=< means equal or less than >= means equal or greater than
Let f(x,y) be a polynomial in x and y.
The curve f(x,y)= 0 divides the xyplane in areas. f(x,y) has a fixed sign in each area. To know the sign in each area, it is sufficient to calculate that sign in a simple point of that area.
Finally, this results in positive and negative areas in the xyplane.
Example:
The parabola y  2x^{2}  x  4 = 0 divides the xyplane in two areas.
f(x,y) = y  2x^{2}  x  4.
The point O(0,0) belongs to a negative area because f(0,0) = 4 < 0.
The point P(0,5) belongs to a positive area because f(0,5) = 1 > 0.
x + 6 y > 4  x <=> x + 3 y  2 > 0Here is f(x,y) = x + 3 y  2
The set of solutions is the set of the coordinates of the points in the positive area.
4 x  2 =< x^{2}  y <=> x^{2}  y  4x + 2 >= 0Here is f(x,y) = x^{2}  y  4x + 2
The set of solutions is the set of the coordinates of the points in the positive area or on the parabola itself.
b + 2  a b < (3b)^{2} <=> (3b)^{2} + a b  2  b > 0 The two variables are b and a We have an inequality of the form f(b,a) > 0. <=> b^{2}  7b + a b + 7 > 0 Now, f(b,a) = b^{2}  7b + a b + 7 The graph of f(b,a) = 0 in the baplane is the graph with equation  b^{2} + 7b 7 a =  b Function investigation shows that it is a hyperbola with a =  b + 7 and b = 0 as asymptotes We plot that graph and there arise three areas. The area with point (0,0) is the positive area. The other areas are negative areas.The set of solutions is the set of the coordinates of the points in the positive area.
Given :

/ m x + n y + 4 = 0 \ y = 1/x The abscissae of the intersection points are the solutions of m x + n/x + 4 = 0 <=> m x^{2} + 4 x + n = 0 So, the hyperbola and the variable line have no common points <=> The discriminant of m x^{2} + 4 x + n = 0 is negative <=> 16  4 m n < 0 <=> m n  4 > 0 m n  4 > 0 is an inequality with two variables m and n. f(m,n) = m n  4 f(m,n) = 0 is, in the mnplane, a hyperbola with equation n = 4/m This hyperbola divides the mnplane in three areas. f(0,0) < 0 ; The point (0,0) is in a negative region f(4,4) > 0 ; The point (4,4) is in a positive region f(4,4) > 0 ; The point (4,4) is in a positive regionConclusion:
Given: The variable line with equation m x + (n1) y + 2 = 0. The line depends on two parameters m and n. Find all the values of m and n such that the distance from point O(0,0) to the given line is greater than 1. 
The distance from point O(0,0) to the given line is greater than 1. 2 <=>  > 1 _________________ V m^{2} + (n1)^{2} _______________ <=> 2 > V m^{2} + (n1)^{2} <=> 4 > m^{2} + (n1)^{2} <=> 4  m^{2}  (n1)^{2} > 0 f(m,n) = 4  m^{2}  (n1)^{2} f(m,n) = 0 <=> m^{2} + (n1)^{2} = 4 In the mnplane, this is a circle with center (0,1) and radius 2. Draw a figure f(0,1) > 0 ; the area inside the circle is a positive area. f(4,0) < 0 ; the area outside the circle is a negative area.Conclusion:
Given :

Solve
(2x + y 5)(x  y + 1)(x + y 1) > 0 (1) 
Method :
Solve the inequalities separately. Find the intersection of the sets of solutions
/ x^{2}  7x + x y + 7 > 0 \ x^{2} + y^{2}  16 > 0
x^{2}  7x + x y + 7 > 0 Now, f(x,y) = x^{2}  7x + y x + 7 The graph of f(x,y) = 0 in the xyplane is the graph with equation  x^{2} + 7x 7 y =  x Function investigation shows that it is a hyperbola with y =  x + 7 and x = 0 as asymptotes We plot that graph and there arise three areas. The area with point (0,0) is the positive area. The other areas are negative areas. The graph and the areas are marked in red.
x^{2} + y^{2}  16 > 0Now, f(x,y) is the equation of a circle with center (0,0) and radius 4. There are two areas. The area inside the circle is a negative area. The area outside the circle is a positive area. The graph and the areas are marked in green.