Graphing Inequalities

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Graphing Inequalities Powered By Docstoc
					Inequalities can be represented very well by
constructing graphs to illustrate their solutions.
EXAMPLE: y < 3
                  The graph for this is similar on
                  one level to the graph for y = 3.



y=3                                         y<3
The graphs for y > 3 and y ≥ 3 are similar to each other.
Notice that the difference between these 2
graphs is the boundary line that separates the
shaded portion from the unshaded portion. The
first boundary does not include y = 3 but the
second boundary does.



 y>3                                             y≥3
The following 5 graphs are similar however they all define a very
distinct set of ordered pairs.
                                       y = 3                  y ≤ 3
When the line is solid, the
points on the line are a
part of the solution.
                                (-2,3)       (5,3)                     (8,3)

When the line is broken,
the points on the line are x                            x                (7,1)
                                                            (-1,0)
NOT a part of the solution.
                                                                        (5,-4)
                                         y                     y
                y > 3                        y < 3                   y ≥ 3
            (8,1)
                                                            (-2,9)     (8,9)
                    (7,5)


                                                                        (6,3)
   x                        x                           x
                                 (0,0)
                                               (5,-2)

           y                         y                        y
Notice that when the equation changes to ‘x =‘
instead of ‘y =‘, the line becomes vertical instead
of horizontal. This change is reflected when
dealing with inequalities (<, >, ≤ or ≥) as well.

               x=3                        x<3
When both variables (x and y) are used, the line
will be neither horizontal nor vertical. It will be
oblique (inclined). That can mean a positive
(inclined up) or negative (inclined down) slope.
When there is an equation with both variables, it
is necessary to make a table of values to find 3
ordered pairs. These ordered pairs will establish
the direction of the oblique line.
                               STEP 1: Isolate y   x+y=5
Example: x + y = 5                                 y = -x + 5
STEP 2: Construct the Table of Values.

       x       y = -x + 5        y       (x,y)
       0       y = -(0) + 5      5       (0,5)
                   y= 5
       2       y = -(2) + 5      3       (2,3)
                   y= 3
      -2       y = -(-2) + 5     7       (-2,7)
                   y= 7
x+y=5                                               x+y>5
                                      If the ‘ = ‘ is changed to a ‘ < ‘, the
                                      solid line will become a broken line.
                                      At that point it is necessary to
                                      decide on shading one of the 2
                                      sides of the line. To do this we can
                                      choose a point from one side and
                                      test it in the inequality.




Let’s test (0,0) in the inequality.

   x + y > 5 (0,0)
      0+0>5
     0 > 5 False
If (0,0) is causes the inequality to
be false then so will all of the
points on that side of the line. We
can test other points to prove this.
  Test (2,1)        Test (-5,6)
   x+y>5              x+y>5
   2+1>5            (-5) + 6 > 5
  3 > 5 False       1 > 5 False
  Test (-3,-5)
    x+y>5
 (-3) + (-5) > 5        The point is that whenever a point on this
  -8 > 5 False          side of the line is tested, the result will be
                        false. This means that these are not
                        ordered pairs that make the statement, ‘x +
                        y > 5’, true.
This leads us to the obvious question:
Which ordered pairs can make this
inequality TRUE.
If we choose ordered pairs from the
other side of the line, we may get a
different result. Let’s choose (5,5).
If (5,5) makes the inequality TRUE,
so should the other points on the
same side of the boundary. We can
prove that this is the case for (0,8),
(6,0) and (10,-3).
We can express the solution for the inequality by shading the side
with the true ordered pairs. Notice that the boundary line is broken.
This indicates that ordered pairs that fall on the line will make the
inequality false. In other words, those ordered pairs are not a part of
the solution.
   Test (5,5)         Test (0,8)      Test (6,0)        Test (10,-3)
    x+y>5              x+y>5           x+y>5               x+y>5
    5+5>5              0+8>5           6+0>5             10 + (-3) > 5
  10 > 5 True         8 > 5 True      6 > 5 True          7 > 5 True
When deciding on which side of the line to shade, it is NOT really
necessary to test a bunch of ordered pairs. You need only test one.
Choose an ordered pair that is clearly NOT on the boundary. Test to
see if it makes the inequality TRUE or FALSE.
If the ordered pair makes the       If the ordered pair makes the
inequality TRUE, shade all of the   inequality FALSE, shade all of
points on that side.                the points on the OTHER side.
This way no matter whether you choose a point from true side or not,
the true side will end up being shaded. (Top side in this case)
With everything that was discussed so far, it is necessary to
demonstrate the steps necessary to construct a linear inequality graph.
                Example 1: 4x + 3y < 12
To graph an inequality, we must determine the boundary We can find
his by changing the inequality to an equation.
 4x + 3y < 12 becomes 4x + 3y = 12 to find the boundary.
 If there are no fractions in the inequality, we can isolate y right away.
                            3y = -4x + 12
                            3y  4x 12
                                    
                            3     3    3
                                 4x
                             y      4
                                  3
 Now it is time to construct a table of values to find 3 ordered
 pairs. Notice that the denominator of the x co-efficient is 3. This
 means that it would be very convenient to choose x-values that are
 multiples of 3 … like –6, -3, 0, 3 , 6, 9, etc.
  x      y
               4x
                   4
                                y   With the 3 points located on the
                3                   graph, we can see the direction of the
           4(0)                    line. We must decide whether the line
  0    y         4        4       should be a solid line (    ) or a broken
             3
         044                     line (    ). To make this decision, it is
  3        4(3)            0       necessary to observe the inequality
       y         4
             3                      sign in the original inequality.
          4  4  0
  6         4(6)           -4
       y         4
             3
          8  4  4

< or > requires a broken line
≤ or ≥ requires a solid line

            4x + 3y < 12
This example has a ‘less than’
sign which means the boundary
will be a broken line
4x + 3y < 12                Test (0,0) in 4x + 3y < 12

Where must we shade the             4(0) + 3(0) < 12
graph to indicate the points           0 + 0 < 12
that will make the equation
true? To find out we must             0 < 12 TRUE
choose and test a non-
boundary ordered pair in the
inequality.

(0,0) makes the inequality true.
All of the points on this side
would do the same. This is the
side that must be shaded.
Example 2: 3x - 5y ≥ 7
     Find boundary of 3x – 5y = 7
                               5 y  3x  7
                          5 y  3x 7           ≤ or ≥ requires a solid
        Isolate y                             line so the boundary
                          5    5 5           line should be solid in
                                3x  7          this case.
                             y
                                  5
 x        y
               3x  7            y
                 5
          3( 0)  7     7
 0   y                       -1.4
               5         5
         1.4
         3( 4)  7 12  7
 4   y                         1
             5          5
       1
         3( 1)  7  3  7
-1   y                        -2
              5          5
         2
                             Test (0,0) in 3x - 5y ≥ 7

(0,0) makes the inequality           3(0) + 5(0) ≥ 7
false. All of the points on this        0+0≥7
side would do the same. The
points on the other side must         0 ≥ 7 FALSE
make it true. It is therefore,
the other side that must be
shaded.
Solve the System of Inequalities:
       -x + 3y ≤ 3              4y + 3x + 9 > 0 
  Change the inequality signs to equal signs to find
the boundary for each.
        -x + 3y = 3             4y + 3x + 9 = 0 
  Isolate y for each.

       3y  x  3                    4 y  3x  9
       3y x 3                       4 y  3x 9
                                            
       3 3 3                        4      4     4
           1                              3x  9
       y  x1                       y
           3                                4
          1                                 3x  9
     y     x1                       y
          3                                   4
x     y
            1
              x1
                        y   x      y
                                         3x  9        y
            3                              4
           1                         3( 0)  9     9
0     y     ( 0)  1   1   0    y                   -2.25
           3                              4          4
          0 1 1                 2.25
          1                          3(1)  9  3  9
-3   y  ( 3)  1      0   1    y                    -3
          3                              4          4
         1  1  0                3
          1                          3( 3)  9 9  9
6     y  (6)  1       3   -3   y                     0
          3                               4          4
          21 3                  0


          (0,1)                       (0,-2.25)

      (-3,0)                            (1,-3)

          (6,3)                         (-3,0)
  -x + 3y ≤ 3                                4y + 3x + 9 > 0 
         Test (0,0) in -x + 3y ≤ 3   Test (0,0) in 4y + 3x + 9 > 0
(0,1)                                                           (0,-2.25)
            -(0) + 3(0) ≤ 3            4(0) + 3(0) + 9 > 0
(-3,0)                                                               (1,-3)
                0+0≤3                      0+0+9>3
(6,3)                                                                (-3,0)
              0 ≤ 3 TRUE                   9 > 0 TRUE
The solution to the system of inequalities is the area on
the grid that contains ordered pairs that makes both
inequalities true. Of the four regions defined by the
intersecting boundary lines, only one of those regions will
be true for BOTH inequalities. Some of the ordered pairs
in this region includes: (0,0), (10,2), (5,-4), (4,1) and (-1,-1)




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