{"id":67077,"date":"2019-01-10T20:10:56","date_gmt":"2019-01-10T20:10:56","guid":{"rendered":"https:\/\/schooltutoring.com\/scholarship\/?p=67077"},"modified":"2019-01-10T20:10:56","modified_gmt":"2019-01-10T20:10:56","slug":"solving-a-system-of-equations-substitution-and-elimination","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/scholarship\/2019\/01\/10\/solving-a-system-of-equations-substitution-and-elimination\/","title":{"rendered":"Solving a System of Equations: Substitution and Elimination"},"content":{"rendered":"

When dealing with a system of linear equations there are two methods to algebraically solve the question. One is substitution and the other is elimination which is meant to be a shortcut. Both methods will bring you to the same solution but with more practice, you will recognize patterns and see which method would work best when given a system. The best way to show how to solve these kinds of questions are by providing an example to work on.<\/span><\/p>\n

Example: <\/span> Line 1: <\/span> 2x + y \u00a0\u00a0= 6<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<\/span>Line 2: <\/span> 6x + 2y = 4<\/span><\/p>\n

Substitution Method:<\/b><\/p>\n

This method involves isolating for one variable (x\/y) of Line 1 then substituting that variable into Line 2. This will allow you to isolate and solve for the other variable (y\/x). Once you have the x- and y- coordinates, you then have the solution which is the point of intersection between the two lines.<\/span><\/p>\n

    \n
  1. Line 1: y = 6 – 2x<\/span><\/li>\n<\/ol>\n

    \u00a02. Substitute into Line 2: <\/span> 6x + 2(6 – 2x) = 4<\/span><\/p>\n

    \u00a0\u00a0\u00a0<\/span> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a06x + 12 – 4x = 4<\/span><\/p>\n

    \u00a0\u00a0\u00a0<\/span> \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 \u00a0 2x = -8<\/span><\/p>\n

    \u00a0\u00a0\u00a0<\/span> \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 \u00a0 \u00a0 x = -4<\/span><\/p>\n

    3. Substitute back into Line 1 or 2: <\/span> y = 6 – 2(-4)<\/span><\/p>\n

    \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<\/span>y = 14<\/span><\/p>\n

    4. Solution: Point of intersection is (-4, 14)<\/span><\/p>\n

     <\/p>\n

    Elimination Method:<\/b><\/p>\n

    This method involves \u201celiminating\u201d one variable by finding the lowest common multiple for a chosen variable. Then you would put both lines as if you are adding or subtracting to find a final line just as you would for adding or subtracting large numbers. This removes leaves one variable to solve. This variable can then be substituted back into the other line to find the remaining variable. <\/span><\/p>\n

     <\/p>\n

      \n
    1. Line 1: 2x + y \u00a0\u00a0= 6<\/span> (x2)<\/span> *The \u2018y\u2019 variable seems like the simplest to <\/span><\/li>\n<\/ol>\n

      \u00a0 \u00a0 \u00a0Line 2: 6x + 2y = 4<\/span> (x1)<\/span> \u00a0\u00a0<\/span> \u00a0\u00a0get the lowest common multiple of. <\/span><\/p>\n

       <\/p>\n

      \u00a0 \u00a02. Line 1: 4x + 2y = 12<\/span><\/p>\n

      \u00a0 \u00a0 \u00a0 Line 2: 6x + 2y = 4<\/span>\u00a0\u00a0 \u00a0\u00a0<\/span><\/p>\n

      \u00a0 \u00a0 \u00a0Final: \u00a0-2x + 0y = 8<\/span><\/p>\n

      *Now that the \u2018y\u2019 variable are the same. We either have to subtract or add the variable to get 0y. In this case, we subtract.<\/span><\/p>\n

      \u00a0 \u00a03. Solve: -2x = 8<\/span><\/p>\n

      \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0x = -4<\/span><\/p>\n

      \u00a0 \u00a04. Substitute back into Line 1 or 2: <\/span> y = 6 – 2(-4)<\/span><\/p>\n

      \u00a0\u00a0\u00a0<\/span>\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 \u00a0 \u00a0y = 14<\/span><\/p>\n

      \u00a0 \u00a05. Solution: Point of intersection is (-4, 14)<\/span><\/p>\n

      There you go, both methods get you the same answer whenever asked to solve a linear system. Notice that when x or y has no coefficient, then substitution would be faster. If the x or y of both lines are the same then elimination would be faster.<\/span><\/p>\n

      SchoolTutoring Academy<\/a>\u00a0is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Markham, Ontario<\/a>: visit: Tutoring in Markham, Ontario<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

      When dealing with a system of linear equations there are two methods to algebraically solve the question. One is substitution and the other is elimination which is meant to be a shortcut. Both methods will bring you to the same solution but with more practice, you will recognize patterns and see which method would work […]<\/p>\n","protected":false},"author":17,"featured_media":67078,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[2,2845,3014,3015,11,3021,13,2850,3016,3017,1],"tags":[],"class_list":["post-67077","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-algebra","category-extra-credit","category-fromulae-and-equations","category-math-act","category-math-fundamentals","category-mathematics-sat","category-pre-algebra","category-psat","category-resources","category-review","category-uncategorized"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/67077","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/users\/17"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/comments?post=67077"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/posts\/67077\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media\/67078"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/media?parent=67077"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/categories?post=67077"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/scholarship\/wp-json\/wp\/v2\/tags?post=67077"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}