{"id":1547,"date":"2023-09-29T10:18:49","date_gmt":"2023-09-29T10:18:49","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1547"},"modified":"2023-09-29T10:22:50","modified_gmt":"2023-09-29T10:22:50","slug":"week-3-and-4-jss-3-second-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-and-4-jss-3-second-term-mathematics-notes\/","title":{"rendered":"Week 3 and 4 &#8211; Jss 3 Second Term Mathematics Notes"},"content":{"rendered":"<p><strong>   WEEK 3<br \/>\n<\/strong><strong> ELIMINATION METHOD<br \/>\n<\/strong>This method is very useful to solve simultaneous equations especially when none of the coefficients of the unknown is 1.<br \/>\nExample III.<br \/>\nSolve the following simultaneous equations by elimination method.<br \/>\n(a)\u00a0\u00a0\u00a0\u00a06x + 5y = 15\u00a0\u00a0\u00a0\u00a0(1)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b)\u00a0\u00a0\u00a0\u00a04c \u2013 4d = 9<br \/>\n\u00a0\u00a0\u00a0\u00a03x + 5y = 12\u00a0\u00a0\u00a0\u00a0(2)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a05c + 4d = 18<br \/>\nOne of the unknown &#8220;Y&#8221; has equal coefficient and with the same signs so we subtract the two equations to eliminate y terms.<br \/>\n6x + 5y \u2013 (3x + 5y) = 15 &#8211; 12<br \/>\n6x + 5y &#8211; 3x &#8211; 5y = 3<br \/>\n6x &#8211; 3x + 5y &#8211; 5y = 3<br \/>\n3x + 0 = 3<br \/>\n=<br \/>\nx=1<\/p>\n<p>\u00a0To find y, substitute x=1 in either (1) or (2) using equation<br \/>\n (1) 6x + 5y = 15<br \/>\n6(i) + 5y = 15<br \/>\n6 + 5y = 15<br \/>\n5y = 15 \u2013 6<br \/>\n =<br \/>\ny=<br \/>\n(b)\u00a0\u00a0\u00a0\u00a04c \u2013 4d = 9 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1)<br \/>\n\u00a0\u00a0\u00a0\u00a05c + 4d = 18 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2)<br \/>\nOne of the unknown &#8220;d&#8221; has equal coefficient but with different sign so we add the two equations to eliminate &#8220;d&#8221;<br \/>\n4c \u2013 4d + 5c + 4d = 9 + 18<br \/>\n4c + 5c \u2013 4d + 4d = 27<br \/>\n  =<br \/>\nC = 3<br \/>\nTo find d substitute c = 3 into (1)<br \/>\n4c \u2013 4d = 9<br \/>\n4(3) \u2013 4d = 9<br \/>\n4d = 12 &#8211; 9<br \/>\n=<br \/>\nD =<br \/>\n<strong>WRAP UP AND ASSESSMENT<br \/>\n<\/strong>In elimination method you may need to multiply one or both of the equations by a number in order to obtain a variable with e same coefficient in both equations.  Then add both equations when the signs of the variables you want to eliminate are opposite but subtract them when the signs are the same.<br \/>\nExercise: 16 4 No 2, 3, 7 \u2013 11.<br \/>\nUse elimination method to solve the following simultaneous equations.<br \/>\n2.\u00a0\u00a0\u00a0\u00a06x + 7y = 15\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x + 3y = 10<br \/>\n\u00a0\u00a0\u00a0\u00a06x \u2013 9y = 31\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x + 5y = 8<\/p>\n<p>\u00a07)\u00a0\u00a0\u00a0\u00a02x + 3y = 8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x + 4y = 10<br \/>\n\u00a0\u00a0\u00a0\u00a03x + 2y = 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x + 5y = 9<\/p>\n<p>\u00a09)\u00a0\u00a0\u00a0\u00a04x + 3y = 11\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04a + 3b = 3<br \/>\n\u00a0\u00a0\u00a0\u00a03x \u2013 4y = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03a + 2b = 1<\/p>\n<p>\u00a0<strong>TICKET OUT<br \/>\n<\/strong>Solve the following Simultaneous equation by Elimination method. Exercise 16.4 No 12 -1<\/p>\n<p>\u00a0<br \/>\n\u00a0<strong>                                                                        WEEK 4<br \/>\n<\/strong><strong>SOLVING SIMULTANEOUS EQUATION GRAPHICALLY<br \/>\n<\/strong>To solve simultaneous equations graphically.<\/p>\n<ol>\n<li>\n<div>Make a table of values for both equations.\n<\/div>\n<\/li>\n<li>\n<div>Draw the graphs for both equations on the same axes\n<\/div>\n<\/li>\n<li>\n<div>Find the co-ordinate (i.e x and y values) where both graphs intersect these values are the solutions of both equations.\n<\/div>\n<\/li>\n<li>\n<div>Check your solutions by putting these values into the original equations to make sure they satisfy them.\n<\/div>\n<\/li>\n<\/ol>\n<p>Example 16.3<br \/>\nSolve the simultaneous equations.<br \/>\nX \u2013 2y = 4 and 2x \u2013 y = 5 graphically<br \/>\nSolution<br \/>\nIn each equation make y the subject of the equation<br \/>\n(i)\u00a0\u00a0\u00a0\u00a0 x \u2013 2y = 4<br \/>\n\u00a0\u00a0\u00a0\u00a0-2y = 4 \u2013 x<br \/>\n\u00a0\u00a0\u00a0\u00a0 y = -2 + 0.5x \u2026\u2026\u2026\u2026\u2026\u2026 (1)<\/p>\n<p>\u00a0<strong><br \/>\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"<p>WEEK 3 ELIMINATION METHOD This method is very useful to solve simultaneous equations especially when&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,142],"tags":[],"class_list":["post-1547","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-jss3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1547","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=1547"}],"version-history":[{"count":2,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1547\/revisions"}],"predecessor-version":[{"id":1549,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1547\/revisions\/1549"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1547"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1547"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1547"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}