Does any body know where i can find a website for adding and subtracting of fractions?
135920
the most important thing to remember when adding and subtracting fractions is to find a common denominator. The only way to get good at doing fractions is to do them.
380314
math.com is an awsome site to go to, i use it for my hw help!
634422
http://www.purplemath.com/lessons.htm
319182
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
482184
Go to Insert, Object, Microsoft Equation 3.0\nThis brings up an equation toolbar.  You can do a reasonable amount of mathematical notation with this tool.  Just press on the relevant tab to insert a fraction (it gives you a few choices as to the particular format) and then just fill in the numbers.\nIt is slightly laborious but works.\nIf you can get your hands on a piece of software called MathType this is easier to use and allows for more options.
70099
http://www.math.com/\nhttp://mathworld.wolfram.com/\nhttp://www.calculus-help.com/\nhttp://math.about.com/od/calculus/
487842
try http://www.webmath.com/\nit gives you help on math on almost any time of math.
76917
Maybe this link will help:\nhttp://www.mathinvestigations.com/
174876
http://www.math.com/ or http://www.gomath.com/algebra.html are good places to start.
617758
http://www.algebra.com
395611
http://mathforum.org/dr.math/\n\nThis is Ask Dr. Math, a website that can offer help for general math through Algebra and beyond. There are other good websites for math out online, type in keywords,  homework  math help and see all that you come up with. Eventually you will find one that can walk you through most problems. Good luck.
528190
Do a Yahoo search.  Enter "Homework help, math."  There are bunches of web sites.  Good luck.
69569
Try this, maybe?\n\nhttp://mathforum.org/library/drmath/drmath.elem.html
584364
The steps for adding two fractions:\n\n1) reduce both fractions  (not strictly necessary, as long as you reduce in the end)\n2) give them both the same demoninator (the bottom number)\n3) add the top numbers (the numerators)\n4) reduce the answer\n\nSo, step one is already done.\n\nStep two is to give them a common (the same) denominator.  So what number should that be?\n\nIt's going to be the product of both denominators.  so 3 * 8 = 24.  So that's our common denominator.\n\nWhat you are going to do is to multiply the top and bottom half of each fraction by the same number (which is okay, it's equivilant to multiplying by 1) so that the bottom number is 24.  So for the left fraction, we will multiply the top and bottom by 3.  So we get (3 *3)/(8 * 3), which is 9/24.\n\nAnd now we multiply the second fraction, top and bottom, by 8.  so (2 * 8)/(3 * 8) = 16/24.  Now that both fractions have the same denominator, we add them by adding the top numbers, and putting that over the common denominator.  So 9 + 16 = 25, so we get 25/24.\n\nThis is technically acceptable, but teachers will probably want you write this as 1 and 1/24th.
240127
In order to add or subtract fractions, you must have common denominators; that is, the bottom number has to be the same. You can change the bottom number of a fraction by multiplying the entire fraction by 1. The fact that n/n=1 for all values of n except 0 makes this a lot easier. So, let's take your problems. The easiest way to get a common denominator is to multiply the first fraction by the second denominator, and the second fraction by the first denominator. (This won't be the Least Common Denominator, though. That involves factoring the numbers.)\n\n1) (4/7) + (7/p)\nSince 4/7 doesn't reduce (we can't tell 7/p), we'll want to get 7p as the common denominator. \n(4/7)(p/p) + (7/p)(7/7)\n(4p/7p) + (49/7p)\nNow that we have the same denominator, we just add the numerators. Whoops. Since we don't know what (p) is, we have to leave it as\n(49 + 4p)/7p for p != 0 . (Oh, yeah. != is one of the ways of saying "does not equal" on the computer. You might also see it as =/=, which looks more like what you would write.)\n\n2) 4/6 - 2/k. First, let's simplify 4/6 to 2/3, this will keep things easier later on. \n2/3 - 2/k. Now, a denominator we can use is 3k\n(2/3)(k/k) - (2/k)(3/3)\n2k/3k - 6/3k\n(2k-6)/3k, for k!=0\n\n3) 2=5 + 6/k. \nOK, this one solves a bit differently. We need to find a value that makes the statement true. First, let's get all the ks on one side, and everything else on the other (remember, a=b therefore b=a)\n6/k = -3\nNow, to get from here to there, we need to get the k out of the denominator. There's a property of multiplication that lets us do this, the shorthand for it is (if a=b, then ac=bc). So, we'll multiply through by k...\n6 = -3k.\nAnd now, we need to get k by itself. Normally, I'd suggest dividing by -3. But I don't particularly like division, so I'm going to tell you to multiply by the reciprocal of -3; Which is 1/(-3) or -1/3.\n6(-1/3) = (-1/3)-3k\n-2 = k
668303
About learning maths this site is a great reference:\n\nhttp://www.mathworld.com\n\nAbout doing home works You simply can use Just Yahoo answers.\n\nBut please send your homeworks one by one.\nYou will find plenty of people here to answer your questions.
487042
Google the word algebra help.  Purple math is great, but there are other sites also.  I am sorry I do not have the names at the moment.