{"id":459,"date":"2019-04-29T13:21:45","date_gmt":"2019-04-29T17:21:45","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=459"},"modified":"2019-12-29T11:11:20","modified_gmt":"2019-12-29T16:11:20","slug":"3-4-graphing-linear-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/3-4-graphing-linear-equations\/","title":{"raw":"3.4 Graphing Linear Equations","rendered":"3.4 Graphing Linear Equations"},"content":{"raw":"[latexpage]\r\n\r\nThere are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.\r\n\r\nIf the equation is given in the form \\(y = mx + b\\), then \\(m\\) gives the rise over run value and the value \\(b\\) gives the point where the line crosses the \\(y\\)-axis, also known as the \\(y\\)-intercept.\r\n
\r\n

Example 3.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nGiven the following equations, identify the slope and the \\(y\\)-intercept.\r\n
    \r\n \t
  1. \\(\\begin{array}{lll} y = 2x - 3\\hspace{0.14in} & \\text{Slope }(m)=2\\hspace{0.1in}&y\\text{-intercept } (b)=-3 \\end{array}\\)<\/li>\r\n \t
  2. \\(\\begin{array}{lll} y = \\dfrac{1}{2}x - 1\\hspace{0.08in} & \\text{Slope }(m)=\\dfrac{1}{2}\\hspace{0.1in}&y\\text{-intercept } (b)=-1 \\end{array}\\)<\/li>\r\n \t
  3. \\(\\begin{array}{lll} y = -3x + 4 & \\text{Slope }(m)=-3 &y\\text{-intercept } (b)=4 \\end{array}\\)<\/li>\r\n \t
  4. \\(\\begin{array}{lll} y = \\dfrac{2}{3}x\\hspace{0.34in} & \\text{Slope }(m)=\\dfrac{2}{3}\\hspace{0.1in} &y\\text{-intercept } (b)=0 \\end{array}\\)<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nWhen graphing a linear equation using the slope-intercept method, start by using the value given for the \\(y\\)-intercept. After this point is marked, then identify other points using the slope.\r\n\r\nThis is shown in the following example.\r\n
    \r\n

    Example 3.4.2<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nGraph the equation \\(y = 2x - 3\\).\r\n\r\nFirst, place a dot on the \\(y\\)-intercept, \\(y = -3\\), which is placed on the coordinate \\((0, -3).\\)\r\n\r\n\"Graph\r\n\r\nNow, place the next dot using the slope of 2.\r\n\r\nA slope of 2 means that the line rises 2 for every 1 across.\r\n\r\nSimply, \\(m = 2\\) is the same as \\(m = \\dfrac{2}{1}\\), where \\(\\Delta y = 2\\) and \\(\\Delta x = 1\\).\r\n\r\nPlacing these points on the graph becomes a simple counting exercise, which is done as follows:\r\n\r\n\"For\r\n\r\n\"From\r\n\r\nOnce several dots have been drawn, draw a line through them, like so:\r\n\r\n\"Line\r\n\r\nNote that dots can also be drawn in the reverse of what has been drawn here.\r\n\r\nSlope is 2 when rise over run is \\(\\dfrac{2}{1}\\) or \\(\\dfrac{-2}{-1}\\), which would be drawn as follows:\r\n\r\n\"For\r\n\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n
    \r\n

    Example 3.4.3<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nGraph the equation \\(y = \\dfrac{2}{3}x\\).\r\n\r\nFirst, place a dot on the \\(y\\)-intercept, \\((0, 0)\\).\r\n\r\nNow, place the dots according to the slope, \\(\\dfrac{2}{3}\\).\r\n\r\n\"When\r\n\r\nThis will generate the following set of dots on the graph. All that remains is to draw a line through the dots.\r\n\r\n\"Line\r\n\r\n<\/div>\r\n<\/div>\r\nThe second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find \\(x\\) when \\(y = 0\\) and find \\(y\\) when \\(x = 0\\).\r\n
    \r\n

    Example 3.4.4<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nGraph the equation \\(2x + y = 6\\).\r\n\r\nTo find the first coordinate, choose \\(x = 0\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{lllll}\r\n2(0)&+&y&=&6 \\\\\r\n&&y&=&6\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((0, 6)\\).\r\n\r\nNow choose \\(y = 0\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{llrll}\r\n2x&+&0&=&6 \\\\\r\n&&2x&=&6 \\\\\r\n&&x&=&\\frac{6}{2} \\text{ or } 3\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((3, 0)\\).\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\"Line\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Example 3.4.5<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nGraph the equation \\(x + 2y = 4\\).\r\n\r\nTo find the first coordinate, choose \\(x = 0\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{llrll}\r\n(0)&+&2y&=&4 \\\\\r\n&&y&=&\\frac{4}{2} \\text{ or } 2\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((0, 2)\\).\r\n\r\nNow choose \\(y = 0\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{llrll}\r\nx&+&2(0)&=&4 \\\\\r\n&&x&=&4\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((4, 0)\\).\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\"Line\r\n\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Example 3.4.6<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nGraph the equation \\(2x + y = 0\\).\r\n\r\nTo find the first coordinate, choose \\(x = 0\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{llrll}\r\n2(0)&+&y&=&0 \\\\\r\n&&y&=&0\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((0, 0)\\).\r\n\r\nSince the intercept is \\((0, 0)\\), finding the other intercept yields the same coordinate. In this case, choose any value of convenience.\r\n\r\nChoose \\(x = 2\\).\r\n\r\nThis yields:\r\n\r\n\\[\\begin{array}{rlrlr}\r\n2(2)&+&y&=&0 \\\\\r\n4&+&y&=&0 \\\\\r\n-4&&&&-4 \\\\\r\n\\midrule\r\n&&y&=&-4\r\n\\end{array}\\]\r\n\r\nCoordinate is \\((2, -4)\\).\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\"Line\r\n\r\n<\/div>\r\n<\/div>\r\n

    Questions<\/h1>\r\nFor questions 1 to 10, sketch each linear equation using the slope-intercept method.\r\n
      \r\n \t
    1. \\(y = -\\dfrac{1}{4}x - 3\\)<\/li>\r\n \t
    2. \\(y = \\dfrac{3}{2}x - 1\\)<\/li>\r\n \t
    3. \\(y = -\\dfrac{5}{4}x - 4\\)<\/li>\r\n \t
    4. \\(y = -\\dfrac{3}{5}x + 1\\)<\/li>\r\n \t
    5. \\(y = -\\dfrac{4}{3}x + 2\\)<\/li>\r\n \t
    6. \\(y = \\dfrac{5}{3}x + 4\\)<\/li>\r\n \t
    7. \\(y = \\dfrac{3}{2}x - 5\\)<\/li>\r\n \t
    8. \\(y = -\\dfrac{2}{3}x - 2\\)<\/li>\r\n \t
    9. \\(y = -\\dfrac{4}{5}x - 3\\)<\/li>\r\n \t
    10. \\(y = \\dfrac{1}{2}x\\)<\/li>\r\n<\/ol>\r\nFor questions 11 to 20, sketch each linear equation using the \\(x\\text{-}\\) and \\(y\\)-intercepts.\r\n
        \r\n \t
      1. \\(x + 4y = -4\\)<\/li>\r\n \t
      2. \\(2x - y = 2\\)<\/li>\r\n \t
      3. \\(2x + y = 4\\)<\/li>\r\n \t
      4. \\(3x + 4y = 12\\)<\/li>\r\n \t
      5. \\(2x - y = 2\\)<\/li>\r\n \t
      6. \\(4x + 3y = -12\\)<\/li>\r\n \t
      7. \\(x + y = -5\\)<\/li>\r\n \t
      8. \\(3x + 2y = 6\\)<\/li>\r\n \t
      9. \\(x - y = -2\\)<\/li>\r\n \t
      10. \\(4x - y = -4\\)<\/li>\r\n<\/ol>\r\nFor questions 21 to 28, sketch each linear equation using any method.\r\n
          \r\n \t
        1. \\(y = -\\dfrac{1}{2}x + 3\\)<\/li>\r\n \t
        2. \\(y = 2x - 1\\)<\/li>\r\n \t
        3. \\(y = -\\dfrac{5}{4}x\\)<\/li>\r\n \t
        4. \\(y = -3x + 2\\)<\/li>\r\n \t
        5. \\(y = -\\dfrac{3}{2}x + 1\\)<\/li>\r\n \t
        6. \\(y = \\dfrac{1}{3}x - 3\\)<\/li>\r\n \t
        7. \\(y = \\dfrac{3}{2}x + 2\\)<\/li>\r\n \t
        8. \\(y = 2x - 2\\)<\/li>\r\n<\/ol>\r\nFor questions 29 to 40, reduce and sketch each linear equation using any method.\r\n
            \r\n \t
          1. \\(y + 3 = -\\dfrac{4}{5}x + 3\\)<\/li>\r\n \t
          2. \\(y - 4 = \\dfrac{1}{2}x\\)<\/li>\r\n \t
          3. \\(x + 5y = -3 + 2y\\)<\/li>\r\n \t
          4. \\(3x - y = 4 + x - 2y\\)<\/li>\r\n \t
          5. \\(4x + 3y = 5 (x + y)\\)<\/li>\r\n \t
          6. \\(3x + 4y = 12 - 2y\\)<\/li>\r\n \t
          7. \\(2x - y = 2 - y \\text{ (tricky)}\\)<\/li>\r\n \t
          8. \\(7x + 3y = 2(2x + 2y) + 6\\)<\/li>\r\n \t
          9. \\(x + y = -2x + 3\\)<\/li>\r\n \t
          10. \\(3x + 4y = 3y + 6\\)<\/li>\r\n \t
          11. \\(2(x + y) = -3(x + y) + 5\\)<\/li>\r\n \t
          12. \\(9x - y = 4x + 5\\)<\/li>\r\n<\/ol>\r\nAnswer Key 3.4<\/a>","rendered":"

            There are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.<\/p>\n

            If the equation is given in the form \"y = mx + b\", then \"m\" gives the rise over run value and the value \"b\" gives the point where the line crosses the \"y\"-axis, also known as the \"y\"-intercept.<\/p>\n

            \n
            \n

            Example 3.4.1<\/p>\n<\/header>\n

            \n

            Given the following equations, identify the slope and the \"y\"-intercept.<\/p>\n

              \n
            1. \"\begin{array}{lll} y = 2x - 3\hspace{0.14in} & \text{Slope }(m)=2\hspace{0.1in}&y\text{-intercept } (b)=-3 \end{array}\"<\/li>\n
            2. \"\begin{array}{lll} y = \dfrac{1}{2}x - 1\hspace{0.08in} & \text{Slope }(m)=\dfrac{1}{2}\hspace{0.1in}&y\text{-intercept } (b)=-1 \end{array}\"<\/li>\n
            3. \"\begin{array}{lll} y = -3x + 4 & \text{Slope }(m)=-3 &y\text{-intercept } (b)=4 \end{array}\"<\/li>\n
            4. \"\begin{array}{lll} y = \dfrac{2}{3}x\hspace{0.34in} & \text{Slope }(m)=\dfrac{2}{3}\hspace{0.1in} &y\text{-intercept } (b)=0 \end{array}\"<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

              When graphing a linear equation using the slope-intercept method, start by using the value given for the \"y\"-intercept. After this point is marked, then identify other points using the slope.<\/p>\n

              This is shown in the following example.<\/p>\n

              \n
              \n

              Example 3.4.2<\/p>\n<\/header>\n

              \n

              Graph the equation \"y = 2x - 3\".<\/p>\n

              First, place a dot on the \"y\"-intercept, \"y = -3\", which is placed on the coordinate \"(0, -3).\"<\/p>\n

              \"Graph<\/p>\n

              Now, place the next dot using the slope of 2.<\/p>\n

              A slope of 2 means that the line rises 2 for every 1 across.<\/p>\n

              Simply, \"m = 2\" is the same as \"m = \dfrac{2}{1}\", where \"\Delta y = 2\" and \"\Delta x = 1\".<\/p>\n

              Placing these points on the graph becomes a simple counting exercise, which is done as follows:<\/p>\n

              \"For<\/p>\n

              \"From<\/p>\n

              Once several dots have been drawn, draw a line through them, like so:<\/p>\n

              \"Line<\/p>\n

              Note that dots can also be drawn in the reverse of what has been drawn here.<\/p>\n

              Slope is 2 when rise over run is \"\dfrac{2}{1}\" or \"\dfrac{-2}{-1}\", which would be drawn as follows:<\/p>\n

              \"For<\/p>\n<\/div>\n

               <\/p>\n<\/div>\n

              \n
              \n

              Example 3.4.3<\/p>\n<\/header>\n

              \n

              Graph the equation \"y = \dfrac{2}{3}x\".<\/p>\n

              First, place a dot on the \"y\"-intercept, \"(0, 0)\".<\/p>\n

              Now, place the dots according to the slope, \"\dfrac{2}{3}\".<\/p>\n

              \"When<\/p>\n

              This will generate the following set of dots on the graph. All that remains is to draw a line through the dots.<\/p>\n

              \"Line<\/p>\n<\/div>\n<\/div>\n

              The second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find \"x\" when \"y = 0\" and find \"y\" when \"x = 0\".<\/p>\n

              \n
              \n

              Example 3.4.4<\/p>\n<\/header>\n

              \n

              Graph the equation \"2x + y = 6\".<\/p>\n

              To find the first coordinate, choose \"x = 0\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{lllll}<\/p>\n

              Coordinate is \"(0, 6)\".<\/p>\n

              Now choose \"y = 0\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{llrll}<\/p>\n

              Coordinate is \"(3, 0)\".<\/p>\n

              Draw these coordinates on the graph and draw a line through them.<\/p>\n

              \"Line<\/p>\n<\/div>\n<\/div>\n

              \n
              \n

              Example 3.4.5<\/p>\n<\/header>\n

              \n

              Graph the equation \"x + 2y = 4\".<\/p>\n

              To find the first coordinate, choose \"x = 0\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{llrll}<\/p>\n

              Coordinate is \"(0, 2)\".<\/p>\n

              Now choose \"y = 0\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{llrll}<\/p>\n

              Coordinate is \"(4, 0)\".<\/p>\n

              Draw these coordinates on the graph and draw a line through them.<\/p>\n

              \"Line<\/p>\n<\/div>\n<\/div>\n

              \n
              \n

              Example 3.4.6<\/p>\n<\/header>\n

              \n

              Graph the equation \"2x + y = 0\".<\/p>\n

              To find the first coordinate, choose \"x = 0\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{llrll}<\/p>\n

              Coordinate is \"(0, 0)\".<\/p>\n

              Since the intercept is \"(0, 0)\", finding the other intercept yields the same coordinate. In this case, choose any value of convenience.<\/p>\n

              Choose \"x = 2\".<\/p>\n

              This yields:<\/p>\n

                <\/span>   <\/span>\"\[\begin{array}{rlrlr}<\/p>\n

              Coordinate is \"(2, -4)\".<\/p>\n

              Draw these coordinates on the graph and draw a line through them.<\/p>\n

              \"Line<\/p>\n<\/div>\n<\/div>\n

              Questions<\/h1>\n

              For questions 1 to 10, sketch each linear equation using the slope-intercept method.<\/p>\n

                \n
              1. \"y = -\dfrac{1}{4}x - 3\"<\/li>\n
              2. \"y = \dfrac{3}{2}x - 1\"<\/li>\n
              3. \"y = -\dfrac{5}{4}x - 4\"<\/li>\n
              4. \"y = -\dfrac{3}{5}x + 1\"<\/li>\n
              5. \"y = -\dfrac{4}{3}x + 2\"<\/li>\n
              6. \"y = \dfrac{5}{3}x + 4\"<\/li>\n
              7. \"y = \dfrac{3}{2}x - 5\"<\/li>\n
              8. \"y = -\dfrac{2}{3}x - 2\"<\/li>\n
              9. \"y = -\dfrac{4}{5}x - 3\"<\/li>\n
              10. \"y = \dfrac{1}{2}x\"<\/li>\n<\/ol>\n

                For questions 11 to 20, sketch each linear equation using the \"x\text{-}\" and \"y\"-intercepts.<\/p>\n

                  \n
                1. \"x + 4y = -4\"<\/li>\n
                2. \"2x - y = 2\"<\/li>\n
                3. \"2x + y = 4\"<\/li>\n
                4. \"3x + 4y = 12\"<\/li>\n
                5. \"2x - y = 2\"<\/li>\n
                6. \"4x + 3y = -12\"<\/li>\n
                7. \"x + y = -5\"<\/li>\n
                8. \"3x + 2y = 6\"<\/li>\n
                9. \"x - y = -2\"<\/li>\n
                10. \"4x - y = -4\"<\/li>\n<\/ol>\n

                  For questions 21 to 28, sketch each linear equation using any method.<\/p>\n

                    \n
                  1. \"y = -\dfrac{1}{2}x + 3\"<\/li>\n
                  2. \"y = 2x - 1\"<\/li>\n
                  3. \"y = -\dfrac{5}{4}x\"<\/li>\n
                  4. \"y = -3x + 2\"<\/li>\n
                  5. \"y = -\dfrac{3}{2}x + 1\"<\/li>\n
                  6. \"y = \dfrac{1}{3}x - 3\"<\/li>\n
                  7. \"y = \dfrac{3}{2}x + 2\"<\/li>\n
                  8. \"y = 2x - 2\"<\/li>\n<\/ol>\n

                    For questions 29 to 40, reduce and sketch each linear equation using any method.<\/p>\n

                      \n
                    1. \"y + 3 = -\dfrac{4}{5}x + 3\"<\/li>\n
                    2. \"y - 4 = \dfrac{1}{2}x\"<\/li>\n
                    3. \"x + 5y = -3 + 2y\"<\/li>\n
                    4. \"3x - y = 4 + x - 2y\"<\/li>\n
                    5. \"4x + 3y = 5 (x + y)\"<\/li>\n
                    6. \"3x + 4y = 12 - 2y\"<\/li>\n
                    7. \"2x - y = 2 - y \text{ (tricky)}\"<\/li>\n
                    8. \"7x + 3y = 2(2x + 2y) + 6\"<\/li>\n
                    9. \"x + y = -2x + 3\"<\/li>\n
                    10. \"3x + 4y = 3y + 6\"<\/li>\n
                    11. \"2(x + y) = -3(x + y) + 5\"<\/li>\n
                    12. \"9x - y = 4x + 5\"<\/li>\n<\/ol>\n

                      Answer Key 3.4<\/a><\/p>\n","protected":false},"author":540,"menu_order":13,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-459","chapter","type-chapter","status-publish","hentry"],"part":360,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/459","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":18,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/459\/revisions"}],"predecessor-version":[{"id":3720,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/459\/revisions\/3720"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/360"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/459\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=459"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=459"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=459"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=459"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}