{"id":779,"date":"2021-09-18T21:55:25","date_gmt":"2021-09-18T21:55:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&p=779"},"modified":"2021-12-02T00:35:56","modified_gmt":"2021-12-02T00:35:56","slug":"6-3-1-graphing-linear-equations-in-two-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/6-3-1-graphing-linear-equations-in-two-variables\/","title":{"raw":"6.3.1: Graphing Linear Equations in Two Variables","rendered":"6.3.1: Graphing Linear Equations in Two Variables"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Plot linear equations in two variables on the coordinate plane.<\/li>\r\n \t
  • Use intercepts to plot lines.<\/li>\r\n \t
  • Use a graphing utility to graph a linear equation on a coordinate plane.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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    Key words<\/h3>\r\n
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    • Graph in two variables<\/strong>: a graph on a 2-dimensional plane<\/li>\r\n \t
    • [latex]x[\/latex]-intercept<\/strong>: the point where the graph crosses the [latex]x[\/latex]-axis<\/li>\r\n \t
    • [latex]y[\/latex]-intercept<\/strong>: the point where the graph crosses the [latex]y[\/latex]-axis<\/li>\r\n<\/ul>\r\n<\/div>\r\n

      Graphing Linear Equations<\/h2>\r\n

      Using Points to Plot Linear Equations<\/h3>\r\nTo graph a linear equation in two variables, we can plot a set of ordered pair solutions as points on a rectangular coordinate system. Its graph is called a\u00a0graph in two variables<\/strong><\/em>. Any graph on a two-dimensional plane is a graph in two variables.\r\n\r\nSuppose we want to graph the equation [latex]y=2x - 1[\/latex]. We can begin by finding solutions for the equation by substituting values for [latex]x[\/latex]\u00a0into the equation and determining the resulting value of [latex]y[\/latex]. Each pair of [latex]x[\/latex]\u00a0and [latex]y[\/latex]-values is an ordered pair that can be plotted. The table below\u00a0lists values of [latex]x[\/latex]\u00a0from \u20133 to 3 and the resulting values for [latex]y[\/latex].\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]y=2x - 1[\/latex]<\/td>\r\n[latex]\\left(x,y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-3[\/latex]<\/td>\r\n[latex]y=2\\left(-3\\right)-1=-7[\/latex]<\/td>\r\n[latex]\\left(-3,-7\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-2[\/latex]<\/td>\r\n[latex]y=2\\left(-2\\right)-1=-5[\/latex]<\/td>\r\n[latex]\\left(-2,-5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1[\/latex]<\/td>\r\n[latex]y=2\\left(-1\\right)-1=-3[\/latex]<\/td>\r\n[latex]\\left(-1,-3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex]y=2\\left(0\\right)-1=-1[\/latex]<\/td>\r\n[latex]\\left(0,-1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex]y=2\\left(1\\right)-1=1[\/latex]<\/td>\r\n[latex]\\left(1,1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex]y=2\\left(2\\right)-1=3[\/latex]<\/td>\r\n[latex]\\left(2,3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]3[\/latex]<\/td>\r\n[latex]y=2\\left(3\\right)-1=5[\/latex]<\/td>\r\n[latex]\\left(3,5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen we plot the points in the table, they form a line, so we can connect them.\r\n\r\nThis is not true for all equations, but the graph of a linear equation is always a line.\r\n\r\n\"This\r\n\r\nNote that the [latex]x[\/latex]-<\/em>values chosen are arbitrary regardless of the type of equation we are graphing. Of course, some situations may require particular values of [latex]x[\/latex]\u00a0<\/em>to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least 2 to graph a line and at least 3 to guarantee the line is correct. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.\r\n
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      How To graph a linear equation<\/h3>\r\n
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      1. Make a solutions table.<\/li>\r\n \t
      2. Plot the ordered pairs on a rectangular coordinate system.<\/li>\r\n \t
      3. Connect the points if they form a line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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        Example<\/h3>\r\nGraph the equation [latex]y=-x+2[\/latex] by plotting points.\r\n

        Solution<\/h4>\r\nFirst, we construct a table by choosing [latex]x[\/latex]-values and calculating the corresponding [latex]y[\/latex]-values.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/td>\r\n[latex]y=-x+2[\/latex]<\/td>\r\n[latex]\\left(x,y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-5[\/latex]<\/td>\r\n[latex]y=-\\left(-5\\right)+2=7[\/latex]<\/td>\r\n[latex]\\left(-5,7\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-3[\/latex]<\/td>\r\n[latex]y=-\\left(-3\\right)+2=5[\/latex]<\/td>\r\n[latex]\\left(-3,5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-1[\/latex]<\/td>\r\n[latex]y=-\\left(-1\\right)+2=3[\/latex]<\/td>\r\n[latex]\\left(-1,3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]0[\/latex]<\/td>\r\n[latex]y=-\\left(0\\right)+2=2[\/latex]<\/td>\r\n[latex]\\left(0,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]1[\/latex]<\/td>\r\n[latex]y=-\\left(1\\right)+2=1[\/latex]<\/td>\r\n[latex]\\left(1,1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]3[\/latex]<\/td>\r\n[latex]y=-\\left(3\\right)+2=-1[\/latex]<\/td>\r\n[latex]\\left(3,-1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]5[\/latex]<\/td>\r\n[latex]y=-\\left(5\\right)+2=-3[\/latex]<\/td>\r\n[latex]\\left(5,-3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow, plot the points. Connect them since they form a line.\r\n\r\n\"This\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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        Try It<\/h3>\r\nConstruct a table and graph the equation by plotting points: [latex]y=\\frac{1}{2}x+2[\/latex].\r\n\r\n[reveal-answer q=\"823002\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"823002\"]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}x+2[\/latex]<\/td>\r\n[latex]\\left(x,y\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-2[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}\\left(-2\\right)+2=1[\/latex]<\/td>\r\n[latex]\\left(-2,1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-1[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}\\left(-1\\right)+2=\\frac{3}{2}[\/latex]<\/td>\r\n[latex]\\left(-1,\\frac{3}{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]0[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}\\left(0\\right)+2=2[\/latex]<\/td>\r\n[latex]\\left(0,2\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]1[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}\\left(1\\right)+2=\\frac{5}{2}[\/latex]<\/td>\r\n[latex]\\left(1,\\frac{5}{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]2[\/latex]<\/td>\r\n[latex]y=\\frac{1}{2}\\left(2\\right)+2=3[\/latex]<\/td>\r\n[latex]\\left(2,3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \"This<\/div>\r\n
        [\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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        TRY IT<\/h3>\r\n