{"id":371,"date":"2024-02-04T13:24:41","date_gmt":"2024-02-04T21:24:41","guid":{"rendered":"https:\/\/www.rightlobemath.com\/blog\/?p=371"},"modified":"2020-02-04T14:07:31","modified_gmt":"2020-02-04T22:07:31","slug":"abacus-math-program-lesson-8-ten-pair-complement-subtraction-part-2","status":"publish","type":"post","link":"https:\/\/sumomath.com\/content\/abacus-math-program-lesson-8-ten-pair-complement-subtraction-part-2\/","title":{"rendered":"Abacus Math Program – Lesson 8 – Ten Pair Complement Subtraction Part 2"},"content":{"rendered":"

Continuing from\u00a0Lesson 7 Subtraction Part 1<\/u><\/a>\u00a0we will look at a few more 10 pair complement\u00a0subtraction examples. Again there are 5 possible 10 pair complements: 9-1, 8-2, 7-3, 6-4, 5-5. All of these pairs are reversible meaning we can use the complement pair, 9-1, for both the problem\u00a010-1 just as we can use the same pair for the problem\u00a010-9. So the following examples will look at some of the reverse pair\u00a0 possibilities such as subtracting -1, -2, -3, -4.<\/p>\n

In the first example we start by adding 9 to the abacus. Next we add 5 to 9 but notice that we will need to use the 10 pair addition rule. So first add 1 to the next rod left, rod A, and then subtract 5, the 10 pair complement of 5, from 9 on rod B. The interim sum is now 14. Next subtract 3 from 4 on rod B for an interim sum of 11. Lastly subtract 2 from 11 by subtracting 2 from 1 on rod B. But since there are not enough beads touching the reckoning bar we will need to use the 10 pair subtraction rule. So first subtract 1 from the next rod left, rod A, and then add 8, the 10 pair complement of 2, to 1 on rod B for a final answer of 9.<\/p>\n

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In the next example we start by adding 8 to the abacus. Next we add 4 to 8 using the 10 pair addition rule. So first add 1 to the next rod left, rod A, and then subtract 6, the 10 pair complement of 4, from 8 on rod B. The interim sum is now 12. Next subtract 3 from 12 by subtracting 3 from 2 on rod B. Again we will need to use the 10 pair subtraction rule. First subtract 1 from the next rod left, rod A, and then add 7, the 10 pair complement of 3, to 2 on rod B. The interim sum is now 9. Finally subtract 7 from 9 on rod B for a final answer of 2.<\/p>\n

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In the last example we start by adding 6 to the abacus. Next add 4 to 6 using the 10 pair addition rule. So first add 1 to the next rod left, rod A, and then subtract 6, the 10 pair complement of 4, from 6 on rod B. The interim sum is now 10. Next subtract 1 from 10 by subtracting 1 from 0 on rod B. Again we will need to use the 10 pair subtraction rule. First subtract 1 from 1 on the next rod left, rod A, and add 9, the 10 pair complement of 1, to 0 on rod B. The interim sum is now 9. Lastly add 8 to 9 on rod B by using the 10 pair addition rule. First add 1 to 0 on next rod left, rod A, and subtract 2, the 10 pair complement of 8, from 9 on rod B. The final answer is 17.<\/p>\n

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Next up Lesson 9 introducing the 5 Pair<\/u><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Continuing from\u00a0Lesson 7 Subtraction Part 1\u00a0we will look at a few more 10 pair complement\u00a0subtraction examples. Again there are 5 possible 10 pair complements: 9-1, 8-2, 7-3, 6-4, 5-5. All of these pairs are reversible meaning we can use the complement pair, 9-1, for both the problem\u00a010-1 just as we can use the same pair<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-371","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/371"}],"collection":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/comments?post=371"}],"version-history":[{"count":0,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/371\/revisions"}],"wp:attachment":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/media?parent=371"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/categories?post=371"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/tags?post=371"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}