Knowing how to solve inequalities is an important skill. The first two sample problems demonstrate how linear inequalities are solved using the rules listed above.
 

Sample Problems:
 

1. 3x + 5 > 2x - 5

        x > -10

Note that the original inequality 3x+5>2x-5 is equivalent to the inequality x>-10. But this is also equivalent to x+10>0. Visually this looks like this:

This number line is showing that the expression x+10 is positive when x >10 , negative when x<-10, and zero when x = -10.

Another way to think about this is to think about the line y = x+10. Where does this line pass through the x-axis? 

Where is this line below the x-axis? Where is this line above the x-axis? The answer to these three questions will give you the same sign study. 

When the line y=x+10 is below the x-axis, the expression x+10 is negative. When the line y=x+10 is above the x-axis, the expression x+10 is positive. When the line y=x+10 passes through the x-axis, the expression x+10 is zero.

2. -5x + 2 > -8

        -5x > -10

        x < 2

Again note that the original inequality -5x+2>-8 is equivalent to x<2. Remember that this is also equivalent to x-2<0. Visually this looks like this:

The number line is showing that the expression x-2 is negative when x<2, positive when x>2 and zero when x = 2. You can create this sign study by thinking about the line y=x-2 as we did in the previous example.

3. (x+2)(x-5)>0

The third sample problem is a little more difficult to analyze. This inequality means that the product of two factors must be positive. This can happen two ways: both factors are positive or both factors are negative.

sign of (x+2)

sign of (x-5)

sign of (x+2)(x-5) or the product of the two factors:

(You can also build the third number line of signs by first selecting the two places where x+2 and x-5 each equal zero. Place these on the number line and then test sample values in each region to decide the sign of (x+2)(x-5).
 

So the solution is x>5 OR x<-2 because this is where the product of (x+2) and (x-5) is positive. (Be careful not to say the solution is x>5 and x<-2.) This answer cannot be combined into a triple inequality (The triple inequality 5<x<-2 is really an empty set.) It must be written as two inequalities joined together with the conjunction OR.

4. 

(This problem can be solved similar to example 3 after 1 is subtracted from each side of the inequality statement.)

To determine where this is true complete the following sign studies:

sign of (x-3)

sign of (x-2)

sign of  or the quotient of the two factors:

 
Again you can create this third line by simply placing a zero above the two locations x-3 and x-2 each equal zero. Then test sample points in each region to decide the sign of  Note that when x<2 or x>3 the fractionis negative or less than zero. You must exclude x=2 because x-2 is an expression in the denominator and a denominator of a fraction cannot equal zero.
 

So the solution is  (This is the same as saying x>2 AND .)
 

Check Your Understanding: (Remember to write down your answers.)
 

1. Which of the rules listed above are new to you? Explain what each of these says.
    
2. Give a numerical example of each rule.
3. Find all values of x where 2 - 5x > 7 and 3x + 1 > -11.
    
4. Find all values of x where 2x + 3 < 4 or 3x + 3 > 9.
    
5. Find all values of x where (x+3)(2x-1)<0
    
6. Find all values of x where 
7. Find all values of x where .

This page was modified on 05/29/10

© Rahn, 2000