Linear Inequalities
© Thomas O'Sullivan 1997


A line is a set of points that obey a certain condition.
e.g. 2x + 3y = 6 is "the set of points such that twice the first ordinate plus three times the second ordinate equals six". (-3,12),  (0,2), (1.5,1) etc. are all points on this line as they obey the fixed condition.
A "half plane" or a "side of a line" is also a set of points that obey a certain condition.
e.g.  2x + 3y
6 is  "the set of points such that twice the first ordinate plus three times the second ordinate is less than or equal to six". (2,1), (2,0), (1,1), (3, -1), (0,1) etc. are all points on this "side of the line" as they all obey the fixed condition.

To sketch the half plane represented by a linear inequality:

  1. Find two points on the line and draw the line.

  2. let x = 0 and get a value for y.

  3. if y0  in (I) then let y =0 and find a value of x.

  4. If y was = 0 let x = 1 and find another value for y. This gives two points which are plotted and the line through them is then drawn).

  1. Find the required side of the line:

  1. Choose a point (called a TEST POINT) clearly NOT on the line. (Preferably a point on one of the axes).

  2. Put this point into the inequality.

  3. If the inequality is true then the point is on the require side and this side should be shaded. (e.g. Above (3,0) could be the test point in the inequality this gives 2(3) + 3(0) 0 i.e. 6 0 which is true. Therefore, (3,0) is on the required side.

  4. If the inequality is false then the point is on the wrong side and the other side of the line should be shaded.


Linear inequalities and modulus (absolute values):


A system of linear inequalities can be written in modulus form.
e.g. Make a sketch of the region represented by  y
| x | and y  2 + | x |
(Note that modulus equations can be written "backwards" e.g. | x |
y is the same as y | x |)

| x | y means both x y and - x y. The associated lines x = y and  -x = y should then be sketched.
Similarily      y 
2 + | x |  gives the lines y  = 2 +  x  and y  = 2 - x  which should be sketched also.

The required area is now found by using  test points from the diagram and substituting these back into the modulus inequalities.
Test points from all possible regions are used viz: (-1,2), (-1,-2), (1,2), (1,-2) ( -3,0) (0,-3), (0,3), (3,0) and (0,1). These are best tested using a table:

The required region is then given by:

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