{"id":1494,"date":"2023-09-29T08:33:38","date_gmt":"2023-09-29T08:33:38","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=1494"},"modified":"2023-09-29T08:34:40","modified_gmt":"2023-09-29T08:34:40","slug":"week-10-and-11-jss-3-first-term-mathematics-lesson-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-10-and-11-jss-3-first-term-mathematics-lesson-notes\/","title":{"rendered":"Week 10 and 11 – Jss 3 First Term Mathematics Lesson Notes"},"content":{"rendered":"

WEEK 10
\n<\/strong>SIMPLE EQUATIONS INVOLVING FRACTIONS
\n<\/strong>Examples: Solve the following equations<\/p>\n

    \n
  1. \n
    \u2013\n<\/div>\n<\/li>\n
  2. \n
    P =\n<\/div>\n<\/li>\n
  3. \n
    = 5 +\n<\/div>\n<\/li>\n<\/ol>\n

    SOLUTION
    \n<\/strong><\/p>\n

      \n
    1. \n
      The L.C.M of 5 and 10 is 10 multiply both sides by 10\n<\/div>\n<\/li>\n<\/ol>\n

      10 x = 10 ()\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0OR<\/strong>\u00a0\u00a0\u00a0\u00a0we cross multiply
      \n2 x = 8 \u2013 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
      \n = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a010 x = 5 x (8 \u2013)
      \n = 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 40 –
      \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
      \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 1
      \n(b) \u00a0\u00a0\u00a0\u00a0P = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) Multiply both sides by 6
      \n2p = 35 \u2013 3p\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06 x = 6 x 5 + 6 x
      \n = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03 (a-5) = 30 + 2a
      \nP = 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03a \u2013 15 = 30 + 2a
      \na = 45
      \nExercise: Solve the following equations
      \n(a) = 5 (b) = (c) = 8 (d) + = 0
      \nFraction with binomial denominator
      \n<\/strong>Examples
      \na) – 4 = 0 (b) – = 0 (c) =
      \nSOLUTION
      \n<\/strong><\/p>\n

        \n
      1. \n
        – 4 = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) – = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) =\n<\/div>\n

        – 4 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 (2y \u2013 1) = 5 (y + 2)
        \n4 (5 -) = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 (2y \u2013 3) = 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6y \u2013 3 = 5y + 10
        \n20 – 4 = 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6y – 9 = 20\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Y = 13
        \n4 = 18\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
        \n = 45\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 y = 4\n<\/li>\n<\/ol>\n

        EXERCISE
        \n<\/strong> a)\u00a0\u00a0\u00a0\u00a0 + 8 = -3 (b) = 4 (c) = 1 (d) =
        \nSimultaneous linear Equations
        \n<\/strong>These are equations such as = 8 and = 6
        \nGraphical Method
        \n<\/strong>To solve simultaneous equations graphically<\/p>\n

          \n
        • \n
          Make a table of values for both equations\n<\/div>\n<\/li>\n
        • \n
          Draw the graphs of both equations on the same axes\n<\/div>\n<\/li>\n
        • \n
          Find the coordinate where both graph interest. This values () are the solutions of both axes\n<\/div>\n<\/li>\n<\/ul>\n

          Examples: Solve the following simultaneous equations graphically (a) and (b) and r
          \nSOLUTION
          \n<\/strong><\/p>\n

            \n
          1. \n
            Y = -2 + 0.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\n<\/div>\n

            \"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0 0 \u00a0\u00a0\u00a0\u00a02 \u00a0\u00a0\u00a0\u00a0 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1 \u00a0\u00a0\u00a0\u00a00 \u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0 = 2
            \n\u00a0\u00a0\u00a0\u00a0 -2\u00a0\u00a0\u00a0\u00a0-1\u00a0\u00a0\u00a0\u00a00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -7\u00a0\u00a0\u00a0\u00a0-5\u00a0\u00a0\u00a0\u00a05\u00a0\u00a0\u00a0\u00a0 = -1<\/p>\n

            \u00a0<\/li>\n

          2. \n
            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\n<\/div>\n

            \"\"\/\"\"\/\"\"\/\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0 -1 \u00a0\u00a0\u00a0\u00a00 \u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -2 \u00a0\u00a0\u00a0\u00a00 \u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0 = 2
            \n\u00a0\u00a0\u00a0\u00a0 12\u00a0\u00a0\u00a0\u00a010\u00a0\u00a0\u00a0\u00a0 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -6\u00a0\u00a0\u00a0\u00a0-2\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0 = -1\n<\/li>\n<\/ol>\n

            \u00a0Exercise<\/p>\n

              \n
            1. \n
              (b)\n<\/div>\n<\/li>\n<\/ol>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
              \nSUBSTITUTION METHOD
              \n<\/strong>Examples (a) , 2X \u2013 Y = 52 (b) <\/p>\n

              \t\tSOLUTION
              \n<\/strong><\/p>\n

                \n
              1. \n
                \u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u20261\n<\/div>\n<\/li>\n<\/ol>\n

                \u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026\u2026.2
                \nStep 1: Rearrange one of the equations so that are variable is made the subject
                \nThat is from eqn I\u00a0\u00a0\u00a0\u00a03
                \nStep 2: Substitute into the second equation. That is substitute and solves the resulting equation.
                \n Into eqn\u2026\u2026\u2026\u20262
                \n – y = 5
                \n =
                \n Y = -1<\/p>\n

                \u00a0
                \n\u00a0
                \n\u00a0Step 3: Substitute your answer into 3 to find the other variable
                \nThat is <\/p>\n

                  \n
                1. \n
                  \n\t\t\t\t<\/div>\n<\/li>\n<\/ol>\n

                  From eqn\u2026\u2026..1 \u2026\u2026… 3
                  \nSubstitute eqn3 into eqn 2 gives<\/p>\n

                  \t\t=
                  \n = 2
                  \n From Eqn 3<\/p>\n

                  \t\tEXERCISE:<\/strong> (b) <\/p>\n

                  \u00a0ELIMINATION METHOD
                  \n<\/strong>Examples: (a)
                  \nWhen one of the unknown has equal coefficient
                  \n<\/strong>SOLUTION
                  \n<\/strong><\/p>\n

                    \n
                  1. \n
                    \u2026\u2026\u2026\u2026. 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\n<\/div>\n

                    +\u2026\u2026\u2026\u2026… 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                    \n = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 – a = -5\n<\/li>\n<\/ol>\n

                    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0a = 5
                    \nFrom eqn 1 \u00a0\u00a0\u00a0\u00a0 from Eqn 1
                    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                    \n\u00a0\u00a0\u00a0\u00a0 =
                    \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
                    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n

                    \u00a0
                    \n\u00a0
                    \n\u00a0Example 2 (When none of the unknown has equal coefficient)
                    \n<\/strong>Example: \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b)
                    \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                    \nSOLUTION<\/strong><\/p>\n

                      \n
                    1. \n
                      To make the coefficients of x equal multiply eqn 1 by 2 and eqn 2 by 1\n<\/div>\n

                      2 x 1\u00a0\u00a0\u00a0\u00a0\u2026\u2026\u2026.. 3
                      \n1 x 2 \u2026\u2026\u2026\u2026. 4
                      \n\u00a0\u00a0\u00a0\u00a0 =
                      \n\u00a0\u00a0\u00a0\u00a0<\/p>\n

                      \u00a0From Eqn 1
                      \n\u00a0\u00a0\u00a0\u00a0
                      \n\u00a0\u00a0\u00a0\u00a0
                      \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\n<\/li>\n

                    2. \n
                      \u00a0\u00a0\u00a0\u00a0——-1\n<\/div>\n<\/li>\n<\/ol>\n

                      \u00a0\u00a0\u00a0\u00a0——-2
                      \n\u00a0\u00a0\u00a0\u00a04 x 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0——–3
                      \n\u00a0\u00a0\u00a0\u00a03 x 2\u00a0\u00a0\u00a0\u00a0 \u2026\u2026\u2026\u2026\u2026.4
                      \n From Eqn (1) <\/p>\n

                      \t\t\u00a0\u00a0\u00a0\u00a0 = <\/p>\n

                      \t\tEXERCISE:<\/strong> ,
                      \nWORD PROBLEMS
                      \n<\/strong>EXAMPLES
                      \n<\/strong><\/p>\n

                        \n
                      1. \n
                        The sum of two numbers is 30 and their difference is 15. Find the two numbers\n<\/div>\n<\/li>\n
                      2. \n
                        3 boxes and 2 packages weigh 1240g while 5 boxes and 7 packages weigh 2800g. What is the weight of a box and a package?\n<\/div>\n<\/li>\n<\/ol>\n

                        SOLUTION<\/strong><\/p>\n

                          \n
                        1. \n
                          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2)\n<\/div>\n<\/li>\n<\/ol>\n

                          \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =
                          \n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \nFrom (1)
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                          \n\u00a0\u00a0\u00a0\u00a0
                          \nExercise:
                          \n<\/strong><\/p>\n

                            \n
                          1. \n
                            The sum of two numbers is 18 and their difference is 12. Find the two numbers\n<\/div>\n<\/li>\n
                          2. \n
                            This shape is an equilateral triangle with dimension show finds its perimeter.\n<\/div>\n

                            \"\"\/\n\t\t\t\t<\/li>\n<\/ol>\n

                            -2 4x \u2013 y + 1<\/p>\n

                            \u00a0
                            \n\u00a0<\/p>\n

                              \n
                            1. \n
                              Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob. How much does each one actually have?\n<\/div>\n<\/li>\n
                            2. \n
                              In a two digit number. The units digit is thrice the tens digit. If 36 is added to the number, the digits interchange their place. Find the number.\n<\/div>\n<\/li>\n
                            3. \n
                              If 2 is added to the numerator and denominator it becomes 9\/10 and if 3 is subtracted from the numerator and denominator it become 4\/5. Find the fractions.\n<\/div>\n<\/li>\n<\/ol>\n

                              ASSIGNMENT:
                              \n<\/strong>EXERCISE 15.5; NO 2 \u2013 5.PAGE 127
                              \n<\/strong>
                              \n\u00a0WEEK 11
                              \n<\/strong>REVISION and EXAMINATION
                              \n<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

                              WEEK 10 SIMPLE EQUATIONS INVOLVING FRACTIONS Examples: Solve the following equations \u2013 P = =…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,138],"tags":[],"class_list":["post-1494","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-jss-3-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1494","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=1494"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1494\/revisions"}],"predecessor-version":[{"id":1495,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/1494\/revisions\/1495"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=1494"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=1494"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=1494"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}