{"id":271396,"date":"2025-07-25T19:04:55","date_gmt":"2025-07-25T19:04:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=271396"},"modified":"2025-07-25T19:04:58","modified_gmt":"2025-07-25T19:04:58","slug":"using-the-elimination-method-solve-each-of-the-following-pairs-of-simultaneous-equations-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/25\/using-the-elimination-method-solve-each-of-the-following-pairs-of-simultaneous-equations-2\/","title":{"rendered":"Using the elimination method, solve each of the following pairs of simultaneous equations"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

Using the elimination method, solve each of the following pairs of simultaneous equations. (a) 4x – y – 7 = 0 4x + 3y – 11 = 0 (c) 5x – 3y – 2 = 0 x + 5y – 6 = 0 (e) 7x + 3y – 8 = 0 3x – 4y – 14 = 0 (b) 7x + 2y – 33 = 0 3y – 7x – 17 = 0 (d) 5x – 3y – 13 = 0 7x – 6y – 20 = 0 (f) 3x + 5y + 8 = 0 4x + 13y – 2 = 0<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s solve the pairs of simultaneous equations using the elimination method<\/strong>. This method involves manipulating the equations to eliminate one of the variables, making it easier to solve for the other.<\/p>\n\n\n\n

(a) 4x – y – 7 = 0<\/h3>\n\n\n\n
CopyEdit  4x + 3y - 11 = 0\n<\/code><\/pre>\n\n\n\n
Step 1:<\/strong> Rearrange the equations into standard form:<\/p>\n\n\n\n
    \n
  1. 4x\u2212y=74x – y = 74x\u2212y=7 (Equation 1)<\/li>\n\n\n\n
  2. 4x+3y=114x + 3y = 114x+3y=11 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
    Step 2:<\/strong> Subtract Equation 1 from Equation 2 to eliminate xxx:(4x+3y)\u2212(4x\u2212y)=11\u22127(4x + 3y) – (4x – y) = 11 – 7(4x+3y)\u2212(4x\u2212y)=11\u22127<\/p>\n\n\n\n
    This simplifies to:4x+3y\u22124x+y=4\u21d24y=4\u21d2y=14x + 3y – 4x + y = 4 \\quad \\Rightarrow \\quad 4y = 4 \\quad \\Rightarrow \\quad y = 14x+3y\u22124x+y=4\u21d24y=4\u21d2y=1<\/p>\n\n\n\n
    Step 3:<\/strong> Substitute y=1y = 1y=1 into Equation 1:4x\u22121=7\u21d24x=8\u21d2x=24x – 1 = 7 \\quad \\Rightarrow \\quad 4x = 8 \\quad \\Rightarrow \\quad x = 24x\u22121=7\u21d24x=8\u21d2x=2<\/p>\n\n\n\n
    Solution for (a):<\/strong>x=2,\u2009y=1x = 2, \\, y = 1x=2,y=1<\/p>\n\n\n\n
    (b) 7x + 2y – 33 = 0<\/h3>\n\n\n\n
    CopyEdit  3y - 7x - 17 = 0\n<\/code><\/pre>\n\n\n\n
    Step 1:<\/strong> Rearrange the equations:<\/p>\n\n\n\n
      \n
    1. 7x+2y=337x + 2y = 337x+2y=33 (Equation 1)<\/li>\n\n\n\n
    2. \u22127x+3y=17-7x + 3y = 17\u22127x+3y=17 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
      Step 2:<\/strong> Add the two equations to eliminate xxx:(7x+2y)+(\u22127x+3y)=33+17(7x + 2y) + (-7x + 3y) = 33 + 17(7x+2y)+(\u22127x+3y)=33+17<\/p>\n\n\n\n
      This simplifies to:5y=50\u21d2y=105y = 50 \\quad \\Rightarrow \\quad y = 105y=50\u21d2y=10<\/p>\n\n\n\n
      Step 3:<\/strong> Substitute y=10y = 10y=10 into Equation 1:7x+2(10)=33\u21d27x+20=33\u21d27x=13\u21d2x=1377x + 2(10) = 33 \\quad \\Rightarrow \\quad 7x + 20 = 33 \\quad \\Rightarrow \\quad 7x = 13 \\quad \\Rightarrow \\quad x = \\frac{13}{7}7x+2(10)=33\u21d27x+20=33\u21d27x=13\u21d2x=713\u200b<\/p>\n\n\n\n
      Solution for (b):<\/strong>x=137,\u2009y=10x = \\frac{13}{7}, \\, y = 10x=713\u200b,y=10<\/p>\n\n\n\n
      (c) 5x – 3y – 2 = 0<\/h3>\n\n\n\n
      nginxCopyEdit  x + 5y - 6 = 0\n<\/code><\/pre>\n\n\n\n
      Step 1:<\/strong> Rearrange the equations:<\/p>\n\n\n\n
        \n
      1. 5x\u22123y=25x – 3y = 25x\u22123y=2 (Equation 1)<\/li>\n\n\n\n
      2. x+5y=6x + 5y = 6x+5y=6 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
        Step 2:<\/strong> Multiply Equation 2 by 5 to align the coefficients of xxx:5(x+5y)=5(6)5(x + 5y) = 5(6)5(x+5y)=5(6)<\/p>\n\n\n\n
        This gives:5x+25y=30(Equation 3)5x + 25y = 30 \\quad \\text{(Equation 3)}5x+25y=30(Equation 3)<\/p>\n\n\n\n
        Step 3:<\/strong> Subtract Equation 1 from Equation 3 to eliminate xxx:(5x+25y)\u2212(5x\u22123y)=30\u22122(5x + 25y) – (5x – 3y) = 30 – 2(5x+25y)\u2212(5x\u22123y)=30\u22122<\/p>\n\n\n\n
        This simplifies to:28y=28\u21d2y=128y = 28 \\quad \\Rightarrow \\quad y = 128y=28\u21d2y=1<\/p>\n\n\n\n
        Step 4:<\/strong> Substitute y=1y = 1y=1 into Equation 2:x+5(1)=6\u21d2x+5=6\u21d2x=1x + 5(1) = 6 \\quad \\Rightarrow \\quad x + 5 = 6 \\quad \\Rightarrow \\quad x = 1x+5(1)=6\u21d2x+5=6\u21d2x=1<\/p>\n\n\n\n
        Solution for (c):<\/strong>x=1,\u2009y=1x = 1, \\, y = 1x=1,y=1<\/p>\n\n\n\n
        (d) 5x – 3y – 13 = 0<\/h3>\n\n\n\n
        CopyEdit  7x - 6y - 20 = 0\n<\/code><\/pre>\n\n\n\n
        Step 1:<\/strong> Rearrange the equations:<\/p>\n\n\n\n
          \n
        1. 5x\u22123y=135x – 3y = 135x\u22123y=13 (Equation 1)<\/li>\n\n\n\n
        2. 7x\u22126y=207x – 6y = 207x\u22126y=20 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
          Step 2:<\/strong> Multiply Equation 1 by 2 to align the coefficients of yyy:2(5x\u22123y)=2(13)2(5x – 3y) = 2(13)2(5x\u22123y)=2(13)<\/p>\n\n\n\n
          This gives:10x\u22126y=26(Equation 3)10x – 6y = 26 \\quad \\text{(Equation 3)}10x\u22126y=26(Equation 3)<\/p>\n\n\n\n
          Step 3:<\/strong> Subtract Equation 2 from Equation 3 to eliminate yyy:(10x\u22126y)\u2212(7x\u22126y)=26\u221220(10x – 6y) – (7x – 6y) = 26 – 20(10x\u22126y)\u2212(7x\u22126y)=26\u221220<\/p>\n\n\n\n
          This simplifies to:3x=6\u21d2x=23x = 6 \\quad \\Rightarrow \\quad x = 23x=6\u21d2x=2<\/p>\n\n\n\n
          Step 4:<\/strong> Substitute x=2x = 2x=2 into Equation 1:5(2)\u22123y=13\u21d210\u22123y=13\u21d2\u22123y=3\u21d2y=\u221215(2) – 3y = 13 \\quad \\Rightarrow \\quad 10 – 3y = 13 \\quad \\Rightarrow \\quad -3y = 3 \\quad \\Rightarrow \\quad y = -15(2)\u22123y=13\u21d210\u22123y=13\u21d2\u22123y=3\u21d2y=\u22121<\/p>\n\n\n\n
          Solution for (d):<\/strong>x=2,\u2009y=\u22121x = 2, \\, y = -1x=2,y=\u22121<\/p>\n\n\n\n
          (e) 7x + 3y – 8 = 0<\/h3>\n\n\n\n
          CopyEdit  3x - 4y - 14 = 0\n<\/code><\/pre>\n\n\n\n
          Step 1:<\/strong> Rearrange the equations:<\/p>\n\n\n\n
            \n
          1. 7x+3y=87x + 3y = 87x+3y=8 (Equation 1)<\/li>\n\n\n\n
          2. 3x\u22124y=143x – 4y = 143x\u22124y=14 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
            Step 2:<\/strong> Multiply Equation 1 by 3 and Equation 2 by 7 to align the coefficients of xxx:3(7x+3y)=3(8)\u21d221x+9y=24(Equation 3)3(7x + 3y) = 3(8) \\quad \\Rightarrow \\quad 21x + 9y = 24 \\quad \\text{(Equation 3)}3(7x+3y)=3(8)\u21d221x+9y=24(Equation 3)7(3x\u22124y)=7(14)\u21d221x\u221228y=98(Equation 4)7(3x – 4y) = 7(14) \\quad \\Rightarrow \\quad 21x – 28y = 98 \\quad \\text{(Equation 4)}7(3x\u22124y)=7(14)\u21d221x\u221228y=98(Equation 4)<\/p>\n\n\n\n
            Step 3:<\/strong> Subtract Equation 3 from Equation 4 to eliminate xxx:(21x\u221228y)\u2212(21x+9y)=98\u221224(21x – 28y) – (21x + 9y) = 98 – 24(21x\u221228y)\u2212(21x+9y)=98\u221224<\/p>\n\n\n\n
            This simplifies to:\u221237y=74\u21d2y=\u22122-37y = 74 \\quad \\Rightarrow \\quad y = -2\u221237y=74\u21d2y=\u22122<\/p>\n\n\n\n
            Step 4:<\/strong> Substitute y=\u22122y = -2y=\u22122 into Equation 1:7x+3(\u22122)=8\u21d27x\u22126=8\u21d27x=14\u21d2x=27x + 3(-2) = 8 \\quad \\Rightarrow \\quad 7x – 6 = 8 \\quad \\Rightarrow \\quad 7x = 14 \\quad \\Rightarrow \\quad x = 27x+3(\u22122)=8\u21d27x\u22126=8\u21d27x=14\u21d2x=2<\/p>\n\n\n\n
            Solution for (e):<\/strong>x=2,\u2009y=\u22122x = 2, \\, y = -2x=2,y=\u22122<\/p>\n\n\n\n
            (f) 3x + 5y + 8 = 0<\/h3>\n\n\n\n
            CopyEdit  4x + 13y - 2 = 0\n<\/code><\/pre>\n\n\n\n
            Step 1:<\/strong> Rearrange the equations:<\/p>\n\n\n\n
              \n
            1. 3x+5y=\u221283x + 5y = -83x+5y=\u22128 (Equation 1)<\/li>\n\n\n\n
            2. 4x+13y=24x + 13y = 24x+13y=2 (Equation 2)<\/li>\n<\/ol>\n\n\n\n
              Step 2:<\/strong> Multiply Equation 1 by 4 and Equation 2 by 3 to align the coefficients of xxx:4(3x+5y)=4(\u22128)\u21d212x+20y=\u221232(Equation 3)4(3x + 5y) = 4(-8) \\quad \\Rightarrow \\quad 12x + 20y = -32 \\quad \\text{(Equation 3)}4(3x+5y)=4(\u22128)\u21d212x+20y=\u221232(Equation 3)3(4x+13y)=3(2)\u21d212x+39y=6(Equation 4)3(4x + 13y) = 3(2) \\quad \\Rightarrow \\quad 12x + 39y = 6 \\quad \\text{(Equation 4)}3(4x+13y)=3(2)\u21d212x+39y=6(Equation 4)<\/p>\n\n\n\n
              Step 3:<\/strong> Subtract Equation 3 from Equation 4 to eliminate xxx:(12x+39y)\u2212(12x+20y)=6\u2212(\u221232)(12x + 39y) – (12x + 20y) = 6 – (-32)(12x+39y)\u2212(12x+20y)=6\u2212(\u221232)<\/p>\n\n\n\n
              This simplifies to:19y=38\u21d2y=219y = 38 \\quad \\Rightarrow \\quad y = 219y=38\u21d2y=2<\/p>\n\n\n\n
              Step 4:<\/strong> Substitute y=2y = 2y=2 into Equation 1:3x+5(2)=\u22128\u21d23x+10=\u22128\u21d23x=\u221218\u21d2x=\u221263x + 5(2) = -8 \\quad \\Rightarrow \\quad 3x + 10 = -8 \\quad \\Rightarrow \\quad 3x = -18 \\quad \\Rightarrow \\quad x = -63x+5(2)=\u22128\u21d23x+10=\u22128\u21d23x=\u221218\u21d2x=\u22126<\/p>\n\n\n\n
              Solution for (f):<\/strong>x=\u22126,\u2009y=2x = -6, \\, y = 2x=\u22126,y=2<\/p>\n\n\n\n
              \n\n\n\n
              Summary of Solutions:<\/h3>\n\n\n\n
                \n
              • (a) x=2,\u2009y=1x = 2, \\, y = 1x=2,y=1<\/li>\n\n\n\n
              • (b) x=137,\u2009y=10x = \\frac{13}{7}, \\, y = 10x=713\u200b,y=10<\/li>\n\n\n\n
              • (c) x=1,\u2009y=1x = 1, \\, y = 1x=1,y=1<\/li>\n\n\n\n
              • (d) x=2,\u2009y=\u22121x = 2, \\, y = -1x=2,y=\u22121<\/li>\n\n\n\n
              • (e) x=2,\u2009y=\u22122x = 2, \\, y = -2x=2,y=\u22122<\/li>\n\n\n\n
              • (f) x=\u22126,\u2009y=2x = -6, \\, y = 2x=\u22126,y=2<\/li>\n<\/ul>\n\n\n\n
                <\/p>\n","protected":false},"excerpt":{"rendered":"
                Using the elimination method, solve each of the following pairs of simultaneous equations. (a) 4x – y – 7 = 0 4x + 3y – 11 = 0 (c) 5x – 3y – 2 = 0 x + 5y – 6 = 0 (e) 7x + 3y – 8 = 0 3x – 4y – […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-271396","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/271396","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=271396"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/271396\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=271396"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=271396"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=271396"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}