Subtracting Fractions

Transcription

1 Lerning Enhncement Tem Model Answers: Adding nd Subtrcting Frctions Adding nd Subtrcting Frctions study guide. When the frctions both hve the sme denomintor (bottom) you cn do them using just simple dding nd subtrcting. Or some of them these cn be done by dding nd subtrcting frctions tht you lredy know. Try sying them out loud (or in your hed if tht s not prcticl) () A third plus third is two thirds or, if you wnt to do it more formlly, the denomintors (bottoms) re both nd so you cn just dd the numertors (tops) to give nd so the new numertor is. Pictorilly this looks like: + = (b) A third plus two thirds is three thirds which is whole or, in other words, one Agin the denomintors re the sme nd so you just dd the numertors to give. Then you hve: It is best not to leve it s / becuse this cn be cncelled down to which is much simpler. Pictorilly this looks like: + =

2 (c) Sy it out loud or in your hed: A hlf plus qurter is three qurters + = (d) A hlf minus qurter is qurter You know tht two qurters mke hlf nd so, if you tke qurter from hlf, you re left with just one qurter. - = (e) Five sixths plus one sixth is six sixths which is whole or Formlly you my write + = (f) or One sixths plus seven sixths is eight sixths So: Then you cn cncel this down by dividing the top nd bottom by to get /. It is preferble to write it s n improper frction s / lthough the mixed frction is lso common. The importnt thing to remember with mixed frctions is tht, even though the numbers re written next to ech other, it is not but.

3 (g) Minus sixth minus sixth is minus two sixths which cncels down to minus third You need to be cler tht negtive number minus nother negtive number is lso negtive number. Then: (h) 0 An eighth minus n eighth is zero Anything minus itself is 0. (i) 0 Minus n eight plus n eight is zero. This question is just the sme s the previous one but with the negtive number before the positive one.. Now you hve to find common denomintor by multiplying the denomintors of the two frctions. Remember you cn only dd or subtrct the numertors of frctions if they hve the sme denomintor. ) or You get the common denomintor by multiplying the two denomintors so Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. Now the denomintors re the sme in both frctions, you cn dd the numertors to give: which is the preferred form for the nswer lthough you my lso see it written s mixed frction s.

4 b) You get the common denomintor by multiplying the two denomintors so. Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: c) You get the common denomintor by multiplying the two denomintors so. Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: Now the denomintors re the sme in both frctions, you cn dd the numertors to give: d) You get the common denomintor by multiplying the two denomintors so. Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht:

5 So / nd / re equivlent frctions. To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: So / nd / re equivlent frctions. Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: e) or 0 0 You get the common denomintor by multiplying the two denomintors so 0. Then, to get 0 to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: 0 To get 0 to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: 0 Now the denomintors re the sme in both frctions, you cn dd the numertors to give: which is the preferred form for the nswer lthough you my lso see it written s mixed frction s. 0 f) 0 00 You get the common denomintor by multiplying the two denomintors so Then, to get 00 to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by 00 so tht: To get 00 to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht:

6 00 00 Now the denomintors re the sme in both frctions, you cn dd the numertors to give: g) 00 You get the common denomintor by multiplying the two denomintors so Then, to get 00 to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: To get 00 to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by 00 so tht: Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: h) You get the common denomintor by multiplying the two denomintors so. Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give:

7 i) 000 You get the common denomintor by multiplying the two denomintors so Then, to get 000 to be the denomintor of the first frction, you multiply both the numertor nd the denomintor by so tht: To get 000 to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by 000 so tht: Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: In these ones you could just multiply the denomintors but it is more efficient to find lower number tht hs both the denomintors s fctors. This is known s the lowest common multiple (see study guide: Lowest Common Multiple) ) Remember hlf plus qurter is three qurters. To do it more formlly you find common denomintor. Now you could multiply the denomintors to get nd then convert the frctions to hve this denomintor which would give And then you cn cncel this down to get /. It is bit quicker nd neter to choose lower common denomintor which is since hs both the originl denomintors ( nd ) s fctors. Then you cn convert the first frction by multiplying both the numertor nd the denomintor by so tht: nd there is no need for ny cncelling down.

8 b) Agin, you could choose common denomintor of but it is quicker to use since is fctor of. Convert the second frction by multiplying both the numertor nd the denomintor by so tht: which cncels down to /. c) Since is fctor of, the lowest common multiple is. Convert the second frction by multiplying both the numertor nd the denomintor by so tht: d) Since is fctor of, the lowest common multiple is. Convert the second frction by multiplying both the numertor nd the denomintor by so tht: e) This is slightly different since is not fctor of. However is not the lowest common multiple. It is becuse hs both nd s fctors. See the study guide Lowest Common Multiple for help with this. Convert the first frction by multiplying both the numertor nd the denomintor by so tht:

9 nd convert the second frction by multiplying both the numertor nd the denomintor by so tht: f) Since is fctor of, the lowest common multiple is. Convert the second frction by multiplying both the numertor nd the denomintor by so tht: g) Since is fctor of, the lowest common multiple is. Convert the first frction by multiplying both the numertor nd the denomintor by so tht: 0 which cncels down to. h) In this one is not fctor of. However is not the lowest common multiple. It is becuse hs both nd s fctors. Convert the first frction by multiplying both the numertor nd the denomintor by so tht:

10 nd convert the second frction by multiplying both the numertor nd the denomintor by so tht: i) In this one is not fctor of. However 0 is not the lowest common multiple. It is becuse hs both nd s fctors. Convert the first frction by multiplying both the numertor nd the denomintor by so tht: nd convert the second frction by multiplying both the numertor nd the denomintor by so tht:. First convert both mixed frctions to improper frctions by multiplying whole number by the denomintor of the frctionl prt nd then dding this to the numertor. Then pply the sme methods s before. Remember tht in mixed frction the whole number is not multiplying the frction prt even though they re written next to ech other. They re ctully being dded. ) Remember the mixed frction does not men but is ctully the sme s sying, two hlves plus one hlf or which is three hlves or /. This is the most common mixed frction nd so it is worth remembering the improper form which is, in generl, more useful. Now you hve converted it to n improper frction, you cn use the sme methods s before. The denomintors re nd which re co-prime nd so you get the common denomintor by multiplying them so. Then, to get to be the denomintor of the

11 first frction, you multiply both the numertor nd the denomintor of the improper frction by so tht: To get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor by so tht: Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: which is the preferred form for the nswer lthough you my lso see it written s mixed frction s. b) The mixed frction is the sme s sying, fifteen thirds plus one third or which is sixteen thirds or /. Now you hve converted it to n improper frction, you cn use the sme methods s before. The denomintors re nd. Since is fctor of, is the lowest common multiple. Then, to get to be the denomintor of the first frction, you multiply both the numertor nd the denomintor of the improper frction by so tht: Now the denomintors re the sme in both frctions, you cn subtrct the numertors to give: which is the preferred form for the nswer lthough you my lso see it written s mixed frction s. c) The mixed frction is the sme s nd the mixed frction is the sme s 0. The denomintors re nd. Since is fctor of, is the lowest common multiple. Then, to get to be the denomintor of the second frction, you multiply both the numertor nd the denomintor of the improper frction by so tht:

12 0 0 Now the denomintors re the sme in both frctions, you cn dd the numertors to give: 0 which is the preferred form for the nswer lthough you my lso see it written s mixed frction s.. The sme techniques pply to dding three or more frctions. Just mke sure tht the denomintor hs ll three (or more) frctions s fctor. ) Since nd re both fctors of. The lowest common multiple is. / cn be written s / nd / cn be written s / so tht: the ddition becomes which simplifies to. b) This cn be rerrnged to: nd so the first two frctions cncel to leve / ; even though there re frctions, the rules of ddition nd subtrction still pply. c) 0 The lowest common multiple of the denomintors is 0. cn be written s 0 cn be written s nd 0 So the nswer is cn be written s nd 0 0

13 d) You hve to relise tht Then, since multiple is 0 nd So the nswer is 0 cn be written s is fctor of 0 the lowest common So to get the deciml nswer, you just move the deciml point seven plces to the left. e) 0. Since both 0 nd 00 re fctors of is the lowest common multiple. So the nswer is 00 cn be written s nd cn be written s which is 0. s deciml. f) recurring This is n exmple of n infinite series. Even though the series goes on forever, it still hs number tht it sums up to. This is generlistion of the previous exmple. You cn choose ny power of 0 to be the denomintor such s 00,000 or 00,000,000. However lrge you mke it the numertor will lwys be s nd so the deciml will be 0. recurring which is the sme s g) + + = See question ) bove. Then, to get generl unit frction /, you cn divide both sides of this eqution by (or, in other words, multiply by / ) so tht:

14 nd so nd then the nswer is = + + This is n exmple of n Egyptin frction. These model nswers re one of series on mthemtics produced by the Lerning Enhncement Tem. Scn the QR-code with smrtphone pp for more resources.