Inequalities by Example 

-- Section 1.7 --

 
  Solve for x:  3(x – 2)  £  2(x – 4)   

Solve

Step

 3(x – 2)  £  2(x – 4)  

 Simplify each side first

3x – 6 £  2x – 8

Remove parenthesis

3x – 6 + 6 £  2x – 8 + 6

 Add 6 to both sides

3x  £  2x – 2

Subtract 2x to both sides

3x – 2x  £  2x – 2x – 2

x  £  – 2

Inequality Notation

Click Link for Inequality Grapher.

(-¥, -2]

Interval Notation

[Click Here]

Graph

     Note:  ] means we include the number. We include -2 since we have the £  symbol.

     Solve for x:  4(x – 3)  £  3(x – 5)  

[Solution]

  Solve for x:  

Solve

Step

Multiply all 3 parts by 3 

Notice the 3's cancel 

6 <  4 – 2x < 9

Now the denominator is gone!

6 – 4 <  4 – 4 – 2x < 9 – 4

Subtract 4 from each part

2 <  – 2x <  5

Look!  If we divide by a negative we have to reverse the inequality signs

Divide all parts by -2

-1 > x > -5/2

Rewrite so we go from smallest to largest value

-5/2 <  x < -1

Inequality Notation

(-5/2, -1) 

Interval Notation

Graph
This means x has to be between -5/2 and -1 (x could not be -5/2 or -1 because of the < symbol)

     Solve for x: 

[Solution]

  Solve for x:     x2 – x  ³ 20  

I will expand upon Method 1 like example 7 on page 137.  

Solve

Step

    Check 

 x2 – x  ³ 20 

First, get zero on one side.  Both methods will fail if you forget this step.
 

  (5 – 5)(5+ 4) ³ 0

          0(9) = 0 ³ 0

x2 – x – 20 ³ 0 

Factor. Click here for help factoring.
 

(-4 – 5)(-4+ 4) ³ 0

(x – 5)(x+ 4) ³ 0

Clearly, 5 and -4 are solutions.

         -9(0) = 0 ³ 0

Since this an inequality, we expect more solutions.
The possible solutions are broken up into these intervals

Use -5,  0,  and 6

Pick points in these intervals to see if they work.  

Test Points

   Solution

if x = -5, then
(-5)2 – (-5) – 20 = 25 + 5 – 20 = 10 ³ 0  TRUE  So (–¥, –4] works.**

     (–¥, –4] 

if x = 0, then
(0)2 – (0) – 20 = 0 + 0 – 20 = –20 ³ 0  FALSE   So [–5, –4] fails.**  
if x = 6, then
(6)2 – (6) – 20 = 36 – 6 – 20 = 10 ³ 0  TRUE   So [5, ¥) works.**  

     [5, ¥)

The solutions is (–¥, –4]  È [5, ¥) 
 
**Important:  How do you know if the Test Point works?  
Check the inequality symbol.  x2 – x – 20 ³ 0  If the statement is true, it works.

    Solve for x:   x2 – 2x  ³ 8

[Solution]

  Solve for x:

I will expand upon Method 1 like example 9 on page 138.  

Solve

Step

Since zero is already on one side, just factor the numerator.

(x + 3)(x – 3) = 0 gives x =3, -3

Clearly, 3 and -3 are solutions.

Check 

Same for -3

x + 1 = 0 gives x = -1

But!!!  Cannot divide by zero

Check 

is undefined.  So -1 is not part of the solution.
Since this an inequality, we expect more solutions.
The possible solutions are broken up into these intervals

Use -4, -2,  0, and 4

Pick points in these intervals to see if they work.

Test Points

if x = -4, then

FALSE    So (–¥, –3] fails.** 
if x = -2, then

TRUE   So [–3, –1) works.**  

if x = 0, then

FALSE    So (–1, 3] fails.** 
if x = 4, then

TRUE   So [3, ¥)  works.**  

The solutions is [–3, –1)  È [3, ¥)
**Important: Remember that -1 is not included because division by zero in not allowed.  

    

 Solve for x: 

[Solution]