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Solving Linear Inequalities (single variable)
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A quick review of working with linear inequalities:

bullet Inequality Notations:
(see more possible notation forms)
a > b ;    a is strictly greater than b
a < b ;    a is striclty less than b
Note: a > b and a < b are called "strict inequalities".
a greatereqal b ;   a is greater than or equal to b
a lessequal 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.
bullet Inequality Important Rule:
The process of solving a linear inequality is the same as solving a linear equation, except ...

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

Check out the "Refresh Section" if you need more review on working with linear inequalities.

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Let's start by justifying why the "Inequality Important Rule" (stated above) is true.
Statement: If you multiply (or divide) both sides of an inequality by a negative value,
you will need to change the direction of the inequality.

1.    If a > b, then -a < -b.
1. Translation of statement (mult. by -1).
2.    a > b
2. Start with the "given" a > b.
3.    a - b > b - b
3. Subtract b from both sides.
4.    a - b > 0
4. Additive Inverse Property b - b = 0
5.    a - a - b > 0 - a
5. Subtract a from both sides.
6.    0 - b > -a
6. Additive Inverse, Additive Identity
7.    -b > -a       
7. Additive Identity Property
8.    -a < -b
8. Read in reverse (converse)

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bullet Basic Properties of Inequalities: (properties apply to symbols )

If a > b, then a + c > b + c.
Addition Property of Inequality
If a > b, then a - c > b - c.
Subtraction Property of Inequality
If a > b and c > 0, then ac > bc.
Multiplication Property of Inequality
Notice that c > 0.
If c < 0, reverse the direction of the final inequality.
If a > b and c > 0, then iqmultrule.
Division Property of Inequality


bullet Solving Linear Inequalities:

      For more basic problems on solving inequalities, see Refresher section.

ex1
ti84c
For calculator help with linear inequalities
click here.

Solve and graph the solution set of:   qmath3a
Proceed as you would when solving a linear equation:
Add 6 to both sides.
Multiply both sides by threehalfs.


Note:
The direction of the inequality stays the same since we did NOT multiply by a negative value.


Graph using a closed circle for 6 (since x can equal 6) and an arrow to the left (since our symbol is less than or equal to).

iqmath5a

circlegraph5

Other notations:
ex1notation



ex2

Solve and graph the solution set of:   5(x - 3) > 10
Distribute (remove the parentheses).

Add 15 to both sides.

Divide both sides by 5.

Note: The direction of the inequality stays the same since we did NOT multiply by a negative value.

Graph using a open circle for 5 (since x can NOT equal 5) and an arrow to the right (since our symbol is greater than).

1qmath9

circlegraph11

Other notations:
ex1notation



ex3

Solve and graph the solution set of:   iqmath10
Multiply both sides by the least common denominator, which is 8.

Divide both sides by -1, and flip the direction of the inequality.


Note: The direction of the inequality was reversed since we divided by a negative value (-1).

Graph using a closed circle for -2 (since x can equal -2) and an arrow to the left (since our symbol is less than or equal to).

iqmath10a

circlegraph12

Other notations:
ex1notation



ex3

Solve and graph the solution set of:    4(x - 1) > 3(x - 2)
Distribute across both sets of parentheses.

Subtract 3x from both sides.
The solution is easier if you move the smaller x value.

Add 4 to both sides.

Note: The direction of the inequality stays the same since we did NOT multiply or divide by a negative value.

Graph using an open circle for -2 (since x can not equal -2) and an arrow to the right (since our symbol is greater than).

iqmath6

circlegraph6

Other notations:
ex1notation



ex3 beware

Solve and graph the solution set of:    4(2x + 1) < 8x + 4
Distribute across the parentheses.

Subtract 8x from both sides.

Notice the FALSE result.

The solution to this inequality is the "empty set" (empty). There are no x-values which will make this inequality true.

There is nothing to be graphed.

iqmath11
This tells us that NO x-values will make this inequality true.

circlegraph13

In this problem, the left side of the inequality is simply another way of writing the right side of the inequality. The two sides are EQUAL to one another. Since a quantity can never be less than itself, this inequality is never true.


ex3

Solve and graph the solution set of:    iqmath12
Proceed as you would when solving a linear equation:

Subtract 3 from both sides.

Multiply both sides by 2.

The solution to this inequality is all real numbers except 2. The number 2 is the only value which makes this inequality false, since:
iqmath12c

igmath12b

circlegraph14a

Other notations:
ex1notation



ex3

Solve and graph the solution set of:   iqmath13
Hint: In this problem, the negative signs appear in front of the two fractional terms. You can place the negative sign in the numerator, or the denomoniator, since
igmath14a

This example places the negative sign in the numerators.

Multiply all terms by the least common denominator (which is 6).

Add 4x to both sides.
The solution is easier if you move the smaller x value.

Be sure to write the 0 on the right side.

Subtract 12 from both sides.

Divide both sides by 3.

 

igmath14

circlegraph15

Other notations:
ex1notation



ex3 beware

Solve for x:    ax > 6a
At first glance, it seems that dividing both sides by a will solve the problem.

BUT ...

We just don't know if a is a positive value or a negative value. If a is positive, our solution of x > 6 will be true.

But if a is negative, we need to reverse the direction of the inequality, and the solution is x < 6.

Since we do not know the value of a, we cannot state a solution to this problem.
Therefore, there is NO SOLUTION!

positive_________________________________

negative

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reminder An inequality cannot be solved by dividing by a "variable" (a letter) if you do not know that the "variable" is either always positive or always negative. (And, of course, division by zero is never allowed.)




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