{"id":1545,"date":"2023-09-29T10:18:09","date_gmt":"2023-09-29T10:18:09","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1545"},"modified":"2023-09-29T10:22:50","modified_gmt":"2023-09-29T10:22:50","slug":"week-2-jss-3-second-term-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-jss-3-second-term-mathematics-notes\/","title":{"rendered":"Week 2 – Jss 3 Second Term Mathematics Notes"},"content":{"rendered":"

WEEK 2
\n<\/strong><\/p>\n

SOLVING OF SIMULTANEOUS LINEAR EQUATIONS
\n<\/h2>\n

A linear equation is an equation with one solution, in equation known as linear there is only one or two variable unknown variables. But in the case where we combine two equations (linear) thereby having what is known as Simultaneous equation. E.g
\n4x + y = 8\u00a0\u00a0\u00a0\u00a0(1)\u00a0\u00a0\u00a0\u00a0are called simultaneous equations
\n3x \u2013 y = 6\u00a0\u00a0\u00a0\u00a0(2)\u00a0\u00a0\u00a0\u00a0<\/p>\n

\u00a0Simultaneous Linear equations can be solved, graphically, algebraically. But in today’s class we shall be considering the algebraically method of solution.<\/p>\n

\u00a0<\/p>\n

ALGEBRAIC METHOD
\n<\/h3>\n

There are two algebraic methods of solving simultaneous equations. These are:
\n(a)\u00a0\u00a0\u00a0\u00a0Substitution method
\n(b)\u00a0\u00a0\u00a0\u00a0Elimination method<\/p>\n

\u00a0<\/p>\n

Substitution Method
\n<\/h3>\n

To use substitution method<\/p>\n

    \n
  1. \n
    Re-arrange one of the equations so that one variable is made the subject of the formula of the equation.\n<\/div>\n<\/li>\n
  2. \n
    Substitute this into the other equations.\n<\/div>\n<\/li>\n
  3. \n
    Solve the resulting equation to obtain one variable.\n<\/div>\n<\/li>\n
  4. \n
    The other variable is found by substituting your answer into the original equation.\n<\/div>\n<\/li>\n
  5. \n
    Check the solutions by substituting the two answers back into the original equation.\n<\/div>\n<\/li>\n<\/ol>\n

    \u00a0WRITE ABOUT
    \nExample 1
    \nSolve the following simultaneous equations by substitution method.<\/p>\n

      \n
    1. \n
      y = 5x + 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0ii.\u00a0\u00a0\u00a0\u00a02x + 3y = 5\n<\/div>\n

      x + 2y = 15\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x + y = 4\n<\/li>\n<\/ol>\n

      iii.\u00a0\u00a0\u00a0\u00a04m \u2013 3n = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0iv.\u00a0\u00a0\u00a0\u00a0x + 6y = -2
      \n\u00a0\u00a0\u00a0\u00a0m + 2n = 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x + 2y = 10<\/p>\n

      \u00a0Solution
      \nII.\u00a0\u00a0\u00a0\u00a02x + 3y = 5\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0(1)
      \n\u00a0\u00a0\u00a0\u00a03x + y = 4\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0(2)<\/p>\n

      \u00a0Step (1)
      \nLabel the 1st<\/sup> equation (1) and the second equation (2) for easy reference later on.
      \nStep (2)
      \nFrom equation (2) make “y” subject of formulae \u00a0\u00a0\u00a0\u00a03x + y = 4
      \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\u00a0\u00a0Y = 4 \u2013 3x \u2026\u2026\u2026\u2026\u2026 (3)
      \nStep (3)
      \nSubstitute y = 4 \u2013 3x into equation (1)
      \n2x + 3y = 5
      \n2x + 3 (4 \u2013 3x) = 5
      \nStep (4)
      \nOpen the brackets and solve for x.
      \n2x + 12 \u2013 9x = 5
      \n12 \u2013 7x = 5
      \n\"\"\/\"\"\/12 \u2013 5 \u2013 7x = 10
      \n\"\"\/\"\"\/ =
      \nx = 1<\/p>\n

      \u00a0\"\"\/Step 5
      \nSubstitute for x = 1 into equation\u2026\u2026\u2026\u2026. (3)
      \ny = 4 \u2013 3x; y = 4 \u2013 3(1), y = 4 \u2013 3
      \ny = 4 \u2013 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0check
      \ny = 1\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
      \nHence: x = 1, y = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02(1) + 3(1)
      \nis the solution to the equation\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02 + 3 = 5
      \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\u00a0\u00a0In Equation (2)
      \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\u00a0\u00a03(1) + (1)
      \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\u00a0\u00a03 + 1 = 4
      \nExample II
      \n4m \u2013 3n = 0
      \nm + 2n = 3
      \nStep (1)
      \nLabel the equations
      \n4m \u2013 3n = 0\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0(1)
      \nm + 2n = 3\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0(2)
      \nStep 2
      \nMake “m” subject of formula in equations (2)
      \nm + 2n = 3
      \nm = 3 \u2013 2n \u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.\u00a0\u00a0\u00a0\u00a0(3)
      \nStep 3
      \nSubstitute m = 3 \u2013 2n into equation \u2026\u2026\u2026. (1)
      \n4m \u2013 3n = 0
      \n4(3 \u2013 2n) \u2013 3n = 0
      \nStep 4
      \nOpen the bracket and solve for “n”
      \n12 \u2013 8n \u2013 3n = 0
      \n\"\"\/12 \u2013 11n = 0
      \n =
      \nn = 1<\/p>\n

      \u00a0Step 5
      \nSubstitute the value on n = into equation\u2026\u2026\u2026\u2026.. (3)
      \nm = 3 \u2013 2
      \nm =
      \n =
      \nHence m =, n = <\/p>\n

      \u00a0WRAP UP AND ASSESSMENT
      \nTwo equations are called simultaneous equations if they are to be solved at the same time. In substitution method make one variable the subject and then substitute this value in the other equation.
      \nSolve the following simultaneously using substitution method.
      \n(1)\u00a0\u00a0\u00a0\u00a0x + 6y = -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2)\u00a0\u00a0\u00a0\u00a0-2 = 5x \u2013 y
      \n\u00a0\u00a0\u00a0\u00a03x + 2y = 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a015 = x + 2y<\/p>\n

      \u00a0(3)\u00a0\u00a0\u00a0\u00a04x + 7y = 20
      \n\u00a0\u00a0\u00a0\u00a03x + y = -2<\/p>\n

      \u00a0TICKET OUT
      \nSolve the following simultaneous Equation by substitution method.
      \nExercise 16.3 pg 149 No 11 \u2013 1
      \n
      \n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

      WEEK 2 SOLVING OF SIMULTANEOUS LINEAR EQUATIONS A linear equation is an equation with one…<\/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-1545","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\/1545","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=1545"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1545\/revisions"}],"predecessor-version":[{"id":1546,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1545\/revisions\/1546"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1545"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1545"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1545"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}