(1)x - 2y = 5. (2) 3x - 2y = 3

Question

 Solve the following systems of equations by drawing a graph. Give the solution and identify the type of lines found in this system.  Give the tables of value

Answer 1

The two equations have variables x and y in the first degree.  Thus, they are equations of lines.

 

Two points determine a line.

 

Usually, it is easy to find the y-intercept [the point where the line crosses the y-axis at x=0; that is, (0,y)] and the x-intercept [the point where the line crosses the x-axis at y=0; that is, (x,0)].

 

For equation (1):    x - 2y = 5    (or y = x/2 - 5/2)

   This is a upward-sloping line (m=+1/2) when going left-to-right.

    y-intercept, at x=0,   point (0,-5/2)

    x-intercept, at y=0,   point (5,0)]

 

    Table of values:

         x      y

         0     -5/2

         5      0

       (you may add more values if you want)

  Draw a line thru these two points and keep going in both directions.

 

For equation (2):   3x - 2y = 3     ([or y = (3/2)x - 3]

     This is an upward-sloping line (m=+3/2) when going left-to-right.

     y-intercept is (0,-3/2)

     x-intercept is (1,0)

 

      Table of values:

          x      y

          0     -3/2

          1      0

   Draw this line.

 

The lines cross where both equation (1) and equation (2) are true.  The graph should show this clearly.

   Algebraically:

       x - 2y = 5

      3x -2y  = 3

     --------------   (Elimination; subtract equations)

      -2x      = 2

        x       =-1

        x - 2y = 5

    -----------------  (Elimination; subtract equations)

            2y  = -6

              y = -3

    The point (-1,-3) satisfies both equation (1) and equation (2).  It is the point where the lines cross.

 

Check:

    Is   -1 -2(-3) = 5   ?

          -1 + 6   = 5   ?

             5   =   5   ?yes

    Is   3(-1) - 2(-3) = 3   ?

           -3    +  6   = 3   ?

                 3    =    3   ?yes