System of linier inequalities is the basic concept of a linear programming. That is, it is an algebraic system of operation under more than one constraints.

System of inequalities

System of inequalities is defined as a set of two or more inequalities. In a way, it is a compound inequality in two variables, related by conjunctions. A single inequality describes a situation under a single constraint. But a system of inequalities mean a situation under more than one constraint. Let us explain the all the situations with an example.
Suppose the specifications of a rectangle is described as follows. The length of the rectangle must be at least twice the width. The perimeter must be minimum of 30 inches and maximum of 60 inches. What could be the possible measures of the rectangle?
This is a case that describes a situation with three constraints. When you algebraically transform the conditions, assuming ‘x’ to be the width and ‘y’ to be the length, the constraints are described by the following inequalities.
y ≥ 2x, 2(x + y) ≥ 30 and  2(x + y) ≤ 60. Simplifying and expressing the inequalities in terms of ‘y’.
y ≥ 2x, y ≥ 15 – x and y ≤ 30 – x
In addition there is a hidden constrain!. Since no dimension of a rectangle can be 0, x > 0
This is nothing but a system of inequalities.        

Solving system of inequalities

The solution to a system of inequalities is the intersection of the sets of solutions of all the inequalities.  Let us try to solve the example given in the previous section. The best method of solution is by graphing.

First step is to assume all the inequalities to be equations and draw their graphs. Since all the equations are linear, the graphs of all the functions will be straight lines. You will find a quadrilateral is formed by these four straight lines as sides of the quadrilateral.
Next , consider the sides of the quadrilateral one by one and hatch the direction in which the corresponding inequality statement is true.
The inequality x > 0 is true to the right of the side representing x = 0.
The inequality y ≥ 2x  is true above the side representing y = 2x.
The inequality y ≥ 15 – x  is true above the side representing y = 15 – x
The inequality y ≤ 30 – x  is true below the side representing y = 30 – x
Thus, you will conclude that any point that lies on interior area of the quadrilateral satisfies all the inequalities. Shade that area to highlight it to be the solution to the given system of inequalities.
The ordered pair of any point inside the shaded region gives the dimensions of the triangle satisfying the given conditions. For example, the ordered pair of point P within the shaded area is (7,16) meaning a rectangle of 7in. wide and 16 in. long is a possible size. Because its length is more than twice the width. Its perimeter is 46 in. which lies between the limits of 30 in. and 60 in.
In some systems of inequalities you might find that the lines may not be able to share a common area. It implies that there is no solution to such systems of inequalities. For example,  the system of inequalities   y > 5x + 5 and y <  5x + 2 has no solution because their graphs do not share a common area.


Find the solution to the system of the inequalities  y  ≥ 3x + 4, y ≤ -2x + 5 and y ≥ 0   

Let us first draw the lines representing the equations y  = 3x + 4 and y ≤ -2x + 5. The corresponding equation for  y ≥ 0 is y = 0, which is nothing but the x-axis.
The solution to the inequality y  ≥ 3x + 4 are all the points that lie above the line  y  = 3x + 4.  The solution to the inequality y ≤ -2x + 5 are all the points that lie below the line  y  = 3x + 4. The solution to the inequality y ≥ 0 are all the points that lie above the x-axis. Therefore, the solution to the system of all these inequalities should therefore lie, above the line y  = 3x + 4 ‘and’ the x-axis ‘but’ below the line            y  = 3x + 4. This condition is satisfied by the area shaded as shown. In other words, the shaded area is the solution to the given system of inequalities. Since none of the inequalities is strict, all the points on all the boundaries of the shaded area are also included in the solution. It may be noted that the solution area has no boundary on the left and also the x-coordinate of the vertex at right is x = 1/5. Hence, the domain of the solution of system is (-infinity, 1/5]