Solving inequalities with models. Inequalities with modulus
Today, friends, there will be no snot or sentimentality. Instead, I will send you, no questions asked, into battle with one of the most formidable opponents in the 8th-9th grade algebra course.
Yes, you understood everything correctly: we are talking about inequalities with modulus. We will look at four basic techniques with which you will learn to solve about 90% of such problems. What about the remaining 10%? Well, we'll talk about them in a separate lesson. :)
However, before analyzing any of the techniques, I would like to remind you of two facts that you already need to know. Otherwise, you risk not understanding the material of today’s lesson at all.
What you already need to know
Captain Obviousness seems to hint that to solve inequalities with modulus you need to know two things:
- How inequalities are resolved;
- What is a module?
Let's start with the second point.
Module Definition
Everything is simple here. There are two definitions: algebraic and graphical. To begin with - algebraic:
Definition. The modulus of a number $x$ is either the number itself, if it is non-negative, or the number opposite to it, if the original $x$ is still negative.
It is written like this:
\[\left| x \right|=\left\( \begin(align) & x,\ x\ge 0, \\ & -x,\ x \lt 0. \\\end(align) \right.\]
In simple terms, a modulus is a “number without a minus.” And it is precisely in this duality (in some places you don’t have to do anything with the original number, but in others you will have to remove some kind of minus) that is where the whole difficulty lies for beginning students.
There is also a geometric definition. It is also useful to know, but we will turn to it only in complex and some special cases, where the geometric approach is more convenient than the algebraic one (spoiler: not today).
Definition. Let point $a$ be marked on the number line. Then the module $\left| x-a \right|$ is the distance from point $x$ to point $a$ on this line.
If you draw a picture, you will get something like this:
Graphical module definition One way or another, from the definition of a module its key property immediately follows: the modulus of a number is always a non-negative quantity. This fact will be a red thread running through our entire narrative today.
Solving inequalities. Interval method
Now let's look at the inequalities. There are a great many of them, but our task now is to be able to solve at least the simplest of them. Those that reduce to linear inequalities, as well as to the interval method.
I have two big lessons on this topic (by the way, very, VERY useful - I recommend studying them):
- Interval method for inequalities (especially watch the video);
- Fractional rational inequalities is a very extensive lesson, but after it you won’t have any questions at all.
If you know all this, if the phrase “let's move from inequality to equation” does not make you vaguely want to kill yourself against the wall, then you are ready: welcome to hell to the main topic of the lesson. :)
1. Inequalities of the form “Modulus is less than function”
This is one of the most common problems with modules. It is required to solve an inequality of the form:
\[\left| f\right| \ltg\]
The functions $f$ and $g$ can be anything, but usually they are polynomials. Examples of such inequalities:
\[\begin(align) & \left| 2x+3 \right| \lt x+7; \\ & \left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0; \\ & \left| ((x)^(2))-2\left| x \right|-3 \right| \lt 2. \\\end(align)\]
All of them can be solved literally in one line according to the following scheme:
\[\left| f\right| \lt g\Rightarrow -g \lt f \lt g\quad \left(\Rightarrow \left\( \begin(align) & f \lt g, \\ & f \gt -g \\\end(align) \right.\right)\]
It is easy to see that we get rid of the module, but instead we get a double inequality (or, which is the same thing, a system of two inequalities). But this transition takes into account absolutely all possible problems: if the number under the module is positive, the method works; if negative, it still works; and even with the most inadequate function in place of $f$ or $g$, the method will still work.
Naturally, the question arises: is it not easier? Unfortunately, it's not possible. This is the whole point of the module.
But enough of the philosophizing. Let's solve a couple of problems:
Task. Solve the inequality:
\[\left| 2x+3 \right| \lt x+7\]
Solution. So, we have a classical inequality of the form “the module is less than” - there is even nothing to transform. We work according to the algorithm:
\[\begin(align) & \left| f\right| \lt g\Rightarrow -g \lt f \lt g; \\ & \left| 2x+3 \right| \lt x+7\Rightarrow -\left(x+7 \right) \lt 2x+3 \lt x+7 \\\end(align)\]
Do not rush to open the brackets that are preceded by a “minus”: it is quite possible that because of the haste you will make an offensive mistake.
\[-x-7 \lt 2x+3 \lt x+7\]
\[\left\( \begin(align) & -x-7 \lt 2x+3 \\ & 2x+3 \lt x+7 \\ \end(align) \right.\]
\[\left\( \begin(align) & -3x \lt 10 \\ & x \lt 4 \\ \end(align) \right.\]
\[\left\( \begin(align) & x \gt -\frac(10)(3) \\ & x \lt 4 \\ \end(align) \right.\]
The problem has been reduced to two elementary inequalities. Let us note their solutions on parallel number lines:
Intersection of many
The intersection of these sets will be the answer.
Answer: $x\in \left(-\frac(10)(3);4 \right)$
Task. Solve the inequality:
\[\left| ((x)^(2))+2x-3 \right|+3\left(x+1 \right) \lt 0\]
Solution. This task is a little more difficult. First, let’s isolate the module by moving the second term to the right:
\[\left| ((x)^(2))+2x-3 \right| \lt -3\left(x+1 \right)\]
Obviously, we again have an inequality of the form “the module is smaller”, so we get rid of the module using the already known algorithm:
\[-\left(-3\left(x+1 \right) \right) \lt ((x)^(2))+2x-3 \lt -3\left(x+1 \right)\]
Now attention: someone will say that I'm a bit of a pervert with all these parentheses. But let me remind you once again that our key goal is correctly solve the inequality and get the answer. Later, when you have perfectly mastered everything that is described in this lesson, you can pervert yourself as you like: open brackets, add minuses, etc.
To begin with, we’ll simply get rid of the double minus on the left:
\[-\left(-3\left(x+1 \right) \right)=\left(-1 \right)\cdot \left(-3 \right)\cdot \left(x+1 \right) =3\left(x+1 \right)\]
Now let's open all the brackets in the double inequality:
Let's move on to the double inequality. This time the calculations will be more serious:
\[\left\( \begin(align) & ((x)^(2))+2x-3 \lt -3x-3 \\ & 3x+3 \lt ((x)^(2))+2x -3 \\ \end(align) \right.\]
\[\left\( \begin(align) & ((x)^(2))+5x \lt 0 \\ & ((x)^(2))-x-6 \gt 0 \\ \end( align)\right.\]
Both inequalities are square and are solved by the interval method (that's why I say: if you don't know what it is, it's better not to take on the modules yet). Let's move on to the equation in the first inequality:
\[\begin(align) & ((x)^(2))+5x=0; \\ & x\left(x+5 \right)=0; \\ & ((x)_(1))=0;((x)_(2))=-5. \\\end(align)\]
As you can see, the output turned out to be an incomplete quadratic equation, which is solved elementarily. Now let's look at the second inequality of the system. There you will have to apply Vieta’s theorem:
\[\begin(align) & ((x)^(2))-x-6=0; \\ & \left(x-3 \right)\left(x+2 \right)=0; \\& ((x)_(1))=3;((x)_(2))=-2. \\\end(align)\]
We mark the obtained numbers on two parallel lines (separate for the first inequality and separate for the second):
Again, since we are solving a system of inequalities, we are interested in the intersection of the shaded sets: $x\in \left(-5;-2 \right)$. This is the answer.
Answer: $x\in \left(-5;-2 \right)$
I think that after these examples the solution scheme is extremely clear:
- Isolate the module by moving all other terms to the opposite side of the inequality. Thus we get an inequality of the form $\left| f\right| \ltg$.
- Solve this inequality by getting rid of the module according to the scheme described above. At some point, it will be necessary to move from double inequality to a system of two independent expressions, each of which can already be solved separately.
- Finally, all that remains is to intersect the solutions of these two independent expressions - and that’s it, we will get the final answer.
A similar algorithm exists for inequalities of the following type, when the modulus is greater than the function. However, there are a couple of serious “buts”. We’ll talk about these “buts” now.
2. Inequalities of the form “Modulus is greater than function”
They look like this:
\[\left| f\right| \gtg\]
Similar to the previous one? It seems. And yet such problems are solved in a completely different way. Formally, the scheme is as follows:
\[\left| f\right| \gt g\Rightarrow \left[ \begin(align) & f \gt g, \\ & f \lt -g \\\end(align) \right.\]
In other words, we consider two cases:
- First, we simply ignore the module and solve the usual inequality;
- Then, in essence, we expand the module with the minus sign, and then multiply both sides of the inequality by −1, while I have the sign.
In this case, the options are combined with a square bracket, i.e. We have a combination of two requirements.
Pay attention again: before us is not a system, but an aggregate, therefore in the answer, the sets are combined, not intersected. This is a fundamental difference from the previous paragraph!
In general, many students are completely confused with unions and intersections, so let’s sort this issue out once and for all:
- "∪" is a concatenation sign. In fact, this is a stylized letter “U”, which came to us from the English language and is an abbreviation for “Union”, i.e. "Associations".
- "∩" is the intersection sign. This crap didn’t come from anywhere, but simply appeared as a counterpoint to “∪”.
To make it even easier to remember, just draw legs to these signs to make glasses (just don’t now accuse me of promoting drug addiction and alcoholism: if you are seriously studying this lesson, then you are already a drug addict):
Difference between intersection and union of sets Translated into Russian, this means the following: the union (totality) includes elements from both sets, therefore it is in no way less than each of them; but the intersection (system) includes only those elements that are simultaneously in both the first set and the second. Therefore, the intersection of sets is never larger than the source sets.
So it became clearer? That is great. Let's move on to practice.
Task. Solve the inequality:
\[\left| 3x+1 \right| \gt 5-4x\]
Solution. We proceed according to the scheme:
\[\left| 3x+1 \right| \gt 5-4x\Rightarrow \left[ \begin(align) & 3x+1 \gt 5-4x \\ & 3x+1 \lt -\left(5-4x \right) \\\end(align) \ right.\]
We solve each inequality in the population:
\[\left[ \begin(align) & 3x+4x \gt 5-1 \\ & 3x-4x \lt -5-1 \\ \end(align) \right.\]
\[\left[ \begin(align) & 7x \gt 4 \\ & -x \lt -6 \\ \end(align) \right.\]
\[\left[ \begin(align) & x \gt 4/7\ \\ & x \gt 6 \\ \end(align) \right.\]
We mark each resulting set on the number line, and then combine them:
Union of sets
It is quite obvious that the answer will be $x\in \left(\frac(4)(7);+\infty \right)$
Answer: $x\in \left(\frac(4)(7);+\infty \right)$
Task. Solve the inequality:
\[\left| ((x)^(2))+2x-3 \right| \gt x\]
Solution. Well? No, it's all the same. We pass from an inequality with a modulus to a set of two inequalities:
\[\left| ((x)^(2))+2x-3 \right| \gt x\Rightarrow \left[ \begin(align) & ((x)^(2))+2x-3 \gt x \\ & ((x)^(2))+2x-3 \lt -x \\\end(align) \right.\]
We solve each inequality. Unfortunately, the roots will not be very good there:
\[\begin(align) & ((x)^(2))+2x-3 \gt x; \\ & ((x)^(2))+x-3 \gt 0; \\&D=1+12=13; \\ & x=\frac(-1\pm \sqrt(13))(2). \\\end(align)\]
In the second inequality, there is also a bit of game:
\[\begin(align) & ((x)^(2))+2x-3 \lt -x; \\ & ((x)^(2))+3x-3 \lt 0; \\&D=9+12=21; \\ & x=\frac(-3\pm \sqrt(21))(2). \\\end(align)\]
Now we need to mark these numbers on two axes - one axis for each inequality. However, you need to mark the points in the correct order: the larger the number, the further the point moves to the right.
And here a setup awaits us. If everything is clear with the numbers $\frac(-3-\sqrt(21))(2) \lt \frac(-1-\sqrt(13))(2)$ (the terms in the numerator of the first fraction are less than the terms in the numerator of the second , so the sum is also less), with the numbers $\frac(-3-\sqrt(13))(2) \lt \frac(-1+\sqrt(21))(2)$ there will also be no difficulties (positive number obviously more negative), then with the last couple everything is not so clear. Which is greater: $\frac(-3+\sqrt(21))(2)$ or $\frac(-1+\sqrt(13))(2)$? The placement of points on the number lines and, in fact, the answer will depend on the answer to this question.
So let's compare:
\[\begin(matrix) \frac(-1+\sqrt(13))(2)\vee \frac(-3+\sqrt(21))(2) \\ -1+\sqrt(13)\ vee -3+\sqrt(21) \\ 2+\sqrt(13)\vee \sqrt(21) \\\end(matrix)\]
We isolated the root, got non-negative numbers on both sides of the inequality, so we have the right to square both sides:
\[\begin(matrix) ((\left(2+\sqrt(13) \right))^(2))\vee ((\left(\sqrt(21) \right))^(2)) \ \ 4+4\sqrt(13)+13\vee 21 \\ 4\sqrt(13)\vee 3 \\\end(matrix)\]
I think it’s a no brainer that $4\sqrt(13) \gt 3$, so $\frac(-1+\sqrt(13))(2) \gt \frac(-3+\sqrt(21)) (2)$, the final points on the axes will be placed like this:
A case of ugly roots
Let me remind you that we are solving a set, so the answer will be a union, not an intersection of shaded sets.
Answer: $x\in \left(-\infty ;\frac(-3+\sqrt(21))(2) \right)\bigcup \left(\frac(-1+\sqrt(13))(2 );+\infty \right)$
As you can see, our scheme works great for both simple and very tough problems. The only “weak point” in this approach is that you need to correctly compare irrational numbers (and believe me: these are not only roots). But a separate (and very serious) lesson will be devoted to comparison issues. And we move on.
3. Inequalities with non-negative “tails”
Now we get to the most interesting part. These are inequalities of the form:
\[\left| f\right| \gt\left| g\right|\]
Generally speaking, the algorithm that we will talk about now is correct only for the module. It works in all inequalities where there are guaranteed non-negative expressions on the left and right:
What to do with these tasks? Just remember:
In inequalities with non-negative “tails”, both sides can be raised to any natural power. There will be no additional restrictions.
First of all, we will be interested in squaring - it burns modules and roots:
\[\begin(align) & ((\left(\left| f \right| \right))^(2))=((f)^(2)); \\ & ((\left(\sqrt(f) \right))^(2))=f. \\\end(align)\]
Just don’t confuse this with taking the root of a square:
\[\sqrt(((f)^(2)))=\left| f \right|\ne f\]
Countless mistakes were made when a student forgot to install a module! But this is a completely different story (these are, as it were, irrational equations), so we will not go into this now. Let's better solve a couple of problems:
Task. Solve the inequality:
\[\left| x+2 \right|\ge \left| 1-2x \right|\]
Solution. Let's immediately notice two things:
- This is a non-strict inequality. Points on the number line will be punctured.
- Both sides of the inequality are obviously non-negative (this is a property of the module: $\left| f\left(x \right) \right|\ge 0$).
Therefore, we can square both sides of the inequality to get rid of the modulus and solve the problem using the usual interval method:
\[\begin(align) & ((\left(\left| x+2 \right| \right))^(2))\ge ((\left(\left| 1-2x \right| \right) )^(2)); \\ & ((\left(x+2 \right))^(2))\ge ((\left(2x-1 \right))^(2)). \\\end(align)\]
At the last step, I cheated a little: I changed the sequence of terms, taking advantage of the evenness of the module (in fact, I multiplied the expression $1-2x$ by −1).
\[\begin(align) & ((\left(2x-1 \right))^(2))-((\left(x+2 \right))^(2))\le 0; \\ & \left(\left(2x-1 \right)-\left(x+2 \right) \right)\cdot \left(\left(2x-1 \right)+\left(x+2 \ right)\right)\le 0; \\ & \left(2x-1-x-2 \right)\cdot \left(2x-1+x+2 \right)\le 0; \\ & \left(x-3 \right)\cdot \left(3x+1 \right)\le 0. \\\end(align)\]
We solve using the interval method. Let's move from inequality to equation:
\[\begin(align) & \left(x-3 \right)\left(3x+1 \right)=0; \\ & ((x)_(1))=3;((x)_(2))=-\frac(1)(3). \\\end(align)\]
We mark the found roots on the number line. Once again: all points are shaded because the original inequality is not strict!
Getting rid of the modulus sign
Let me remind you for those who are especially stubborn: we take the signs from the last inequality, which was written down before moving on to the equation. And we paint over the areas required in the same inequality. In our case it is $\left(x-3 \right)\left(3x+1 \right)\le 0$.
OK it's all over Now. The problem is solved.
Answer: $x\in \left[ -\frac(1)(3);3 \right]$.
Task. Solve the inequality:
\[\left| ((x)^(2))+x+1 \right|\le \left| ((x)^(2))+3x+4 \right|\]
Solution. We do everything the same. I won't comment - just look at the sequence of actions.
Square it:
\[\begin(align) & ((\left(\left| ((x)^(2))+x+1 \right| \right))^(2))\le ((\left(\left | ((x)^(2))+3x+4 \right| \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))\le ((\left(((x)^(2))+3x+4 \right))^(2)); \\ & ((\left(((x)^(2))+x+1 \right))^(2))-((\left(((x)^(2))+3x+4 \ right))^(2))\le 0; \\ & \left(((x)^(2))+x+1-((x)^(2))-3x-4 \right)\times \\ & \times \left(((x) ^(2))+x+1+((x)^(2))+3x+4 \right)\le 0; \\ & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)\le 0. \\\end(align)\]
Interval method:
\[\begin(align) & \left(-2x-3 \right)\left(2((x)^(2))+4x+5 \right)=0 \\ & -2x-3=0\ Rightarrow x=-1.5; \\ & 2((x)^(2))+4x+5=0\Rightarrow D=16-40 \lt 0\Rightarrow \varnothing . \\\end(align)\]
There is only one root on the number line:
The answer is a whole interval
Answer: $x\in \left[ -1.5;+\infty \right)$.
A small note about the last task. As one of my students accurately noted, both submodular expressions in this inequality are obviously positive, so the modulus sign can be omitted without harm to health.
But this is a completely different level of thinking and a different approach - it can conditionally be called the method of consequences. About it - in a separate lesson. Now let’s move on to the final part of today’s lesson and look at a universal algorithm that always works. Even when all previous approaches were powerless. :)
4. Method of enumeration of options
What if all these techniques don't help? If the inequality cannot be reduced to non-negative tails, if it is impossible to isolate the module, if in general there is pain, sadness, melancholy?
Then the “heavy artillery” of all mathematics comes onto the scene—the brute force method. In relation to inequalities with modulus it looks like this:
- Write out all submodular expressions and set them equal to zero;
- Solve the resulting equations and mark the roots found on one number line;
- The straight line will be divided into several sections, within which each module has a fixed sign and therefore is uniquely revealed;
- Solve the inequality on each such section (you can separately consider the roots-boundaries obtained in step 2 - for reliability). Combine the results - this will be the answer. :)
So how? Weak? Easily! Only for a long time. Let's see in practice:
Task. Solve the inequality:
\[\left| x+2 \right| \lt \left| x-1 \right|+x-\frac(3)(2)\]
Solution. This crap doesn't boil down to inequalities like $\left| f\right| \lt g$, $\left| f\right| \gt g$ or $\left| f\right| \lt \left| g \right|$, so we act ahead.
We write out submodular expressions, equate them to zero and find the roots:
\[\begin(align) & x+2=0\Rightarrow x=-2; \\ & x-1=0\Rightarrow x=1. \\\end(align)\]
In total, we have two roots that divide the number line into three sections, within which each module is revealed uniquely:
Partitioning the number line by zeros of submodular functions
Let's consider each section separately.
1. Let $x \lt -2$. Then both submodular expressions are negative, and the original inequality will be rewritten as follows:
\[\begin(align) & -\left(x+2 \right) \lt -\left(x-1 \right)+x-1,5 \\ & -x-2 \lt -x+1+ x-1.5 \\ & x \gt 1.5 \\\end(align)\]
We got a fairly simple constraint. Let's intersect it with the original assumption that $x \lt -2$:
\[\left\( \begin(align) & x \lt -2 \\ & x \gt 1,5 \\\end(align) \right.\Rightarrow x\in \varnothing \]
Obviously, the variable $x$ cannot simultaneously be less than −2 but greater than 1.5. There are no solutions in this area.
1.1. Let's separately consider the boundary case: $x=-2$. Let's just substitute this number into the original inequality and check: is it true?
\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=-2) ) \\ & 0 \lt \left| -3\right|-2-1.5; \\ & 0 \lt 3-3.5; \\ & 0 \lt -0.5\Rightarrow \varnothing . \\\end(align)\]
Obviously, the chain of calculations has led us to the wrong inequality. Therefore, the original inequality is also false, and $x=-2$ is not included in the answer.
2. Let now $-2 \lt x \lt 1$. The left module will already open with a "plus", but the right one is still with a "minus". We have:
\[\begin(align) & x+2 \lt -\left(x-1 \right)+x-1.5 \\ & x+2 \lt -x+1+x-1.5 \\& x \lt -2.5 \\\end(align)\]
Again we intersect with the original requirement:
\[\left\( \begin(align) & x \lt -2.5 \\ & -2 \lt x \lt 1 \\\end(align) \right.\Rightarrow x\in \varnothing \]
And again, the set of solutions is empty, since there are no numbers that are both less than −2.5 and greater than −2.
2.1. And again a special case: $x=1$. We substitute into the original inequality:
\[\begin(align) & ((\left. \left| x+2 \right| \lt \left| x-1 \right|+x-1.5 \right|)_(x=1)) \\ & \left| 3\right| \lt \left| 0\right|+1-1.5; \\ & 3 \lt -0.5; \\ & 3 \lt -0.5\Rightarrow \varnothing . \\\end(align)\]
Similar to the previous “special case”, the number $x=1$ is clearly not included in the answer.
3. The last piece of the line: $x \gt 1$. Here all modules are opened with a plus sign:
\[\begin(align) & x+2 \lt x-1+x-1.5 \\ & x+2 \lt x-1+x-1.5 \\ & x \gt 4.5 \\ \end(align)\]
And again we intersect the found set with the original constraint:
\[\left\( \begin(align) & x \gt 4.5 \\ & x \gt 1 \\\end(align) \right.\Rightarrow x\in \left(4.5;+\infty \right)\]
Finally! We have found an interval that will be the answer.
Answer: $x\in \left(4,5;+\infty \right)$
Finally, one remark that may save you from stupid mistakes when solving real problems:
Solutions to inequalities with moduli usually represent continuous sets on the number line - intervals and segments. Isolated points are much less common. And even less often, it happens that the boundary of the solution (the end of the segment) coincides with the boundary of the range under consideration.
Consequently, if boundaries (the same “special cases”) are not included in the answer, then the areas to the left and right of these boundaries will almost certainly not be included in the answer. And vice versa: the border entered into the answer, which means that some areas around it will also be answers.
Keep this in mind when reviewing your solutions.
There are several ways to solve inequalities containing a modulus. Let's look at some of them.
1) Solving the inequality using the geometric property of the module.
Let me remind you what the geometric property of a modulus is: the modulus of a number x is the distance from the origin to the point with coordinate x.
When solving inequalities using this method, two cases may arise:
1. |x| ≤ b, 
And the inequality with modulus obviously reduces to a system of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
2. |x| ≥ b, then the solution picture looks like this:
And the inequality with modulus obviously reduces to a combination of two inequalities. Here the sign can be strict, in which case the dots in the picture will be “punctured”.
Example 1.
Solve the inequality |4 – |x|| ≥ 3.
Solution.
This inequality is equivalent to the following set:
U [-1;1] U
Example 2.
Solve the inequality ||x+2| – 3| ≤ 2.
Solution.
This inequality is equivalent to the following system.
(|x + 2| – 3 ≥ -2
(|x + 2| – 3 ≤ 2,
(|x + 2| ≥ 1
(|x + 2| ≤ 5.
Let us solve separately the first inequality of the system. It is equivalent to the following set:
U[-1; 3].
2) Solving inequalities using the definition of the modulus.
Let me remind you first module definition.
|a| = a if a ≥ 0 and |a| = -a if a< 0.
For example, |34| = 34, |-21| = -(-21) = 21.
Example 1.
Solve the inequality 3|x – 1| ≤ x+3.
Solution.
Using the module definition, we get two systems:
(x – 1 ≥ 0
(3(x – 1) ≤ x + 3
(x – 1< 0
(-3(x – 1) ≤ x + 3.
Solving the first and second systems separately, we get:
(x ≥ 1
(x ≤ 3,
(x< 1
(x ≥ 0.
The solution to the original inequality will be all solutions of the first system and all solutions of the second system.
Answer: x € .
3) Solving inequalities by squaring.
Example 1.
Solve the inequality |x 2 – 1|< | x 2 – x + 1|.
Solution.
Let's square both sides of the inequality. Let me note that it is possible to square both sides of the inequality only if they are both positive. In this case, we have modules on both the left and right, so we can do this.
(|x 2 – 1|) 2< (|x 2 – x + 1|) 2 .
Now let's use the following property of the module: (|x|) 2 = x 2 .
(x 2 – 1) 2< (x 2 – x + 1) 2 ,
(x 2 - 1) 2 - (x 2 - x + 1) 2< 0.
(x 2 – 1 – x 2 + x – 1)(x 2 – 1 + x 2 – x + 1)< 0,
(x – 2)(2x 2 – x)< 0,
x(x – 2)(2x – 1)< 0.
We solve using the interval method.
Answer: x € (-∞; 0) U (1/2; 2)
4) Solving inequalities by changing variables.
Example.
Solve the inequality (2x + 3) 2 – |2x + 3| ≤ 30.
Solution.
Note that (2x + 3) 2 = (|2x + 3|) 2 . Then we get the inequality
(|2x + 3|) 2 – |2x + 3| ≤ 30.
Let's make the change y = |2x + 3|.
Let's rewrite our inequality taking into account the replacement.
y 2 – y ≤ 30,
y 2 – y – 30 ≤ 0.
Let's factorize the quadratic trinomial on the left.
y1 = (1 + 11) / 2,
y2 = (1 – 11) / 2,
(y – 6)(y + 5) ≤ 0.
Let's solve using the interval method and get:
Back to replacement:
5 ≤ |2x + 3| ≤ 6.
This double inequality is equivalent to the system of inequalities:
(|2x + 3| ≤ 6
(|2x + 3| ≥ -5.
Let's solve each of the inequalities separately.
The first is equivalent to the system
(2x + 3 ≤ 6
(2x + 3 ≥ -6.
Let's solve it.
(x ≤ 1.5
(x ≥ -4.5.
The second inequality obviously holds for all x, since the modulus is, by definition, a positive number. Since the solution to the system is all x that simultaneously satisfy both the first and second inequality of the system, then the solution to the original system will be the solution to its first double inequality (after all, the second is true for all x).
Answer: x € [-4.5; 1.5].
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Modulus of numbers this number itself is called if it is non-negative, or the same number with the opposite sign if it is negative.
For example, the modulus of the number 6 is 6, and the modulus of the number -6 is also 6.
That is, the modulus of a number is understood as the absolute value, the absolute value of this number without taking into account its sign.
It is designated as follows: |6|, | X|, |A| etc.
(More details in the “Number module” section).
Equations with modulus.
Example 1 . Solve the equation|10 X - 5| = 15.
Solution.
According to the rule, the equation is equivalent to the combination of two equations:
10X - 5 = 15
10X - 5 = -15
We decide:
10X = 15 + 5 = 20
10X = -15 + 5 = -10
X = 20: 10
X = -10: 10
X = 2
X = -1
Answer: X 1 = 2, X 2 = -1.
Example 2 . Solve the equation|2 X + 1| = X + 2.
Solution.
Since the modulus is a non-negative number, then X+ 2 ≥ 0. Accordingly:
X ≥ -2.
Let's make two equations:
2X + 1 = X + 2
2X + 1 = -(X + 2)
We decide:
2X + 1 = X + 2
2X + 1 = -X - 2
2X - X = 2 - 1
2X + X = -2 - 1
X = 1
X = -1
Both numbers are greater than -2. So both are roots of the equation.
Answer: X 1 = -1, X 2 = 1.
Example 3
. Solve the equation
|X + 3| - 1
————— = 4
X - 1
Solution.
The equation makes sense if the denominator is not zero - that means if X≠ 1. Let's take this condition into account. Our first action is simple - we don’t just get rid of the fraction, but transform it so as to obtain the module in its pure form:
|X+ 3| - 1 = 4 · ( X - 1),
|X + 3| - 1 = 4X - 4,
|X + 3| = 4X - 4 + 1,
|X + 3| = 4X - 3.
Now we have only an expression under the modulus on the left side of the equation. Go ahead.
The modulus of a number is a non-negative number - that is, it must be greater than zero or equal to zero. Accordingly, we solve the inequality:
4X - 3 ≥ 0
4X ≥ 3
X ≥ 3/4
Thus, we have a second condition: the root of the equation must be at least 3/4.
In accordance with the rule, we compose a set of two equations and solve them:
X + 3 = 4X - 3
X + 3 = -(4X - 3)
X + 3 = 4X - 3
X + 3 = -4X + 3
X - 4X = -3 - 3
X + 4X = 3 - 3
X = 2
X = 0
We received two answers. Let's check whether they are roots of the original equation.
We had two conditions: the root of the equation cannot be equal to 1, and it must be at least 3/4. That is X ≠ 1, X≥ 3/4. Both of these conditions correspond to only one of the two answers received - the number 2. This means that only this is the root of the original equation.
Answer: X = 2.
Inequalities with modulus.
Example 1 . Solve the inequality| X - 3| < 4
Solution.
The module rule states:
|A| = A, If A ≥ 0.
|A| = -A, If A < 0.
The module can have both non-negative and negative numbers. So we have to consider both cases: X- 3 ≥ 0 and X - 3 < 0.
1) When X- 3 ≥ 0 our original inequality remains as it is, only without the modulus sign:
X - 3 < 4.
2) When X - 3 < 0 в исходном неравенстве надо поставить знак минус перед всем подмодульным выражением:
-(X - 3) < 4.
Opening the brackets, we get:
-X + 3 < 4.
Thus, from these two conditions we came to the unification of two systems of inequalities:
X - 3 ≥ 0
X - 3 < 4
X - 3 < 0
-X + 3 < 4
Let's solve them:
X ≥ 3
X < 7
X < 3
X > -1
So, our answer is a union of two sets:
3 ≤ X < 7 U -1 < X < 3.
Determine the smallest and largest values. These are -1 and 7. Moreover X greater than -1 but less than 7.
Besides, X≥ 3. This means that the solution to the inequality is the entire set of numbers from -1 to 7, excluding these extreme numbers.
Answer: -1 < X < 7.
Or: X ∈ (-1; 7).
Add-ons.
1) There is a simpler and shorter way to solve our inequality - graphically. To do this, you need to draw a horizontal axis (Fig. 1).
Expression | X - 3| < 4 означает, что расстояние от точки X to point 3 is less than four units. We mark the number 3 on the axis and count 4 divisions to the left and to the right of it. On the left we will come to point -1, on the right - to point 7. Thus, the points X we just saw them without calculating them.
Moreover, according to the inequality condition, -1 and 7 themselves are not included in the set of solutions. Thus, we get the answer:
1 < X < 7.
2) But there is another solution that is simpler even than the graphical method. To do this, our inequality must be presented in the following form:
4 < X - 3 < 4.
After all, this is how it is according to the modulus rule. The non-negative number 4 and the similar negative number -4 are the boundaries for solving the inequality.
4 + 3 < X < 4 + 3
1 < X < 7.
Example 2 . Solve the inequality| X - 2| ≥ 5
Solution.
This example is significantly different from the previous one. The left side is greater than 5 or equal to 5. From a geometric point of view, the solution to the inequality is all numbers that are at a distance of 5 units or more from point 2 (Fig. 2). The graph shows that these are all numbers that are less than or equal to -3 and greater than or equal to 7. This means that we have already received the answer.
Answer: -3 ≥ X ≥ 7.
Along the way, we solve the same inequality by rearranging the free term to the left and to the right with the opposite sign:
5 ≥ X - 2 ≥ 5
5 + 2 ≥ X ≥ 5 + 2
The answer is the same: -3 ≥ X ≥ 7.
Or: X ∈ [-3; 7]
The example is solved.
Example 3 . Solve the inequality 6 X 2 - | X| - 2 ≤ 0
Solution.
Number X can be a positive number, negative number, or zero. Therefore, we need to take into account all three circumstances. As you know, they are taken into account in two inequalities: X≥ 0 and X < 0. При X≥ 0 we simply rewrite our original inequality as is, only without the modulus sign:
6x 2 - X - 2 ≤ 0.
Now about the second case: if X < 0. Модулем отрицательного числа является это же число с противоположным знаком. То есть пишем число под модулем с обратным знаком и опять же освобождаемся от знака модуля:
6X 2 - (-X) - 2 ≤ 0.
Expanding the brackets:
6X 2 + X - 2 ≤ 0.
Thus, we received two systems of equations:
6X 2 - X - 2 ≤ 0
X ≥ 0
6X 2 + X - 2 ≤ 0
X < 0
We need to solve inequalities in systems - and this means we need to find the roots of two quadratic equations. To do this, we equate the left-hand sides of the inequalities to zero.
Let's start with the first one:
6X 2 - X - 2 = 0.
How to solve a quadratic equation - see the section “Quadratic Equation”. We will immediately name the answer:
X 1 = -1/2, x 2 = 2/3.
From the first system of inequalities we obtain that the solution to the original inequality is the entire set of numbers from -1/2 to 2/3. We write the union of solutions at X ≥ 0:
[-1/2; 2/3].
Now let's solve the second quadratic equation:
6X 2 + X - 2 = 0.
Its roots:
X 1 = -2/3, X 2 = 1/2.
Conclusion: when X < 0 корнями исходного неравенства являются также все числа от -2/3 до 1/2.
Let's combine the two answers and get the final answer: the solution is the entire set of numbers from -2/3 to 2/3, including these extreme numbers.
Answer: -2/3 ≤ X ≤ 2/3.
Or: X ∈ [-2/3; 2/3].
Methods (rules) for revealing inequalities with modules consist in sequential disclosure of modules, using intervals of constant sign of submodular functions. In the final version, several inequalities are obtained from which intervals or intervals are found that satisfy the conditions of the problem.
Let's move on to solving examples that are common in practice.
Linear inequalities with modules
By linear we mean equations in which a variable enters the equation linearly.
Example 1. Find a solution to the inequality
Solution:
It follows from the condition of the problem that the modules turn into zero at x=-1 and x=-2. These points divide the numerical axis into intervals
In each of these intervals, we solve the given inequality. To do this, first of all, we draw up graphical drawings of areas of constant sign of submodular functions. They are depicted as areas with signs of each of the functions.
or intervals with signs of all functions. 
On the first interval, open the modules
We multiply both sides by minus one, and the sign in the inequality will change to the opposite. If this rule is difficult for you to get used to, you can move each of the parts behind the sign to get rid of the minus. In the end, you will receive
The intersection of the set x>-3 with the area on which the equations were solved will be the interval (-3;-2). For those who find it easier to find solutions, you can graphically draw the intersection of these areas
General intersection of areas will be the solution. If strictly uneven, the edges are not included. If not strict, check by substitution.
On the second interval we get 
The cross section will be the interval (-2;-5/3). Graphically the solution will look like
On the third interval we get
This condition does not provide solutions in the desired region.
Since the two solutions found (-3;-2) and (-2;-5/3) border on the point x=-2, we check it too.
Thus the point x=-2 is the solution. The general solution taking this into account will look like (-3;5/3).
Example 2. Find a solution to the inequality
|x-2|-|x-3|>=|x-4|
Solution:
The zeros of the submodular functions will be the points x=2, x=3, x=4. For argument values less than these points, the submodular functions are negative, and for larger values, they are positive.
The points divide the real axis into four intervals. We open the modules according to the intervals of constancy of sign and solve the inequalities.
1) In the first interval, all submodular functions are negative, so when expanding the modules, we change the sign to the opposite one.
The intersection of the found x values with the considered interval will be a set of points
2) On the interval between points x=2 and x=3, the first submodular function is positive, the second and third are negative. Expanding the modules, we get
an inequality that, when intersected with the interval on which we are solving, gives one solution – x=3.
3) On the interval between points x=3 and x=4, the first and second submodular functions are positive, and the third is negative. Based on this we get
This condition shows that the whole interval will satisfy the inequality with moduli.
4) For values of x>4 all functions have positive signs. When expanding modules, we do not change their sign.
The found condition at the intersection with the interval gives the following set of solutions
Since the inequality is solved on all intervals, it remains to find the common value of all found values of x. The solution will be two intervals 
This concludes the example.
Example 3. Find a solution to the inequality
||x-1|-5|>3-2x
Solution:
We have an inequality with a module from a module. Such inequalities are revealed as modules are nested, starting with those that are located deeper.
The submodule function x-1 is converted to zero at the point x=1 . For smaller values beyond 1 it is negative and positive for x>1 . Based on this, we expand the internal module and consider the inequality on each of the intervals.
First consider the interval from minus infinity to one 
The submodule function is zero at the point x=-4 . At smaller values it is positive, at larger values it is negative. Expand the module for x<-4:
At the intersection with the area in which we are considering, we obtain a set of solutions
The next step is to expand the module on the interval (-4;1) 
Taking into account the expansion area of the module, we obtain the solution interval
REMEMBER: if in such irregularities with modules you get two intervals bordering a common point, then, as a rule, this is also a solution.
To do this, you just need to check.
In this case, we substitute the point x=-4.
So x=-4 is the solution.
Let's expand the internal module for x>1
Submodular function negative for x<6.
Expanding the module, we get
This condition in the section with the interval (1;6) gives an empty set of solutions.
For x>6 we get the inequality
Also solving we got an empty set.
Taking into account all of the above, the only solution to the inequality with modules will be the following interval.
Inequalities with moduli containing quadratic equations
Example 4. Find a solution to the inequality
|x^2+3x|>=2-x^2 
Solution:
The submodular function vanishes at points x=0, x=-3. Simple substitution of minus one 
we establish that it is less than zero in the interval (-3;0) and positive beyond it.
Let us expand the module in areas where the submodular function is positive 
It remains to determine the regions where the square function is positive. To do this, we determine the roots of the quadratic equation 
For convenience, we substitute the point x=0, which belongs to the interval (-2;1/2). The function is negative in this interval, which means the solution will be the following sets x 
Here the edges of the areas with solutions are indicated by brackets; this was done deliberately, taking into account the following rule.
REMEMBER: If an inequality with moduli, or a simple inequality is strict, then the edges of the found areas are not solutions, but if the inequalities are not strict (), then the edges are solutions (denoted by square brackets).
This rule is used by many teachers: if a strict inequality is given, and during calculations you write a square bracket ([,]) in the solution, they will automatically consider this to be an incorrect answer. Also, when testing, if a non-strict inequality with modules is given, then look for areas with square brackets among the solutions.
On the interval (-3;0), expanding the module, we change the sign of the function to the opposite one 
Taking into account the area of inequality disclosure, the solution will have the form
Together with the previous area this will give two half-intervals
Example 5. Find a solution to the inequality
9x^2-|x-3|>=9x-2 
Solution:
A non-strict inequality is given whose submodular function is equal to zero at the point x=3. For smaller values it is negative, for larger values it is positive. Expand the module on the interval x<3.
Finding the discriminant of the equation
and roots 
Substituting point zero, we find out that on the interval [-1/9;1] the quadratic function is negative, therefore the interval is a solution. Next we expand the module at x>3 
