Answers<\/p>\n
1. \u00a0 \u00a0 [pmath]1\/6[\/pmath] +\u00a0 [pmath]3\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]4\/6[\/pmath] , \u00a0 \u00a0 \u00a0 \u00a0\u00a0 [pmath]6\/6[\/pmath] – \u00a0 [pmath]4\/6[\/pmath]\u00a0 = \u00a0 [pmath]2\/6[\/pmath]<\/p>\n
2.\u00a0\u00a0\u00a0\u00a0 [pmath]4\/8[\/pmath] +\u00a0 [pmath]1\/8[\/pmath] = \u00a0 [pmath]5\/8[\/pmath], \u00a0 \u00a0 \u00a0 \u00a0\u00a0 [pmath]8\/8[\/pmath] –\u00a0 [pmath]5\/8[\/pmath] =\u00a0 [pmath]3\/8[\/pmath]<\/p>\n
3.\u00a0\u00a0\u00a0 [pmath]3\/8[\/pmath] +\u00a0 [pmath]4\/8[\/pmath] =\u00a0 [pmath]7\/8[\/pmath] , \u00a0 \u00a0 \u00a0\u00a0 [pmath]8\/8[\/pmath] –\u00a0 [pmath]7\/8[\/pmath] =\u00a0 [pmath]1\/8[\/pmath]<\/p>\n
4.\u00a0\u00a0\u00a0 [pmath]2\/12[\/pmath] +\u00a0 [pmath]4\/12[\/pmath] +\u00a0 [pmath]5\/12[\/pmath] =\u00a0 [pmath]11\/12[\/pmath] ,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [pmath]12\/12[\/pmath] –\u00a0 [pmath]11\/12[\/pmath] =\u00a0 [pmath]1\/12[\/pmath]<\/p>\n
5a. \u00a0 [pmath]5\/8[\/pmath]\u00a0\u00a0\u00a0\u00a0 5b. \u00a0 [pmath]2\/8[\/pmath]\u00a0\u00a0\u00a0\u00a0 5c. \u00a0 [pmath]3\/8[\/pmath]\u00a0\u00a0\u00a0\u00a0 5d. \u00a0 [pmath]4\/12[\/pmath]<\/p>\n
5e.\u00a0\u00a0 [pmath]14\/14[\/pmath] = 1\u00a0\u00a0\u00a0\u00a0 5f.\u00a0\u00a0 [pmath]6\/7[\/pmath]\u00a0\u00a0\u00a0 5g.\u00a0\u00a0 [pmath]3\/5[\/pmath]\u00a0\u00a0\u00a0\u00a0 5h.\u00a0\u00a0 [pmath]5\/12[\/pmath]<\/p>\n
5i.\u00a0\u00a0 [pmath]9\/10[\/pmath]\u00a0\u00a0\u00a0\u00a0 5j.\u00a0\u00a0 [pmath]3\/12[\/pmath]\u00a0\u00a0\u00a0\u00a0 5k.\u00a0\u00a0 [pmath]3\/10[\/pmath]\u00a0\u00a0\u00a0\u00a0 5l.\u00a0\u00a0 [pmath]4\/12[\/pmath]<\/p>\n
Did you notice that 5b is the inverse of 5a? Inverses \u201cundo\u201d each other. That sounds funny, so let me explain: In 5a, you added two-eighths to three-eighths to get five-eighths. In 5b, you took the answer to 5a (five-eighths) and subtracted one of the things you added (three-eighths) to get the other thing you added (two-eighths). So we can say that 5b \u201cundid\u201d 5a. That\u2019s pretty cool, isn\u2019t it? I\u2019ve got a lot more cool stuff in my other books. Oh, in my next book, you\u2019ll meet my friend, Cleveland. He\u2019s cool, too!<\/p>\n","protected":false},"excerpt":{"rendered":"
Answers 1. \u00a0 \u00a0 [pmath]1\/6[\/pmath] +\u00a0 [pmath]3\/6[\/pmath]\u00a0 =\u00a0\u00a0 [pmath]4\/6[\/pmath] , \u00a0 \u00a0 \u00a0 \u00a0\u00a0 [pmath]6\/6[\/pmath] – \u00a0 [pmath]4\/6[\/pmath]\u00a0 = \u00a0 [pmath]2\/6[\/pmath] 2.\u00a0\u00a0\u00a0\u00a0 [pmath]4\/8[\/pmath] +\u00a0 [pmath]1\/8[\/pmath] = \u00a0 [pmath]5\/8[\/pmath], \u00a0 \u00a0 \u00a0 \u00a0\u00a0 [pmath]8\/8[\/pmath] –\u00a0 [pmath]5\/8[\/pmath] =\u00a0 [pmath]3\/8[\/pmath] 3.\u00a0\u00a0\u00a0 [pmath]3\/8[\/pmath] +\u00a0 [pmath]4\/8[\/pmath] =\u00a0 [pmath]7\/8[\/pmath] , \u00a0 \u00a0 \u00a0\u00a0 [pmath]8\/8[\/pmath] –\u00a0 [pmath]7\/8[\/pmath] =\u00a0 [pmath]1\/8[\/pmath] 4.\u00a0\u00a0\u00a0 [pmath]2\/12[\/pmath] +\u00a0 … Continue reading Answers to Book 1 Problems<\/span>