work for all the problems.  Just remember to subtract the numerators in the last two problems rather than adding them.  Getting good at working with fractions prepares you for some of the harder work in algebra. It is a very important topic to get good at doing.  Good luck!

Answer 3

to add fractions you have to have a common denominator...there are two ways you can solve this..haha i should get best answer because this is going to be long haha jk...

way one: find the greatest common multiple (gcm) by multiplying all the denominators together.

way two: count in increments of the largest denominator until you reach a number that is divisible by all the denominators... sounds confusing but its really easy least common multiple (lcm)

im going to use way 2.

in the first problem the biggest denominator is 15 and you have to find an number that is divisible by 3, 4,6, and 15 so 15 is not divisible by 6 so try one increment up 30.  

30 is divisible by 3,6, and 15 but not four.

45 is divisible by 3, and 15 but not 4 or 6

and finally you reach 60 which is divisible by all of the numbers.

so now 60 becomes the number that you want to change all the denominators to.