When solving a linear inequality, treat it just like you were solving an equation with a few exceptions.

- When you multiply or divide both sides of an inequality by a negative constant, change the sense (direction) of the inequality.
- When both sides of an inequality are the same sign, change the sense of the inequality when you take the reciprocal of both sides.
- You may chain like inequalities together:
- If a<b and b<c, then a<c (transitive property)
- If a<b and c<d, then a+c<b+d. In English, that means that if you take two things that are smaller and put them together, it will be smaller than the two larger things put together.

- You can not combine mixed inequalities.
- If a<b and b>c, then you can't say for sure that a<c or that c<a.
- If a<b and c>d, then you can't say for sure that a+c<b+d or that a+c>b+d

- If you rewrite the entire problem, just switching sides, make sure you change the sense of the inequality so that it still points to the same quantity.
- The following operations do
**not**change the sense of the inequality- Adding a constant to both sides of an inequality
- Subtracting a constant from both sides of an inequality
- Multiplying or dividing both sides of an inequality by a positive constant

Sometimes, two inequalities are combined into one. However, you need to be careful:

If x>3 and x<6, then you can write 3<x<6. But, if x<3 or x>6, you can **not** write 3>x>6 because
that would imply that 3>6 and that would be a false statement and the set would be empty.

To solve a double inequality, just apply the operations to all three portions: If 3<x+2<6, then subtract 2 from all three parts to get 1<x<4.

These are going to give you trouble. They don't have to, but they will. People just don't get absolute values.

I could tell you that the major reason people don't get it is because they don't understand restrictions. That would be the truth, but just telling you that isn't going to help you. The book is going to give a short cut and people are going to ask "why are there so many rules?". There aren't! There is one definition of absolute value, and if you know it, and apply the restrictions like you're supposed to, then the problems work out every time without a need for additional rules. The additional rules from the book are used to make the problems go quicker, just like when we omit the absolute value when taking the square root of both sides and just stick in the plus/minus instead. It sounds like an extra rule, but it's just an application of absolute value that we try to get around because we don't like them (and we wonder why we have difficulty with them).

Whether the original inequality was a > or < doesn't affect how the problem is done when you're doing it the correct way. I could show an example of both, but the technique is similar.

Consider the absolute value inequality | (x - 3) / 4 | ≤ 5

Begin by splitting the absolute value up into its two cases. Remember that the absolute value can be found by dropping the absolute value signs when the argument is non-negative (case 1) and by taking the opposite of what was in absolute values when the argument is negative (case 2).

- (x-3)/4 ≤ 5 if (x-3)/4 ≥ 0
- - (x-3)/4 ≤ 5 if (x-3)/4 < 0

(x-3)/4 ≤ 5 if (x-3)/4 ≥ 0

First the function portion

(x-3) / 4 ≤ 5

x-3 ≤ 20

x ≤ 23

Now the restriction

(x-3)/4 ≥ 0

x-3 ≥ 0

x ≥ 3

Be sure to pay special attention to the restrictions about what's in absolute value being non-negative and negative. Inequalities that occur at the same time like the function and its restriction must both be satisfied. That's why the two portions from solving the first case can be combined so that x is between 3 and 23 inclusive. It must satisfy both x≥3 and x≤23.

3 ≤ x ≤ 23

- (x-3)/4 ≤ 5 if (x-3)/4 < 0

First the function. We'll start off by multiplying by -1 to get rid of the negative sign, but then we have to change the sense (direction) of the inequality.

- (x-3) / 4 ≤ 5

(x-3) / 4 ≥ -5

x-3 ≥ -20

x ≥ -17

Now the restriction.

(x-3)/4 < 0

x-3 < 0

x < 3

Likewise, for the second case, the two portions must both be satisfied at the same time. Also notice that when I multiplied both sides of the inequality by -4, I changed the sense of the inequality so that it was no longer ≤, but now ≥. You could have simplified this in other ways, but I usually find it easier to get rid of the negative right away.

-17 ≤ x < 3

The two inequalities within a particular case must both be satisfied at the same time. However, the inequalities that occur between the two sides may be combined with an "or". Just like if (x-2)(x+4)=0, you would say the answer is x=2 or x=-4, you wouldn't insist that it be both 2 and -4 and the same time. The same principle holds with the two parts of the absolute value. I told you it all fits together. I know you're going to be tired of hearing me say it, but it is much easier to understand if you see the big picture.

So, put the two answers from Case 1 and Case 2 together

-17 ≤ x < 3 or 3 ≤ x ≤ 23

These two intervals can be unioned together to get the final answer.

-17 ≤ x ≤ 23

In interval notation, that would be [ -17, 23 ]

- Less Than
- When the absolute value is less than the right hand side, then the answer will be between the opposite of the right hand side and the right hand side.
- The absolute value inequality |x-2|<3 can be written as -3<x-2<3. Applying techniques discussed in the double inequality section above, this can be solved by adding 2 to each part to arrive at -1<x<5.
- Greater Than
- When the absolute value is greater than the right hand side, then the answer will be in two parts, with an or separating them. It will be greater than the right hand side or less than the opposite of the right hand side.
- The absolute value inequality |x-2|>3 can be written as x-2<-3 or x-2>3. Solving this arrives at x<-1 or x>5.
- Note that this cannot be combined to be -1>x>5 because that implies that 1>5 which is just plain false and would be the empty set.

I have no problem with using the geometric approach to solving absolute value inequalities. Not the geometric approach where you put it into the calculator, but the geometric approach where you use the geometric definition of absolute value. The geometric definition of absolute value is the distance from 0 on the number line. If you modify it slightly, then |x-a| is the distance from x=a on the number line.

However, this technique requires that you know some of the properties of absolute values. These are good things to know anyway (isn't it all?), so it would be good if you learned them.

Let's consider the same inequality that we solved before: | (x - 3) / 4 | ≤ 5

Multiply both sides by 4 to get |x-3| ≤ 20.

This says that the distance from 3 on the number line is less than or equal to 20. So, start at 3 and go 20 to the left (-17) and 20 to the right (23). You need the distance to be less than (or equal to) 20 units away. This would include values between -17 and 23. Since the equal to is included, you include the endpoints to get the answer (interval notation) of [-17,23].

Wow! That was easy, you say. Yep.

A little bit harder problem: | (3-5x) / 2| ≥ 3

Multiply both sides by 2 to get | 3-5x | ≥ 6.

Here's where those properties of absolute values come into play. The absolute value of the opposite of a number is the same as the absolute value of the number. That means that we can say | 3-5x | = | 5x-3 | and |5x-3| ≥ 6.

Now divide both sides by 5 to get | x- 3/5 | ≥ 6/5.

This says that the distance from 3/5 is at least 6/5. So, start at 3/5 and go 6/5 to the left (-3/5) and 6/5 to the right (9/5). Since you need the distance to be more than 6/5 units away from 3/5, you need the values to the left of -3/5 and to the right of 9/5. The answer is therefore x≤-3/5 or x≥9/5. In interval notation, you would have to use the union of two intervals (-∞,-3/5] U [9/5, +∞).

Polynomials are continuous. That means that you can draw them without picking up your pencil
(there's a more rigorous definition in calculus, but that definition will work for us, now). If
you're going to change from being less than zero to being greater than zero *and* you can't pick up
your pencil, then at some point, you must cross the x-axis. That means that the only place the
inequality can change is at an x-intercept, a zero, a root, a solution.

The key then, to finding the solution set for a polynomial inequality, is to find the zeros of the inequality (pretend it was an equation), putting them on the number line, and picking a test point in each region.

- Write the polynomial inequality in standard form so that the right side is zero.
- Find the real solutions (ignore complex solutions involving
*i*) to the inequality any way that you want to. Factoring is preferred, but you can use the quadratic formula if you can get it down to a quadratic factor. These values are called "critical numbers" (not to be confused with the similar but slightly different "critical values" from calculus). - Put the zeros of the polynomial (critical numbers) on the number line. Be sure to put them in order from smallest to largest. It is not important to label any other value on the number line. Some people were taught that you always put zero on the number line. That is not necessary here.
- Pick a test point in each interval. You will have one more interval than the number of test points. Plug that test point back into either the factored inequality or the original inequality. If the test point gives you a true statement, then any point in that interval will work, and you want to include that interval in the answer. If the test point gives you a false statement, then all points in that interval will not work and you do not want to include that interval.
- Include the endpoints if the inequality includes the equal to and do not include the endpoints if the inequality does not include the equal to.

Rational inequalities are similar to polynomials, but there is an extra temptation and an extra place where critical numbers could occur.

**Temptation:** Yield not to temptation, for yielding is sin.

I know you don't like fractions. You try to get rid of them at every chance you get. But, fractions are your friends and I really hope you don't treat all of your friends that way.

Here's the problem. If you multiply by a constant, it's pretty obvious whether it is positive or
negative so you know whether or not to change the inequality. However, with rational
expressions, there's going to be a variable in the denominator. Variables can take on different
values - that's why they're called variable (and you thought math made no sense). Sometimes an
expression is positive, but for other values of x, the expression is negative. So, if you multiply
by the least common denominator, you don't know whether your multiplying by a positive
number or a negative number *unless you keep track of the restrictions!* So, you have your choice
- fractions that you hate or restrictions that you hate? In this case, fractions are the lesser of the
two evils (in your mind - neither one is really evil in reality)

Polynomial functions are continuous everywhere. However, rational functions are not. They are continuous everywhere except where undefined and that occurs when the denominator is zero.

If you're moving along and you can't pick up your pencil and you change from negative to positive, then it has to happen at a zero of the function, same as with a polynomial. But, if the function is undefined, then you have to pick up your pencil. While the pencil is up, there is nothing that says you can't move to a completely different location, perhaps on the other side of the x-axis when you put it back down.

We now have two places that critical numbers can occur. One is at a zero of the function, the other is where the function is undefined.

- Write the rational inequality in standard form so that the right side is zero.
- Get a common denominator by multiplying top and bottom of terms. Remember you can not multiply through and get rid of the fractions because you don't know if what you're multiplying by is negative or positive (unless you want to mess with restrictions - and you don't in this case)
- Find the critical numbers. In simple terms a critical number is anything that
makes the numerator or denominator zero.
- Find the places where the function is undefined because of division by zero. These will be critical points that cannot be included in the final answer even if the equal to is included because it would cause division by zero.
- Find the real solutions
(ignore complex solutions involving
*i*) to the function any way that you want to. To find the zeros, all you need to worry about is the numerator.

- Put the critical numbers on the number line.
- Pick a test point in each interval and determine if the interval works or doesn't work.
- Include the endpoints if the inequality includes the equal to and do not include the endpoints if the inequality does not include the equal to. Be sure you do not include any endpoint that would cause division by zero if included.

Actually, converting the whole thing to an equation and solving to find the critical numbers isn't that bad of a route. The book, and most mathematics teachers assume that you're going to keep the inequality in the problem in all the way to the end. If you're willing to plug the values back into the original problem, you can change over to an equal sign, get rid of the fractions, and find the critical numbers. You still have to watch the restrictions, but now the restrictions are of the form x≠2 instead of x<2, and much easier to deal with.

However, learn it this way, because when we get to chapter 3, there is going to be a fundamental concept which will greatly speed finding the solutions to these problems, and it requires (sort of) that you look at positives and negatives.

Another way to solve inequalities is to graph the left hand side of the inequality as y_{1} and the
right hand side as y_{2} and then find the intersection point. The intersection point will be your
critical number. By looking at the graph, you can tell when the inequality is satisfied and record
that interval. Remember that when you're giving intervals, only the x-coordinate is necessary.
There is no y in the original problem, it was something that was added to make the graphing
convenient.