Inequalities in two variables




Conventions

 
  =<    means equal or less than
  >=    means equal or greater than

Positive and negative areas in a plane

An x-y-plane is defined as a plane in which the x-axis, the y-axis, and the units are chosen in such a way that they form an orthonormal coordinate system.

Let f(x,y) be a polynomial in x and y.

The curve f(x,y)= 0 divides the x-y-plane 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 x-y-plane.

Example:

The parabola y - 2x2 - x - 4 = 0 divides the x-y-plane in two areas.
f(x,y) = y - 2x2 - 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.

Inequalities with two variables

General method

In the description of the method we use an x-y-plane.
The names of the two variables, x and y, are actually less important and they depend on the the nature of the application.

  1. Write the inequality in the form f(x,y) > 0 or f(x,y) >= 0
  2. Plot the graph of f(x,y)=0 in the x-y-plane
  3. The graph divides the x-y-plane in different areas.
  4. f(x,y) has a fixed sign in each area. Mark that area with this sign. This results in positive and negative areas in the plane.
  5. If the inequality has the form f(x,y) > 0, then the set of solutions is the set of the coordinates of the points in the positive areas.
  6. If the inequality has the form f(x,y) >= 0, then the set of solutions is the set of the coordinates of the points in the positive areas and the points on the boundary line.

Example 1

 
    x + 6 y >  4 - x

<=> x + 3 y - 2 > 0
Here is f(x,y) = x + 3 y - 2
We draw the line in the x-y-plane and two areas arise.
The area including point (0,0) is the negative area.
The area including point (0,1) is the positive area.
 
        
The set of solutions is the set of the coordinates of the points in the positive area.

Example 2

 
       4 x - 2 =< x2 - y

<=>    x2 - y - 4x + 2 >= 0
Here is f(x,y) = x2 - y - 4x + 2
The graph of f(x,y)= 0 is the parabola with equation y = x2 - 4x + 2
We draw the graph in the x-y-plane and two areas arise.
The area with point (0,0) is the positive area.
The area with point (1,0) is the negative area.
 
       
The set of solutions is the set of the coordinates of the points in the positive area or on the parabola itself.

Example 3

 
    b + 2 - a b < (3-b)2

<=> (3-b)2  + a b - 2 - b > 0

   The two variables are b and a
   We have an inequality of the form f(b,a) > 0.

<=>  b2 - 7b + a b + 7 > 0

   Now, f(b,a) = b2 - 7b + a b + 7
   The graph of f(b,a) = 0 in the b-a-plane is the graph with equation

        - b2 + 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.

Application 1

Given :
  • The hyperbola with equation y = 1/x.
  • The variable line with equation m x + n y + 4 = 0. The line depends on two parameters m and n.
Find all the values of m and n such that the hyperbola and the variable line have no common points.

The intersection points of the line and the hyperbola are the solutions of the system
 
  / 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 x2 + 4 x + n = 0

So,
    the hyperbola and the variable line have no common points

<=> The discriminant of m x2 + 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 m-n-plane, a hyperbola with equation  n = 4/m


       

  This hyperbola  divides the m-n-plane 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 region
Conclusion:
The variable line m x + n y + 4 = 0 and the hyperbola y = 1/x have no common points if and only if the point (m,n) is in the positive area.

Application 2

Given:
The variable line with equation m x + (n-1) 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 m2 + (n-1)2

              _______________
<=>     2 >  V m2 + (n-1)2

<=>     4 > m2 + (n-1)2

<=>     4 - m2 - (n-1)2 > 0

  f(m,n) =  4 - m2 - (n-1)2

  f(m,n) = 0  <=> m2 + (n-1)2 = 4

  In the m-n-plane, 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:
The distance from point O(0,0) to the variable line m x + (n-1) y + 2 = 0 is greater than 1 if and only if the point (m,n), in the m-n-plane , is inside the circle m2 + (n-1)2 = 4

Exercise 1

Given :
  • Parabola P1 with equation y = x2 + m x
  • Parabola P2 with equation y = - x2 + 2x - n
Find all the values of m and n such that the two parabolas are not intersecting.

Exercise 2

Solve
 
(2x + y -5)(x - y + 1)(x + y -1) > 0     (1)

Systems of inequalities with two variables

Method :

 
  Solve the inequalities separately.

  Find the intersection of the sets of solutions

Example

 
  / x2 - 7x + x y + 7 > 0
  \ x2 + y2 - 16 > 0



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