{"id":1228,"date":"2021-12-02T19:37:02","date_gmt":"2021-12-03T00:37:02","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/3-4-graphing-linear-equations\/"},"modified":"2023-08-30T12:31:05","modified_gmt":"2023-08-30T16:31:05","slug":"graphing-linear-equations","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/graphing-linear-equations\/","title":{"raw":"3.4 Graphing Linear Equations","rendered":"3.4 Graphing Linear Equations"},"content":{"raw":"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.\r\n\r\nIf the equation is given in the form [latex]y = mx + b[\/latex], then [latex]m[\/latex] gives the rise over run value and the value [latex]b[\/latex] gives the point where the line crosses the [latex]y[\/latex]-axis, also known as the [latex]y[\/latex]-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 [latex]y[\/latex]-intercept.\r\n
    \r\n \t
  1. [latex]\\begin{array}{lll} y = 2x - 3\\hspace{0.14in} & \\text{Slope }(m)=2\\hspace{0.1in}&y\\text{-intercept } (b)=-3 \\end{array}[\/latex]<\/li>\r\n \t
  2. [latex]\\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}[\/latex]<\/li>\r\n \t
  3. [latex]\\begin{array}{lll} y = -3x + 4 & \\text{Slope }(m)=-3 &y\\text{-intercept } (b)=4 \\end{array}[\/latex]<\/li>\r\n \t
  4. [latex]\\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}[\/latex]<\/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 [latex]y[\/latex]-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 [latex]y = 2x - 3[\/latex].\r\n\r\nFirst, place a dot on the [latex]y[\/latex]-intercept, [latex]y = -3[\/latex], which is placed on the coordinate [latex](0, -3).[\/latex]\r\n\r\n\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, [latex]m = 2[\/latex] is the same as [latex]m = \\dfrac{2}{1}[\/latex], where [latex]\\Delta y = 2[\/latex] and [latex]\\Delta x = 1[\/latex].\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\r\n\r\nOnce several dots have been drawn, draw a line through them, like so:\r\n\r\n\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 [latex]\\dfrac{2}{1}[\/latex] or [latex]\\dfrac{-2}{-1}[\/latex], 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 [latex]y = \\dfrac{2}{3}x[\/latex].\r\n\r\nFirst, place a dot on the [latex]y[\/latex]-intercept, [latex](0, 0)[\/latex].\r\n\r\nNow, place the dots according to the slope, [latex]\\dfrac{2}{3}[\/latex].\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 [latex]x[\/latex] when [latex]y = 0[\/latex] and find [latex]y[\/latex] when [latex]x = 0[\/latex].\r\n
    \r\n

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

    \r\n\r\nGraph the equation [latex]2x + y = 6[\/latex].\r\n\r\nTo find the first coordinate, choose [latex]x = 0[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{lllll}\r\n2(0)&+&y&=&6 \\\\\r\n&&y&=&6\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](0, 6)[\/latex].\r\n\r\nNow choose [latex]y = 0[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{llrll}\r\n2x&+&0&=&6 \\\\\r\n&&2x&=&6 \\\\\r\n&&x&=&\\frac{6}{2} \\text{ or } 3\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](3, 0)[\/latex].\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\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 [latex]x + 2y = 4[\/latex].\r\n\r\nTo find the first coordinate, choose [latex]x = 0[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{llrll}\r\n(0)&+&2y&=&4 \\\\\r\n&&y&=&\\frac{4}{2} \\text{ or } 2\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](0, 2)[\/latex].\r\n\r\nNow choose [latex]y = 0[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{llrll}\r\nx&+&2(0)&=&4 \\\\\r\n&&x&=&4\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](4, 0)[\/latex].\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\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 [latex]2x + y = 0[\/latex].\r\n\r\nTo find the first coordinate, choose [latex]x = 0[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{llrll}\r\n2(0)&+&y&=&0 \\\\\r\n&&y&=&0\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](0, 0)[\/latex].\r\n\r\nSince the intercept is [latex](0, 0)[\/latex], finding the other intercept yields the same coordinate. In this case, choose any value of convenience.\r\n\r\nChoose [latex]x = 2[\/latex].\r\n\r\nThis yields:\r\n\r\n[latex]\\begin{array}{rlrlr}\r\n2(2)&+&y&=&0 \\\\\r\n4&+&y&=&0 \\\\\r\n-4&&&&-4 \\\\\r\n\\hline\r\n&&y&=&-4\r\n\\end{array}[\/latex]\r\n\r\nCoordinate is [latex](2, -4)[\/latex].\r\n\r\nDraw these coordinates on the graph and draw a line through them.\r\n\r\n\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. [latex]y = -\\dfrac{1}{4}x - 3[\/latex]<\/li>\r\n \t
    2. [latex]y = \\dfrac{3}{2}x - 1[\/latex]<\/li>\r\n \t
    3. [latex]y = -\\dfrac{5}{4}x - 4[\/latex]<\/li>\r\n \t
    4. [latex]y = -\\dfrac{3}{5}x + 1[\/latex]<\/li>\r\n \t
    5. [latex]y = -\\dfrac{4}{3}x + 2[\/latex]<\/li>\r\n \t
    6. [latex]y = \\dfrac{5}{3}x + 4[\/latex]<\/li>\r\n \t
    7. [latex]y = \\dfrac{3}{2}x - 5[\/latex]<\/li>\r\n \t
    8. [latex]y = -\\dfrac{2}{3}x - 2[\/latex]<\/li>\r\n \t
    9. [latex]y = -\\dfrac{4}{5}x - 3[\/latex]<\/li>\r\n \t
    10. [latex]y = \\dfrac{1}{2}x[\/latex]<\/li>\r\n<\/ol>\r\nFor questions 11 to 20, sketch each linear equation using the [latex]x\\text{-}[\/latex] and [latex]y[\/latex]-intercepts.\r\n
        \r\n \t
      1. [latex]x + 4y = -4[\/latex]<\/li>\r\n \t
      2. [latex]2x - y = 2[\/latex]<\/li>\r\n \t
      3. [latex]2x + y = 4[\/latex]<\/li>\r\n \t
      4. [latex]3x + 4y = 12[\/latex]<\/li>\r\n \t
      5. [latex]2x - y = 2[\/latex]<\/li>\r\n \t
      6. [latex]4x + 3y = -12[\/latex]<\/li>\r\n \t
      7. [latex]x + y = -5[\/latex]<\/li>\r\n \t
      8. [latex]3x + 2y = 6[\/latex]<\/li>\r\n \t
      9. [latex]x - y = -2[\/latex]<\/li>\r\n \t
      10. [latex]4x - y = -4[\/latex]<\/li>\r\n<\/ol>\r\nFor questions 21 to 28, sketch each linear equation using any method.\r\n
          \r\n \t
        1. [latex]y = -\\dfrac{1}{2}x + 3[\/latex]<\/li>\r\n \t
        2. [latex]y = 2x - 1[\/latex]<\/li>\r\n \t
        3. [latex]y = -\\dfrac{5}{4}x[\/latex]<\/li>\r\n \t
        4. [latex]y = -3x + 2[\/latex]<\/li>\r\n \t
        5. [latex]y = -\\dfrac{3}{2}x + 1[\/latex]<\/li>\r\n \t
        6. [latex]y = \\dfrac{1}{3}x - 3[\/latex]<\/li>\r\n \t
        7. [latex]y = \\dfrac{3}{2}x + 2[\/latex]<\/li>\r\n \t
        8. [latex]y = 2x - 2[\/latex]<\/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. [latex]y + 3 = -\\dfrac{4}{5}x + 3[\/latex]<\/li>\r\n \t
          2. [latex]y - 4 = \\dfrac{1}{2}x[\/latex]<\/li>\r\n \t
          3. [latex]x + 5y = -3 + 2y[\/latex]<\/li>\r\n \t
          4. [latex]3x - y = 4 + x - 2y[\/latex]<\/li>\r\n \t
          5. [latex]4x + 3y = 5 (x + y)[\/latex]<\/li>\r\n \t
          6. [latex]3x + 4y = 12 - 2y[\/latex]<\/li>\r\n \t
          7. [latex]2x - y = 2 - y \\text{ (tricky)}[\/latex]<\/li>\r\n \t
          8. [latex]7x + 3y = 2(2x + 2y) + 6[\/latex]<\/li>\r\n \t
          9. [latex]x + y = -2x + 3[\/latex]<\/li>\r\n \t
          10. [latex]3x + 4y = 3y + 6[\/latex]<\/li>\r\n \t
          11. [latex]2(x + y) = -3(x + y) + 5[\/latex]<\/li>\r\n \t
          12. [latex]9x - y = 4x + 5[\/latex]<\/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 [latex]y = mx + b[\/latex], then [latex]m[\/latex] gives the rise over run value and the value [latex]b[\/latex] gives the point where the line crosses the [latex]y[\/latex]-axis, also known as the [latex]y[\/latex]-intercept.<\/p>\n

            \n
            \n

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

            \n

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

              \n
            1. [latex]\\begin{array}{lll} y = 2x - 3\\hspace{0.14in} & \\text{Slope }(m)=2\\hspace{0.1in}&y\\text{-intercept } (b)=-3 \\end{array}[\/latex]<\/li>\n
            2. [latex]\\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}[\/latex]<\/li>\n
            3. [latex]\\begin{array}{lll} y = -3x + 4 & \\text{Slope }(m)=-3 &y\\text{-intercept } (b)=4 \\end{array}[\/latex]<\/li>\n
            4. [latex]\\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}[\/latex]<\/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 [latex]y[\/latex]-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 [latex]y = 2x - 3[\/latex].<\/p>\n

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

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

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

              \"For<\/p>\n

              \"image\"<\/p>\n

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

              \"image\"<\/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 [latex]\\dfrac{2}{1}[\/latex] or [latex]\\dfrac{-2}{-1}[\/latex], 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 [latex]y = \\dfrac{2}{3}x[\/latex].<\/p>\n

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

              Now, place the dots according to the slope, [latex]\\dfrac{2}{3}[\/latex].<\/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 [latex]x[\/latex] when [latex]y = 0[\/latex] and find [latex]y[\/latex] when [latex]x = 0[\/latex].<\/p>\n

              \n
              \n

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

              \n

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

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

              This yields:<\/p>\n

              [latex]\\begin{array}{lllll} 2(0)&+&y&=&6 \\\\ &&y&=&6 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](0, 6)[\/latex].<\/p>\n

              Now choose [latex]y = 0[\/latex].<\/p>\n

              This yields:<\/p>\n

              [latex]\\begin{array}{llrll} 2x&+&0&=&6 \\\\ &&2x&=&6 \\\\ &&x&=&\\frac{6}{2} \\text{ or } 3 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](3, 0)[\/latex].<\/p>\n

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

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

              \n
              \n

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

              \n

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

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

              This yields:<\/p>\n

              [latex]\\begin{array}{llrll} (0)&+&2y&=&4 \\\\ &&y&=&\\frac{4}{2} \\text{ or } 2 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](0, 2)[\/latex].<\/p>\n

              Now choose [latex]y = 0[\/latex].<\/p>\n

              This yields:<\/p>\n

              [latex]\\begin{array}{llrll} x&+&2(0)&=&4 \\\\ &&x&=&4 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](4, 0)[\/latex].<\/p>\n

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

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

              \n
              \n

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

              \n

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

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

              This yields:<\/p>\n

              [latex]\\begin{array}{llrll} 2(0)&+&y&=&0 \\\\ &&y&=&0 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](0, 0)[\/latex].<\/p>\n

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

              Choose [latex]x = 2[\/latex].<\/p>\n

              This yields:<\/p>\n

              [latex]\\begin{array}{rlrlr} 2(2)&+&y&=&0 \\\\ 4&+&y&=&0 \\\\ -4&&&&-4 \\\\ \\hline &&y&=&-4 \\end{array}[\/latex]<\/p>\n

              Coordinate is [latex](2, -4)[\/latex].<\/p>\n

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

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

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

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

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

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

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

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

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