
Not all statements in mathematics involve dealing with equal quantities.
Sometimes, we may only know that something is "greater than" a certain value,
or "less than or equal to" another value.
These situations are referred to as inequalities (because they are not simply "equal").
Inequality Notations: (see other notation forms at Notations for Solutions) 
a > b ; a is strictly greater than b 
a b ; a is greater than or equal to b 
a < b ; a is striclty less than b 
a b ; a is less than or equal to b 
a ≠ b ; a is not equal to b 
Hint: The "open" (larger) part of the inequality symbol always faces the larger quantity. 

If you can solve a linear equation, you can solve a linear inequality. The process is the same, with one important exception ... 

... when you multiply (or divide) an inequality by a negative value,
you must change the direction of the inequality. 

Let's see why this "exception" is actually needed.
We know that 3 is less than 7.
Now, lets
multiply both sides by 1.
Examine the results (the products). 
... written 3 < 7.
... written (1)(3) ? (1)(7)
... written 3 ? 7

On a number line, 3 is to the right of 7, making 3 greater than 7.
3 > 7
We have to reverse the direction of the inequality,
when we multiply by a negative value, in order to maintain a "true" statement.


When graphing a linear inequality on a number line, use an open circle for "less than" or "greater than", and a closed circle for "less than or equal to" or "greater than or equal to". 
To CHECK an inequaltiy, it is not possible to test every value.
So check a value in each shaded region to see if it is TRUE.
Then check a value in each nonshaded region to see if it is FALSE.
Graph the solution set of: x 3 
The solution set for this problem will be all values that are graphed to the right of 3, and including 3.
Graph using closed circle for 3 since x = 3 is among the solutions. 
CHECK:
A number in the shaded region = TRUE.
A number in the nonshaded area = FALSE.
Pick 0: 0 > 3 TRUE
Pixk 4: 4 > 3 FALSE

Graph the solution set of: 3 < x < 4 
AND 

The solution set for this problem will be all values that satisfy both 3 < x and x < 4. Look for where the two inequalities overlap.
The solution will be all x's that are greater than 3 and at the time are less than +4.
Graph using open circles for 3 and 4 (since x can not equal 3 nor 4), and a bar to show the overlapping section.

This type of inequality may be referred to as a "sandwich" since the solutions are sandwiched between 3 and +4.
CHECK: pick 1: 3 < 1 < 4 TRUE
pick 4: 3 < 4 < 4 FALSE
pick 5: 3 < 5 < 4 FALSE

Graph the solution set of: x < 3 or x 1 
OR 

The solution set for this problem will be the full graph of both inequalities, since the two inequalities do not overlap.
Your answers can lie in either section. The solution could be:
in x < 3 or in x > 1.

Notice that there is one open circle
(for 3) and one closed circle (for 1).

CHECK: Remember "OR" is true when either (or both) section is true, and false when both are false.
pick 4: 4 < 3 OR 4 > 1 This is TRUE since 4 < 3 is true.
pick 3: 3 < 3 OR 3 > 1 This is TRUE since 3 > 1 is true.
pick 1: 1 < 3 OR 1 > 1 This is FALSE since both sections are false. 
Solve and graph the solution set of: 4x < 24 
Proceed as you would when solving a linear equation:
Divide both sides by 4.
Note: The direction of the inequality stays the same since we did NOT divide by a negative value.
Graph using an open circle for 6 (since x can not equal 6) and an arrow to the left (since our symbol is less than). 
CHECK: pick 1: 1 < 6 TRUE
pick 7: 7 < 6 FALSE

Solve and graph the solution set of: 5 x 25 
Divide both sides by 5.
Note: The direction of the inequality was reversed since we divided by a negative value.
Graph using a closed circle for 5 (since x can equal 5) and an arrow to the left (since our final symbol is less than or equal to). 
CHECK: pick 7: 7 < 5 TRUE
pick 0: 0 < 5 FALSE

Solve and graph the solution set of: 3x + 4 > 13 
Proceed as you would when solving a linear equation:
Subtract 4 from both sides.
Divide both sides by 3.
Note: The direction of the inequality stays the same since we did NOT divide by a negative value.
Graph using an open circle for 3 (since x can not equal 3) and an arrow to the right (since our symbol is greater than). 
CHECK: pick 4: 4 > 3 TRUE
pick 0: 0 > 3 FALSE

Solve and graph the solution set of: 9  2 x 3 
Subtract 9 from both sides.
Divide both sides by 2.
Note: The direction of the inequality was reversed since we divided by a negative value.
Graph using a closed circle for 3 (since x can equal 3) and an arrow to the right (since our symbol is greater than or equal to). 
CHECK: pick 4: 4 > 3 TRUE
pick 0: 0 > 3 FALSE

Solve and graph the solution set of: 5  2 x 13 + 2 x 
Add 2x to both sides.
Subtract 13 from both sides.
Divide both sides by 4.
Note: There was no multiplication or division by a negative value, so the inequality symbol did not get reversed.
Graph using a closed circle for 2 (since x can equal 2).
CHECK: pick 0: 0 > 2 TRUE
pick 4: 4 > 2 FALSE 

Solve and graph the solution set of: 4x + 10 < 3(2x + 4) 
Distribute across parentheses.
Subtract 4x from both sides.
Add 12 to both sides.
Divide both sides by 2.
Note: There was no multiplication or division by a negative value, so the inequality symbol did not get reversed.
Graph using a closed circle for 2 (since x can equal 2).
CHECK: pick 0: 0 > 1 TRUE
pick 3: 3 > 1 FALSE 



Yes, there is a way to determine solutions for inequalities on your graphing calculator. Click the calculator at the right to see how to use the calculator with single variable inequalities.




For calculator help with inequalities
(single variable)
click here. 


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