We know that the expressions: 3x + 1 = x - 3  and  x² - 3x = 0 are equations. But expressions like these are called "inequations" when instead of being joined together by an equals sign (=), the two sides are joined together by one of the following:

> (greater than); < (less than); # (different to); ³ (greater than or equal to); £ (less than or equal to)

For example: 3x + 1 > x - 3  or x² - 3x £ 0

As is the case with equations, they are ordered by degree and by the number of unknowns.

In the two examples the first will be a linear inequality with one unknown and the second will be quadratic inequality with one unknown.

As with equations, we solve an inequation by finding the x value or values which satisfy the relation.

Inequations where the two sides are joined by a # sign (different to), can be considered as solved as we just need to solve the equation and the solution to the inequation is any other real number except that of the solution given.

Example: Solve the inequation: x² + x - 2 # 0 .

The solutions to the equation: x² + x - 2 = 0 are x = -2 and x = 1, so the solutions to the inequation are all the real number values except 1 and -2 (i.e.) R - {-2, 1}

We need to draw the graphs which correspond to each side of the inequation. In this example we have two straight lines as shown below: 

2.- Note which line is 3x + 1 and which is x - 3. Note for which values of x the points of the LHS are "above those of the RHS".

Both expressions are written in the top left-hand corner of the window together with their value for each value of x.

The "zone" on the X-axis which represents all the values of x which satisfy the inequation is highlighted in green and are all values greater than -2.

The solution can be written as the interval: (-2, ¥). 

3.- Note that the equation is satisfied when x = -2 , so x = -2 "does not satisfy" the inequation. The inequation is satisfied by any value greater than -2, i.e. the "open" interval from -2.

NUMERICAL SOLUTION

Inequations are solved numerically by using similar methods to solving equations but with differences depending on the degree of the inequation or the complexity of the expression. One example is: 3x + 1 > x - 3 which we found the graphical solution to earlier. To solve this inequation numerically we use the same method as with equations (getting x on its own): 3x - x > -3 - 1. ; 2x > -4; x > -4/2 ; x > -2, which gives us the same solution as before..  Careful: When getting "x" on its own we have to divide by a negative number, which is the same as changing the sign of all the terms in the inequality, the inequality sign should "be reversed". For example:

We are going to use this window to solve inequations in two exercises.

1.- Solve the inequation x - 1 ³ 5x + 3

This gives -4x ³ -4; x £ -4/-4 ; x £ 1

2.- Use the window to check the solution, writing in the corresponding equations. Change the values of x until you reach the point where the two straight lines cross each other and:

Careful: "You must decide if solution is in the blue zone or the green zone"

3.- Find both the graphical solutions, changing the equations appropriately, and numerical solutions in your exercise book for the following inequations:

a) 1-3x £ 2x - 9

b) x/2 - x/3 > 1

Any inequation where the expressions on both sides are polynomials of a second or higher degree are known as quadratic or higher degree inequations. They are all solved in the same way as a quadratic one.

2.- Use the arrows to change the values of x and look carefully at the red and blue points and at the numerical values of the LHS and RHS given in the top left-hand corner of the window.

3.- How would the solution be written? Note that it is linked to the green part of the X-axis.

In terms of intervals it can be written as: (-¥, -2) U (2, ¥)¥, -2) U (2, ¥) where U means "union".  

We can "try" any value from each interval in the inequation, which if true, makes the whole interval valid; for example, let's try:

x = -3 ( 9 - 4 > 0 "true" ) ; x = 0 ( -4 > 0 "false" ) and x = 4 ( 16 - 4 > 0 " true" ) , which gives us the intervals (-¥, -2) and (2, ¥) we found graphically as being valid for the solution

NUMERICAL SOLUTION

The easiest way to solve these equations numerically is to express the inequation so that the RHS is equal to 0. The previous example: x² + 2x - 1 > 2x + 3 gives

x² - 4 > 0 . We then solve the equation: x² - 4 = 0 , which has the obvious solutions x = 2 and x = -2. We just need to bear in mind now that the possible intervals of the solution are: (-¥, -2), (-2 , 2) and (2, ¥).

If we bring everything over to the LHS so that the RHS of the inequation is equal to 0, graphically speaking we would get the following graph shown in the window. We can see that the solution of the inequation can be obtained when the graph corresponding to the LHS (parabola) is "above" the X-axis (y = 0).

a) x² - 3x + 2 ³ 0. . . . . . . . . . . b) 1 - 2x < x - x² + 1

Here we have to find x values which satisfy two inequations at the same time.

In this case it is advisable to make the RHS of both inequations equal to 0 and solve them separately. Then, we must find the x values which belong to the intervals of both solutions.

The graphs corresponding to the RHS of both inequations are drawn (the LHS equal to 0) and we note for which x values both graphs are above or below the X-axis at the same time, according to the inequality sign.

x - 2 < 0

x² +4x - 5 > 0

2.- Look at the solution in the window, changing the x values and observing when the red graph is below and the blue graph above the X-axis.

What is the solution?

Surely you can see that the solution to this exercise is: (- ¥, -5) U (1 , 2)

3.- Solve the simultaneous inequations numerically bearing in mind that you need to solve both inequations separately, as we saw earlier, and that the intervals which satisfy both equations at the same time will be the solution of the simultaneous inequations.

Sometimes we need to solve inequations where the expressions are what we call "rational" or quotients of polynomials.

One way is: The graphical solution can be found by drawing the graph which corresponds to the LHS and seeing for which intervals the inequation is satisfied. In this case the graph can be drawn easily in the window (using the LHS expression of the inequation as the first equation and y = 0 as the second) and we can see that the solution is found at the values -2, -1 and from 2 to infinity, indicating when the intervals are closed or open:       Solution: (-2 , -1] U (2, ¥)

Another way is to solve the simultaneous inequations separately:

x + 1 ³ 0 , x² - 4 > 0  

x + 1 < 0 , x² - 4 < 0

and find the intervals where both solutions are satisfied.

2.- Solve the inequation: (x - 2) / (x + 1) > 0 numerically in your exercise book. Check the result graphically by writing the appropriate equation in the window above.

In general, any inequation can be solved graphically, using the methods we have seen here, and numerically depending on the type of expression.