Diagram in Figure 5:

And the following example: Solve the inequality 0.6 2x - 3< 0,36 .

Following the diagram in Figure 5, we get
2x - 3 > log 0.6 0.36,
2x - 3 > 2,
2x > 5,
x > 2.5

Answer: (2,5; +∞) .

Consider the last scheme for solving an inequality of the form a f(x)< b , at a>0 And b<0 shown in Figure 6:

For example, let's solve the inequality:

We notice that no matter what number we substitute for x, the left side of the inequality is always greater than zero, and in our case this expression is less than -8, i.e. and zero means there are no solutions.

Answer: no solutions.

Knowing how the simplest exponential inequalities are solved, we can proceed to solving exponential inequalities.

Example 1

Find the largest integer value of x that satisfies the inequality

Since 6 x is greater than zero (for no x does the denominator go to zero), we multiply both sides of the inequality by 6 x, we get:

440 - 2 6 2x > 8, then
– 2 6 2x > 8 – 440,
– 2 6 2x > – 332,
6 2x< 216,
2x< 3,

x< 1,5. Наибольшее целое число из помежутка (–∞; 1,5) это число 1.

Answer: 1.

Example 2.

Solve the inequality 2 2 x – 3 2 x + 2 ≤ 0

Denote 2 x by y, we get the inequality y 2 - 3y + 2 ≤ 0, we solve this quadratic inequality.

y 2 - 3y +2 = 0,
y 1 = 1 and y 2 = 2.

The branches of the parabola are directed upwards, let's draw a graph:

Then the solution of the inequality will be the inequality 1< у < 2, вернемся к нашей переменной х и получим неравенство 1< 2 х < 2, решая которое и найдем ответ 0 < x < 1.

Answer: (0; 1) .

Example 3. Solve the inequality 5x+1 – 3x+2< 2·5 x – 2·3 x –1
Collect expressions with the same bases in one part of the inequality

5 x +1 – 2 5 x< 3 x +2 – 2·3 x –1

Let's take out the inequality on the left side of the brackets 5 x , and on the right side of the inequality 3 x and get the inequality

5 x (5 - 2)< 3 х (9 – 2/3),
3 5 x< (25/3)·3 х

We divide both parts of the inequality by the expression 3 3 x, the inequality sign will not change, since 3 3 x is a positive number, we get the inequality:

X< 2 (так как 5/3 > 1).

Answer: (–∞; 2) .

If you have any questions about solving exponential inequalities or want to practice solving similar examples, sign up for my lessons. Tutor Valentina Galinevskaya.

site, with full or partial copying of the material, a link to the source is required.

Spacing Method is a simple way to solve fractional rational inequalities. This is the name of inequalities containing rational (or fractional-rational) expressions that depend on a variable.

1. Consider, for example, the following inequality

The interval method allows you to solve it in a couple of minutes.

On the left side of this inequality is a fractional rational function. Rational, because it does not contain any roots, sines, or logarithms - only rational expressions. On the right is zero.

The interval method is based on the following property of a fractional rational function.

A fractional-rational function can change sign only at those points where it is equal to zero or does not exist.

Recall how a square trinomial is factored, that is, an expression of the form .

Where and are the roots of the quadratic equation.

We draw an axis and arrange the points at which the numerator and denominator vanish.

The zeros of the denominator and are punctured points, since at these points the function on the left side of the inequality is not defined (you cannot divide by zero). The zeros of the numerator and - are shaded because the inequality is not strict. For and our inequality is satisfied, since both its parts are equal to zero.

These points break the axis into intervals.

Let us determine the sign of the fractional-rational function on the left side of our inequality on each of these intervals. We remember that a fractional rational function can change sign only at those points where it is equal to zero or does not exist. This means that on each of the intervals between the points where the numerator or denominator vanishes, the sign of the expression on the left side of the inequality will be constant - either "plus" or "minus".

And therefore, to determine the sign of the function on each such interval, we take any point belonging to this interval. The one that suits us.
. Take, for example, and check the sign of the expression on the left side of the inequality. Each of the "brackets" is negative. The left side has a sign.

Next interval: . Let's check the sign for . We get that the left side has changed sign to .

Let's take . When the expression is positive - therefore, it is positive on the entire interval from to .

For , the left side of the inequality is negative.

And finally class="tex" alt="x>7"> . Подставим и проверим знак выражения в левой части неравенства. Каждая "скобочка" положительна. Следовательно, левая часть имеет знак .!}

We have found on what intervals the expression is positive. It remains to write the answer:

Answer: .

Please note: the signs on the intervals alternate. This happened because when passing through each point, exactly one of the linear factors changed sign, and the rest kept it unchanged.

We see that the interval method is very simple. To solve a fractional-rational inequality by the method of intervals, we bring it to the form:

Or class="tex" alt="\genfrac()()()(0)(\displaystyle P\left(x \right))(\displaystyle Q\left(x \right)) > 0"> !}, or or .

(on the left side - a fractional-rational function, on the right side - zero).

Then - we mark on the number line the points at which the numerator or denominator vanishes.
These points divide the entire number line into intervals, on each of which the fractional-rational function retains its sign.
It remains only to find out its sign on each interval.
We do this by checking the sign of the expression at any point within the given interval. After that, we write down the answer. That's all.

But the question arises: do the signs always alternate? No not always! We must be careful not to place signs mechanically and thoughtlessly.

2. Let's look at another inequality.

Class="tex" alt="\genfrac()()()(0)(\displaystyle \left(x-2 \right)^2)(\displaystyle \left(x-1 \right) \left(x-3\right))>0"> !}

We again place points on the axis. The points and are punctured because they are the zeros of the denominator. The dot is also punctured, since the inequality is strict.

When the numerator is positive, both factors in the denominator are negative. This is easy to check by taking any number from a given interval, for example, . The left side has the sign:

When the numerator is positive; the first factor in the denominator is positive, the second factor is negative. The left side has the sign:

When the situation is the same! The numerator is positive, the first factor in the denominator is positive, the second is negative. The left side has the sign:

Finally, with class="tex" alt="x>3"> все множители положительны, и левая часть имеет знак :!}

Answer: .

Why was the alternation of characters broken? Because when passing through the point, the multiplier "responsible" for it did not change sign. Consequently, the entire left-hand side of our inequality did not change sign either.

Conclusion: if the linear factor is in an even power (for example, in a square), then when passing through a point, the sign of the expression on the left side does not change. In the case of an odd degree, the sign, of course, changes.

3. Let's consider a more complicated case. It differs from the previous one in that the inequality is not strict:

The left side is the same as in the previous problem. The picture of signs will be the same:

Maybe the answer will be the same? No! The solution is added This is because at , both the left and right sides of the inequality are equal to zero - therefore, this point is a solution.

Answer: .

In the problem on the exam in mathematics, this situation is often encountered. Here, applicants fall into a trap and lose points. Be careful!

4. What if the numerator or denominator cannot be factored into linear factors? Consider this inequality:

The square trinomial cannot be factorized: the discriminant is negative, there are no roots. But this is good! This means that the sign of the expression is the same for all, and specifically, it is positive. You can read more about this in the article on the properties of a quadratic function.

And now we can divide both sides of our inequality by a value that is positive for all . We arrive at an equivalent inequality:

Which is easily solved by the interval method.

Pay attention - we divided both sides of the inequality by the value, which we knew for sure that it was positive. Of course, in the general case, you should not multiply or divide an inequality by a variable whose sign is unknown.

5 . Consider another inequality, seemingly quite simple:

So I want to multiply it by . But we are already smart, and we will not do this. After all, it can be both positive and negative. And we know that if both parts of the inequality are multiplied by a negative value, the sign of the inequality changes.

We will act differently - we will collect everything in one part and bring it to a common denominator. Zero will remain on the right side:

Class="tex" alt="\genfrac()()()(0)(\displaystyle x-2)(\displaystyle x)>0"> !}

And after that - applicable interval method.

And today not everyone can solve rational inequalities. More precisely, not only everyone can decide. Few people can do it.
Klitschko

This lesson is going to be tough. So tough that only the Chosen will reach the end of it. Therefore, before reading, I recommend removing women, cats, pregnant children and ...

Okay, it's actually quite simple. Suppose you have mastered the interval method (if you haven’t mastered it, I recommend you go back and read it) and learned how to solve inequalities of the form $P\left(x \right) \gt 0$, where $P\left(x \right)$ is some polynomial or product of polynomials.

I believe that it will not be difficult for you to solve, for example, such a game (by the way, try it for a warm-up):

\[\begin(align) & \left(2((x)^(2))+3x+4 \right)\left(4x+25 \right) \gt 0; \\ & x\left(2((x)^(2))-3x-20 \right)\left(x-1 \right)\ge 0; \\ & \left(8x-((x)^(4)) \right)((\left(x-5 \right))^(6))\le 0. \\ \end(align)\]

Now let's complicate the task a little and consider not just polynomials, but the so-called rational fractions of the form:

where $P\left(x \right)$ and $Q\left(x \right)$ are the same polynomials of the form $((a)_(n))((x)^(n))+(( a)_(n-1))((x)^(n-1))+...+((a)_(0))$, or the product of such polynomials.

This will be a rational inequality. The fundamental point is the presence of the variable $x$ in the denominator. For example, here are rational inequalities:

\[\begin(align) & \frac(x-3)(x+7) \lt 0; \\ & \frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0; \\ & \frac(3((x)^(2))+10x+3)(((\left(3-x \right))^(2))\left(4-((x)^( 2)) \right))\ge 0. \\ \end(align)\]

And this is not a rational, but the most common inequality, which is solved by the interval method:

\[\frac(((x)^(2))+6x+9)(5)\ge 0\]

Looking ahead, I’ll say right away: there are at least two ways to solve rational inequalities, but all of them in one way or another are reduced to the method of intervals already known to us. Therefore, before analyzing these methods, let's recall the old facts, otherwise there will be no sense from the new material.

What you already need to know

There are not many important facts. We really only need four.

Abbreviated multiplication formulas

Yes, yes: they will haunt us throughout the school math curriculum. And at the university too. There are quite a few of these formulas, but we only need the following:

\[\begin(align) & ((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2)); \\ & ((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right); \\ & ((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b) ^(2))\right); \\ & ((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^( 2))\right). \\ \end(align)\]

Pay attention to the last two formulas - this is the sum and difference of cubes (and not the cube of the sum or difference!). They are easy to remember if you notice that the sign in the first bracket is the same as the sign in the original expression, and in the second bracket it is opposite to the sign in the original expression.

Linear equations

These are the simplest equations of the form $ax+b=0$, where $a$ and $b$ are ordinary numbers, and $a\ne 0$. This equation is easy to solve:

\[\begin(align) & ax+b=0; \\ &ax=-b; \\ & x=-\frac(b)(a). \\ \end(align)\]

I note that we have the right to divide by the coefficient $a$, because $a\ne 0$. This requirement is quite logical, since with $a=0$ we get this:

First, there is no $x$ variable in this equation. This, generally speaking, should not confuse us (this happens, say, in geometry, and quite often), but still we are no longer a linear equation.

Secondly, the solution of this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation is $0=0$. This equality is always true; hence $x$ is any number (usually written as $x\in \mathbb(R)$). If the coefficient $b$ is not equal to zero, then the equality $b=0$ is never satisfied, i.e. no answers (written $x\in \varnothing $ and read "solution set is empty").

To avoid all these complexities, we simply assume $a\ne 0$, which does not in any way restrict us from further reflections.

Quadratic equations

Let me remind you that this is called a quadratic equation:

Here on the left is a polynomial of the second degree, and again $a\ne 0$ (otherwise, instead of a quadratic equation, we get a linear one). The following equations are solved through the discriminant:

1. If $D \gt 0$, we get two different roots;
2. If $D=0$, then the root will be one, but of the second multiplicity (what kind of multiplicity it is and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
3. For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. This, by the way, is a very useful fact, which for some reason is forgotten to be told in algebra classes.

The roots themselves are calculated according to the well-known formula:

\[((x)_(1,2))=\frac(-b\pm \sqrt(D))(2a)\]

Hence, by the way, the restrictions on the discriminant. After all, the square root of a negative number does not exist. As for the roots, many students have a terrible mess in their heads, so I specially recorded a whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)

Operations with rational fractions

Everything that was written above, you already know if you studied the method of intervals. But what we will analyze now has no analogues in the past - this is a completely new fact.

Definition. A rational fraction is an expression of the form

\[\frac(P\left(x \right))(Q\left(x \right))\]

where $P\left(x \right)$ and $Q\left(x \right)$ are polynomials.

It is obvious that it is easy to obtain an inequality from such a fraction - it is enough just to attribute the sign “greater than” or “less than” to the right. And a little further we will find that solving such problems is a pleasure, everything is very simple there.

Problems begin when there are several such fractions in one expression. They have to be reduced to a common denominator - and it is at this moment that a large number of offensive mistakes are made.

Therefore, in order to successfully solve rational equations, it is necessary to firmly master two skills:

1. Factorization of the polynomial $P\left(x \right)$;
2. Actually, bringing fractions to a common denominator.

How to factorize a polynomial? Very simple. Let we have a polynomial of the form

Let's equate it to zero. We get the $n$-th degree equation:

\[((a)_(n))((x)^(n))+((a)_(n-1))((x)^(n-1))+...+(( a)_(1))x+((a)_(0))=0\]

Let's say we solved this equation and got the roots $((x)_(1)),\ ...,\ ((x)_(n))$ (don't worry: in most cases there will be no more than two of these roots) . In this case, our original polynomial can be rewritten like this:

\[\begin(align) & P\left(x \right)=((a)_(n))((x)^(n))+((a)_(n-1))((x )^(n-1))+...+((a)_(1))x+((a)_(0))= \\ & =((a)_(n))\left(x -((x)_(1)) \right)\cdot \left(x-((x)_(2)) \right)\cdot ...\cdot \left(x-((x)_( n)) \right) \end(align)\]

That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate factor in front of the brackets, and if necessary, it can be inserted into any of these brackets (practice shows that with $((a)_ (n))\ne \pm 1$ there are almost always fractions among the roots).

Task. Simplify the expression:

\[\frac(((x)^(2))+x-20)(x-4)-\frac(2((x)^(2))-5x+3)(2x-3)-\ frac(4-8x-5((x)^(2)))(x+2)\]

Solution. First, let's look at the denominators: they are all linear binomials, and there is nothing to factorize here. So let's factorize the numerators:

\[\begin(align) & ((x)^(2))+x-20=\left(x+5 \right)\left(x-4 \right); \\ & 2((x)^(2))-5x+3=2\left(x-\frac(3)(2) \right)\left(x-1 \right)=\left(2x- 3\right)\left(x-1\right); \\ & 4-8x-5((x)^(2))=-5\left(x+2 \right)\left(x-\frac(2)(5) \right)=\left(x +2 \right)\left(2-5x \right). \\\end(align)\]

Please note: in the second polynomial, the senior coefficient "2", in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since a fraction got out there.

The same thing happened in the third polynomial, only there the order of the terms is also confused. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter a factor in one and only one bracket!), which saved us from the inconvenience associated with fractional roots.

As for the first polynomial, everything is simple there: its roots are sought either in the standard way through the discriminant, or using the Vieta theorem.

Let's go back to the original expression and rewrite it with the numerators decomposed into factors:

\[\begin(matrix) \frac(\left(x+5 \right)\left(x-4 \right))(x-4)-\frac(\left(2x-3 \right)\left( x-1 \right))(2x-3)-\frac(\left(x+2 \right)\left(2-5x \right))(x+2)= \\ =\left(x+5 \right)-\left(x-1 \right)-\left(2-5x \right)= \\ =x+5-x+1-2+5x= \\ =5x+4. \\ \end(matrix)\]

Answer: $5x+4$.

As you can see, nothing complicated. A bit of 7th-8th grade math and that's it. The point of all transformations is to turn a complex and scary expression into something simple and easy to work with.

However, this will not always be the case. So now we will consider a more serious problem.

But first, let's figure out how to bring two fractions to a common denominator. The algorithm is extremely simple:

1. Factorize both denominators;
2. Consider the first denominator and add to it the factors present in the second denominator, but not in the first. The resulting product will be the common denominator;
3. Find out what factors each of the original fractions lacks so that the denominators become equal to the common one.

Perhaps this algorithm will seem to you just a text in which there are “a lot of letters”. So let's take a look at a specific example.

Task. Simplify the expression:

\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]

Solution. Such voluminous tasks are best solved in parts. Let's write out what is in the first bracket:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3))-8 )-\frac(1)(x-2)\]

Unlike the previous problem, here the denominators are not so simple. Let's factorize each of them.

The square trinomial $((x)^(2))+2x+4$ cannot be factorized because the equation $((x)^(2))+2x+4=0$ has no roots (the discriminant is negative). We leave it unchanged.

The second denominator, the cubic polynomial $((x)^(3))-8$, upon closer examination is the difference of cubes and can be easily decomposed using the abbreviated multiplication formulas:

\[((x)^(3))-8=((x)^(3))-((2)^(3))=\left(x-2 \right)\left(((x) ^(2))+2x+4 \right)\]

Nothing else can be factored, since the first bracket contains a linear binomial, and the second one contains a construction already familiar to us, which has no real roots.

Finally, the third denominator is a linear binomial that cannot be decomposed. Thus, our equation will take the form:

\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))-\frac(1)(x-2)\]

It is quite obvious that $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$ will be the common denominator, and to reduce all fractions to it, you need to multiply the first fraction to $\left(x-2 \right)$, and the last one to $\left(((x)^(2))+2x+4 \right)$. Then it remains only to bring the following:

\[\begin(matrix) \frac(x\cdot \left(x-2 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \ right))+\frac(((x)^(2))+8)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))- \frac(1\cdot \left(((x)^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x +4 \right))= \\ =\frac(x\cdot \left(x-2 \right)+\left(((x)^(2))+8 \right)-\left(((x )^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \\ =\frac (((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))= \\ =\frac(((x)^(2))-4x+4)(\left(x-2 \right)\ left(((x)^(2))+2x+4 \right)). \\ \end(matrix)\]

Pay attention to the second line: when the denominator is already common, i.e. instead of three separate fractions, we wrote one large one, you should not immediately get rid of the brackets. It is better to write an extra line and note that, say, there was a minus before the third fraction - and it will not go anywhere, but will “hang” in the numerator in front of the bracket. This will save you a lot of mistakes.

Well, in the last line it is useful to factorize the numerator. Moreover, this is an exact square, and the abbreviated multiplication formulas again come to our aid. We have:

\[\frac(((x)^(2))-4x+4)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right) )=\frac(x-2)(((x)^(2))+2x+4)\]

Now let's deal with the second bracket in the same way. Here I will simply write a chain of equalities:

\[\begin(matrix) \frac(((x)^(2)))(((x)^(2))-4)-\frac(2)(2-x)=\frac((( x)^(2)))(\left(x-2 \right)\left(x+2 \right))-\frac(2)(2-x)= \\ =\frac(((x) ^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)= \\ =\frac(((x)^( 2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2\cdot \left(x+2 \right))(\left(x-2 \right )\cdot \left(x+2 \right))= \\ =\frac(((x)^(2))+2\cdot \left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right) ). \\ \end(matrix)\]

We return to the original problem and look at the product:

\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]

Answer: \[\frac(1)(x+2)\].

The meaning of this problem is the same as the previous one: to show how much rational expressions can be simplified if you approach their transformation wisely.

And now, when you know all this, let's move on to the main topic of today's lesson - solving fractional rational inequalities. Moreover, after such preparation, the inequalities themselves will click like nuts. :)

The main way to solve rational inequalities

There are at least two approaches to solving rational inequalities. Now we will consider one of them - the one that is generally accepted in the school mathematics course.

But first, let's note an important detail. All inequalities are divided into two types:

1. Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
2. Nonstrict: $f\left(x \right)\ge 0$ or $f\left(x \right)\le 0$.

Inequalities of the second type are easily reduced to the first, as well as the equation:

This small "addition" $f\left(x \right)=0$ leads to such an unpleasant thing as filled points - we met them back in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's analyze the universal algorithm:

1. Collect all non-zero elements on one side of the inequality sign. For example, on the left;
2. Bring all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factorize into the numerator and denominator. One way or another, we get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the tick is the inequality sign.
3. Equate the numerator to zero: $P\left(x \right)=0$. We solve this equation and get the roots $((x)_(1))$, $((x)_(2))$, $((x)_(3))$, ... Then we require that the denominator was not equal to zero: $Q\left(x \right)\ne 0$. Of course, in essence, we have to solve the equation $Q\left(x \right)=0$, and we get the roots $x_(1)^(*)$, $x_(2)^(*)$, $x_(3 )^(*)$, ... (in real problems there will hardly be more than three such roots).
4. We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punched out.
5. We place the plus and minus signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a "plus". If $f\left(x \right) \lt 0$, then we look at intervals with "minuses".

Practice shows that points 2 and 4 cause the greatest difficulties - competent transformations and the correct arrangement of numbers in ascending order. Well, at the last step, be extremely careful: we always place signs based on the last inequality written before moving on to the equations. This is a universal rule inherited from the interval method.

So, there is a scheme. Let's practice.

Task. Solve the inequality:

\[\frac(x-3)(x+7) \lt 0\]

Solution. We have a strict inequality of the form $f\left(x \right) \lt 0$. Obviously, points 1 and 2 from our scheme have already been completed: all the elements of inequality are collected on the left, nothing needs to be reduced to a common denominator. So let's move on to the third point.

Set the numerator to zero:

\[\begin(align) & x-3=0; \\ &x=3. \end(align)\]

And the denominator:

\[\begin(align) & x+7=0; \\ & ((x)^(*))=-7. \\ \end(align)\]

In this place, many people get stuck, because in theory you need to write down $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But after all, in the future we will poke out the points that came from the denominator, so you shouldn’t complicate your calculations once again - write an equal sign everywhere and don’t worry. No one will deduct points for this. :)

Fourth point. We mark the obtained roots on the number line:

All points are punctured because the inequality is strict

Note: all points are punctured because the original inequality is strict. And here it doesn’t matter anymore: these points came from the numerator or from the denominator.

Well, look at the signs. Take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but you could have just as well taken $((x)_(0))=3.1$ or $((x)_(0)) =1\000\000$). We get:

So, to the right of all the roots we have a positive area. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, we proceed to the fifth point: we place the signs and choose the right one:

We return to the last inequality, which was before solving the equations. Actually, it coincides with the original one, because we did not perform any transformations in this task.

Since it is necessary to solve an inequality of the form $f\left(x \right) \lt 0$, I shaded the interval $x\in \left(-7;3 \right)$ - it is the only one marked with a minus sign. This is the answer.

Answer: $x\in \left(-7;3 \right)$

That's all! Is it difficult? No, it's not difficult. Indeed, it was an easy task. Now let's complicate the mission a little and consider a more "fancy" inequality. When solving it, I will no longer give such detailed calculations - I will simply outline the key points. In general, we will arrange it the way we would have done it on an independent work or exam. :)

Task. Solve the inequality:

\[\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\ge 0$. All non-zero elements are collected on the left, there are no different denominators. Let's move on to equations.

Numerator:

\[\begin(align) & \left(7x+1 \right)\left(11x+2 \right)=0 \\ & 7x+1=0\Rightarrow ((x)_(1))=-\ frac(1)(7); \\ & 11x+2=0\Rightarrow ((x)_(2))=-\frac(2)(11). \\ \end(align)\]

Denominator:

\[\begin(align) & 13x-4=0; \\ & 13x=4; \\ & ((x)^(*))=\frac(4)(13). \\ \end(align)\]

I don’t know what kind of pervert made up this problem, but the roots didn’t turn out very well: it will be difficult to arrange them on a number line. And if everything is more or less clear with the root $((x)^(*))=(4)/(13)\;$ (this is the only positive number - it will be on the right), then $((x)_(1 ))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\;$ require further study: which one is larger?

You can find this out, for example:

\[((x)_(1))=-\frac(1)(7)=-\frac(2)(14) \gt -\frac(2)(11)=((x)_(2 ))\]

I hope there is no need to explain why the numeric fraction $-(2)/(14)\; \gt -(2)/(11)\;$? If necessary, I recommend remembering how to perform actions with fractions.

And we mark all three roots on the number line:

The points from the numerator are shaded, from the denominator they are cut out

We put up signs. For example, you can take $((x)_(0))=1$ and find out the sign at this point:

\[\begin(align) & f\left(x \right)=\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4); \\ & f\left(1 \right)=\frac(\left(7\cdot 1+1 \right)\left(11\cdot 1+2 \right))(13\cdot 1-4)=\ frac(8\cdot 13)(9) \gt 0. \\\end(align)\]

The last inequality before the equations was $f\left(x \right)\ge 0$, so we are interested in the plus sign.

We got two sets: one is an ordinary segment, and the other is an open ray on the number line.

Answer: $x\in \left[ -\frac(2)(11);-\frac(1)(7) \right]\bigcup \left(\frac(4)(13);+\infty \right )$

An important note about the numbers that we substitute to find out the sign on the rightmost interval. It is not necessary to substitute a number close to the rightmost root. You can take billions or even "plus-infinity" - in this case, the sign of the polynomial in the bracket, numerator or denominator is determined solely by the sign of the leading coefficient.

Let's take another look at the $f\left(x \right)$ function from the last inequality:

It contains three polynomials:

\[\begin(align) & ((P)_(1))\left(x \right)=7x+1; \\ & ((P)_(2))\left(x \right)=11x+2; \\ & Q\left(x\right)=13x-4. \end(align)\]

All of them are linear binomials, and all of them have positive coefficients (numbers 7, 11 and 13). Therefore, when substituting very large numbers, the polynomials themselves will also be positive. :)

This rule may seem overly complicated, but only at first, when we analyze very easy problems. In serious inequalities, the "plus-infinity" substitution will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.

We will face such challenges very soon. But first, let's look at an alternative way to solve fractional rational inequalities.

Alternative way

This technique was suggested to me by one of my students. I myself have never used it, but practice has shown that it is really more convenient for many students to solve inequalities in this way.

So, the original data is the same. We need to solve a fractional rational inequality:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\]

Let's think: why is the polynomial $Q\left(x \right)$ "worse" than the polynomial $P\left(x \right)$? Why do we have to consider separate groups of roots (with and without an asterisk), think about punched points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is different from zero.

Otherwise, there are no differences between the numerator and the denominator: we also equate it to zero, look for the roots, then mark them on the number line. So why not replace the fractional bar (in fact, the division sign) with the usual multiplication, and write all the requirements of the DHS as a separate inequality? For example, like this:

\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\Rightarrow \left\( \begin(align) & P\left(x \right)\cdot Q \left(x \right) \gt 0, \\ & Q\left(x \right)\ne 0. \\ \end(align) \right.\]

Please note: this approach will allow you to reduce the problem to the method of intervals, but it will not complicate the solution at all. After all, anyway, we will equate the polynomial $Q\left(x \right)$ to zero.

Let's see how it works on real tasks.

Task. Solve the inequality:

\[\frac(x+8)(x-11) \gt 0\]

Solution. So, let's move on to the interval method:

\[\frac(x+8)(x-11) \gt 0\Rightarrow \left\( \begin(align) & \left(x+8 \right)\left(x-11 \right) \gt 0 , \\ & x-11\ne 0. \\ \end(align) \right.\]

The first inequality is solved elementarily. Just set each parenthesis to zero:

\[\begin(align) & x+8=0\Rightarrow ((x)_(1))=-8; \\ & x-11=0\Rightarrow ((x)_(2))=11. \\ \end(align)\]

With the second inequality, everything is also simple:

We mark the points $((x)_(1))$ and $((x)_(2))$ on the real line. All of them are punctured because the inequality is strict:

The right point turned out to be punctured twice. This is fine.

Pay attention to the point $x=11$. It turns out that it is “twice gouged out”: on the one hand, we gouge it out because of the severity of inequality, on the other hand, because of the additional requirement of ODZ.

In any case, it will be just a punctured point. Therefore, we put signs for the inequality $\left(x+8 \right)\left(x-11 \right) \gt 0$ - the last one we saw before we started solving the equations:

We are interested in positive regions, since we are solving an inequality of the form $f\left(x \right) \gt 0$, and we will color them. It remains only to write down the answer.

Answer. $x\in \left(-\infty ;-8 \right)\bigcup \left(11;+\infty \right)$

Using this solution as an example, I would like to warn you against a common mistake among novice students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you a lot of problems.

Now let's try something more difficult.

Task. Solve the inequality:

\[\frac(\left(2x-13 \right)\left(12x-9 \right))(15x+33)\le 0\]

Solution. This is a non-strict inequality of the form $f\left(x \right)\le 0$, so here you need to carefully monitor the filled points.

Let's move on to the interval method:

\[\left\( \begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)\le 0, \\ & 15x+33\ ne 0. \\ \end(align) \right.\]

Let's move on to the equation:

\[\begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0 \\ & 2x-13=0\Rightarrow ((x )_(1))=6.5; \\ & 12x-9=0\Rightarrow ((x)_(2))=0.75; \\ & 15x+33=0\Rightarrow ((x)_(3))=-2,2. \\ \end(align)\]

We take into account the additional requirement:

We mark all the obtained roots on the number line:

If a point is both punched out and filled in at the same time, it is considered punched out.

Again, two points "overlap" each other - this is normal, it will always be so. It is only important to understand that a point marked both as punched out and filled in is actually a punched out point. Those. "Gouging" is a stronger action than "painting over".

This is absolutely logical, because by puncturing we mark points that affect the sign of the function, but do not themselves participate in the answer. And if at some point the number ceases to suit us (for example, it does not fall into the ODZ), we delete it from consideration until the very end of the task.

In general, stop philosophizing. We arrange the signs and paint over those intervals that are marked with a minus sign:

Answer. $x\in \left(-\infty ;-2,2 \right)\bigcup \left[ 0,75;6,5 \right]$.

And again I wanted to draw your attention to this equation:

\[\left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0\]

Once again: never open parentheses in such equations! You're only making it harder for yourself. Remember: the product is zero when at least one of the factors is zero. Consequently, this equation simply “falls apart” into several smaller ones, which we solved in the previous problem.

Taking into account the multiplicity of roots

From the previous problems, it is easy to see that it is precisely the non-strict inequalities that are most difficult, because in them you have to keep track of the filled points.

But there is an even greater evil in the world - these are multiple roots in inequalities. Here it is already necessary to follow not some filled points there - here the inequality sign may not suddenly change when passing through these same points.

We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). So let's introduce a new definition:

Definition. The root of the equation $((\left(x-a \right))^(n))=0$ is equal to $x=a$ and is called the root of the $n$th multiplicity.

Actually, we are not particularly interested in the exact value of the multiplicity. The only important thing is whether this very number $n$ is even or odd. Because:

1. If $x=a$ is a root of even multiplicity, then the sign of the function does not change when passing through it;
2. And vice versa, if $x=a$ is a root of odd multiplicity, then the sign of the function will change.

A special case of a root of odd multiplicity are all the previous problems considered in this lesson: there the multiplicity is equal to one everywhere.

And further. Before we start solving problems, I would like to draw your attention to one subtlety that seems obvious to an experienced student, but drives many beginners into a stupor. Namely:

The multiplicity root $n$ occurs only when the entire expression is raised to this power: $((\left(x-a \right))^(n))$, and not $\left(((x)^( n))-a\right)$.

Once again: the bracket $((\left(x-a \right))^(n))$ gives us the root $x=a$ of multiplicity $n$, but the bracket $\left(((x)^(n)) -a \right)$ or, as often happens, $(a-((x)^(n)))$ gives us a root (or two roots, if $n$ is even) of the first multiplicity, no matter what is equal to $n$.

Compare:

\[((\left(x-3 \right))^(5))=0\Rightarrow x=3\left(5k \right)\]

Everything is clear here: the whole bracket was raised to the fifth power, so at the output we got the root of the fifth degree. And now:

\[\left(((x)^(2))-4 \right)=0\Rightarrow ((x)^(2))=4\Rightarrow x=\pm 2\]

We got two roots, but both of them have the first multiplicity. Or here's another one:

\[\left(((x)^(10))-1024 \right)=0\Rightarrow ((x)^(10))=1024\Rightarrow x=\pm 2\]

And do not be confused by the tenth degree. The main thing is that 10 is an even number, so we have two roots at the output, and both of them again have the first multiplicity.

In general, be careful: multiplicity occurs only when the degree applies to the entire bracket, not just the variable.

Task. Solve the inequality:

\[\frac(((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right))(((\left(x+7 \right))^(5)))\ge 0\]

Solution. Let's try to solve it in an alternative way - through the transition from the particular to the product:

\[\left\( \begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ( (\left(x+7 \right))^(5))\ge 0, \\ & ((\left(x+7 \right))^(5))\ne 0. \\ \end(align )\right.\]

We deal with the first inequality using the interval method:

\[\begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ((\left( x+7 \right))^(5))=0; \\ & ((x)^(2))=0\Rightarrow x=0\left(2k \right); \\ & ((\left(6-x \right))^(3))=0\Rightarrow x=6\left(3k \right); \\ & x+4=0\Rightarrow x=-4; \\ & ((\left(x+7 \right))^(5))=0\Rightarrow x=-7\left(5k \right). \\ \end(align)\]

Additionally, we solve the second inequality. In fact, we have already solved it, but so that the reviewers do not find fault with the solution, it is better to solve it again:

\[((\left(x+7 \right))^(5))\ne 0\Rightarrow x\ne -7\]

Note that there are no multiplicities in the last inequality. Indeed: what difference does it make how many times to cross out the point $x=-7$ on the number line? At least once, at least five times - the result will be the same: a punctured point.

Let's note everything that we got on the number line:

As I said, the $x=-7$ point will eventually be punched out. The multiplicities are arranged based on the solution of the inequality by the interval method.

It remains to place the signs:

Since the point $x=0$ is a root of even multiplicity, the sign does not change when passing through it. The remaining points have an odd multiplicity, and everything is simple with them.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$

Pay attention to $x=0$ again. Because of the even multiplicity, an interesting effect arises: everything to the left of it is painted over, to the right - too, and the point itself is completely painted over.

As a consequence, it does not need to be isolated when recording a response. Those. you don't have to write something like $x\in \left[ -4;0 \right]\bigcup \left[ 0;6 \right]$ (although formally such an answer would also be correct). Instead, we immediately write $x\in \left[ -4;6 \right]$.

Such effects are possible only for roots of even multiplicity. And in the next task, we will encounter the reverse "manifestation" of this effect. Ready?

Task. Solve the inequality:

\[\frac(((\left(x-3 \right))^(4))\left(x-4 \right))(((\left(x-1 \right))^(2)) \left(7x-10-((x)^(2)) \right))\ge 0\]

Solution. This time we will follow the standard scheme. Set the numerator to zero:

\[\begin(align) & ((\left(x-3 \right))^(4))\left(x-4 \right)=0; \\ & ((\left(x-3 \right))^(4))=0\Rightarrow ((x)_(1))=3\left(4k \right); \\ & x-4=0\Rightarrow ((x)_(2))=4. \\ \end(align)\]

And the denominator:

\[\begin(align) & ((\left(x-1 \right))^(2))\left(7x-10-((x)^(2)) \right)=0; \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(1)^(*)=1\left(2k \right); \\ & 7x-10-((x)^(2))=0\Rightarrow x_(2)^(*)=5;\ x_(3)^(*)=2. \\ \end(align)\]

Since we are solving a non-strict inequality of the form $f\left(x \right)\ge 0$, the roots from the denominator (which have asterisks) will be cut out, and those from the numerator will be painted over.

We arrange the signs and stroke the areas marked with a "plus":

The point $x=3$ is isolated. This is part of the answer

Before writing down the final answer, take a close look at the picture:

1. The point $x=1$ has an even multiplicity, but is itself punctured. Therefore, it will have to be isolated in the answer: you need to write $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left(-\ infty ;2\right)$.
2. The point $x=3$ also has even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step to the left and right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written as $x\in \left\( 3 \right\)$.

We combine all the obtained pieces into a common set and write down the answer.

Answer: $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;5 \right) $

Definition. Solving the inequality means find the set of all its solutions, or prove that this set is empty.

It would seem: what can be incomprehensible here? Yes, the fact of the matter is that sets can be specified in different ways. Let's rewrite the answer to the last problem:

We literally read what is written. The variable "x" belongs to a certain set, which is obtained by the union (symbol "U") of four separate sets:

• The interval is $\left(-\infty ;1 \right)$, which literally means "all numbers less than one, but not one itself";
• The interval is $\left(1;2 \right)$, i.e. "all numbers between 1 and 2, but not the numbers 1 and 2 themselves";
• The set $\left\( 3 \right\)$, consisting of a single number - three;
• The interval $\left[ 4;5 \right)$ containing all numbers between 4 and 5, plus 4 itself, but not 5.

The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only denote only the boundaries of these sets, the set $\left\( 3 \right\)$ defines exactly one number by enumeration.

To understand that we are listing the specific numbers included in the set (and not setting boundaries or anything else), curly braces are used. For example, the notation $\left\( 1;2 \right\)$ means exactly "a set consisting of two numbers: 1 and 2", but not a segment from 1 to 2. In no case do not confuse these concepts.

Multiplicity addition rule

Well, at the end of today's lesson, a little tin from Pavel Berdov. :)

Attentive students have probably already asked themselves the question: what will happen if the same roots are found in the numerator and denominator? So the following rule works:

Multiplicities of identical roots are added. Always. Even if this root occurs in both the numerator and the denominator.

Sometimes it's better to decide than to talk. Therefore, we solve the following problem:

Task. Solve the inequality:

\[\frac(((x)^(2))+6x+8)(\left(((x)^(2))-16 \right)\left(((x)^(2))+ 9x+14 \right))\ge 0\]

\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -4. \\ \end(align)\]

So far, nothing special. Set the denominator to zero:

\[\begin(align) & \left(((x)^(2))-16 \right)\left(((x)^(2))+9x+14 \right)=0 \\ & ( (x)^(2))-16=0\Rightarrow x_(1)^(*)=4;\ x_(2)^(*)=-4; \\ & ((x)^(2))+9x+14=0\Rightarrow x_(3)^(*)=-7;\ x_(4)^(*)=-2. \\ \end(align)\]

Two identical roots are found: $((x)_(1))=-2$ and $x_(4)^(*)=-2$. Both have the first multiplicity. Therefore, we replace them with one root $x_(4)^(*)=-2$, but with a multiplicity of 1+1=2.

In addition, there are also identical roots: $((x)_(2))=-4$ and $x_(2)^(*)=-4$. They are also of the first multiplicity, so only $x_(2)^(*)=-4$ of multiplicity 1+1=2 remains.

Please note: in both cases, we left exactly the “cut out” root, and threw out the “painted over” one from consideration. Because even at the beginning of the lesson, we agreed: if a point is both punched out and painted over at the same time, then we still consider it punched out.

As a result, we have four roots, and all of them turned out to be gouged out:

\[\begin(align) & x_(1)^(*)=4; \\ & x_(2)^(*)=-4\left(2k \right); \\ & x_(3)^(*)=-7; \\ & x_(4)^(*)=-2\left(2k \right). \\ \end(align)\]

We mark them on the number line, taking into account the multiplicity:

We place the signs and paint over the areas of interest to us:

All. No isolated points and other perversions. You can write down the answer.

Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left(4;+\infty \right)$.

multiplication rule

Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to a certain power. This changes the multiplicities of all the original roots.

This is rare, so most students do not have experience in solving such problems. And the rule here is:

When an equation is raised to a power $n$, the multiplicity of all its roots also increases by a factor of $n$.

In other words, raising to a power results in multiplying multiplicities by the same power. Let's take this rule as an example:

Task. Solve the inequality:

\[\frac(x((\left(((x)^(2))-6x+9 \right))^(2))((\left(x-4 \right))^(5)) )(((\left(2-x \right))^(3))((\left(x-1 \right))^(2)))\le 0\]

Solution. Set the numerator to zero:

The product is equal to zero when at least one of the factors is equal to zero. Everything is clear with the first multiplier: $x=0$. And here's where the problems start:

\[\begin(align) & ((\left(((x)^(2))-6x+9 \right))^(2))=0; \\ & ((x)^(2))-6x+9=0\left(2k \right); \\ & D=((6)^(3))-4\cdot 9=0 \\ & ((x)_(2))=3\left(2k \right)\left(2k \right) \ \ & ((x)_(2))=3\left(4k \right) \\ \end(align)\]

As you can see, the equation $((x)^(2))-6x+9=0$ has a unique root of the second multiplicity: $x=3$. The whole equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which we finally wrote down.

\[((\left(x-4 \right))^(5))=0\Rightarrow x=4\left(5k \right)\]

No problem with the denominator either:

\[\begin(align) & ((\left(2-x \right))^(3))((\left(x-1 \right))^(2))=0; \\ & ((\left(2-x \right))^(3))=0\Rightarrow x_(1)^(*)=2\left(3k \right); \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(2)^(*)=1\left(2k \right). \\ \end(align)\]

In total, we got five points: two punched out and three filled in. There are no coinciding roots in the numerator and denominator, so we just mark them on the number line:

We arrange the signs taking into account the multiplicities and paint over the intervals of interest to us:

Again one isolated point and one punctured

Because of the roots of even multiplicity, we again received a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, not $x\in \left[ 0;2 \right)$, and also an isolated point $ x\in \left\( 3 \right\)$.

Answer. $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;+\infty \right)$

As you can see, everything is not so difficult. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the very ones that we discussed at the very beginning.

Preconversions

The inequalities we will discuss in this section are not complex. However, unlike the previous tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.

We discussed this issue in detail at the very beginning of today's lesson. If you are not sure that you understand what it is about, I strongly recommend that you go back and repeat. Because there is no point in cramming the methods for solving inequalities if you "swim" in the conversion of fractions.

In homework, by the way, there will also be many similar tasks. They are placed in a separate subsection. And there you will find very non-trivial examples. But this will be in the homework, but now let's analyze a couple of such inequalities.

Task. Solve the inequality:

\[\frac(x)(x-1)\le \frac(x-2)(x)\]

Solution. Moving everything to the left:

\[\frac(x)(x-1)-\frac(x-2)(x)\le 0\]

We reduce to a common denominator, open the brackets, give like terms in the numerator:

\[\begin(align) & \frac(x\cdot x)(\left(x-1 \right)\cdot x)-\frac(\left(x-2 \right)\left(x-1 \ right))(x\cdot \left(x-1 \right))\le 0; \\ & \frac(((x)^(2))-\left(((x)^(2))-2x-x+2 \right))(x\left(x-1 \right)) \le0; \\ & \frac(((x)^(2))-((x)^(2))+3x-2)(x\left(x-1 \right))\le 0; \\ & \frac(3x-2)(x\left(x-1 \right))\le 0. \\\end(align)\]

Now we have a classical fractional rational inequality, the solution of which is no longer difficult. I propose to solve it by an alternative method - through the method of intervals:

\[\begin(align) & \left(3x-2 \right)\cdot x\cdot \left(x-1 \right)=0; \\ & ((x)_(1))=\frac(2)(3);\ ((x)_(2))=0;\ ((x)_(3))=1. \\ \end(align)\]

Don't forget the constraint that comes from the denominator:

We mark all the numbers and restrictions on the number line:

All roots have first multiplicity. No problem. We just place the signs and paint over the areas we need:

This is all. You can write down the answer.

Answer. $x\in \left(-\infty ;0 \right)\bigcup \left[ (2)/(3)\;;1 \right)$.

Of course, this was a very simple example. So now let's take a closer look at the problem. And by the way, the level of this task is quite consistent with independent and control work on this topic in the 8th grade.

Task. Solve the inequality:

\[\frac(1)(((x)^(2))+8x-9)\ge \frac(1)(3((x)^(2))-5x+2)\]

Solution. Moving everything to the left:

\[\frac(1)(((x)^(2))+8x-9)-\frac(1)(3((x)^(2))-5x+2)\ge 0\]

Before bringing both fractions to a common denominator, we decompose these denominators into factors. Suddenly the same brackets will come out? With the first denominator it's easy:

\[((x)^(2))+8x-9=\left(x-1 \right)\left(x+9 \right)\]

The second one is a little more difficult. Feel free to add a constant multiplier to the bracket where the fraction was found. Remember: the original polynomial had integer coefficients, so it is highly likely that the factorization will also have integer coefficients (in fact, it always will, except when the discriminant is irrational).

\[\begin(align) & 3((x)^(2))-5x+2=3\left(x-1 \right)\left(x-\frac(2)(3) \right)= \\ & =\left(x-1 \right)\left(3x-2 \right) \end(align)\]

As you can see, there is a common bracket: $\left(x-1 \right)$. We return to the inequality and bring both fractions to a common denominator:

\[\begin(align) & \frac(1)(\left(x-1 \right)\left(x+9 \right))-\frac(1)(\left(x-1 \right)\ left(3x-2\right))\ge 0; \\ & \frac(1\cdot \left(3x-2 \right)-1\cdot \left(x+9 \right))(\left(x-1 \right)\left(x+9 \right )\left(3x-2 \right))\ge 0; \\ & \frac(3x-2-x-9)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ & \frac(2x-11)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ \end(align)\]

Set the denominator to zero:

\[\begin(align) & \left(x-1 \right)\left(x+9 \right)\left(3x-2 \right)=0; \\ & x_(1)^(*)=1;\ x_(2)^(*)=-9;\ x_(3)^(*)=\frac(2)(3) \\ \end( align)\]

No multiplicities and no coinciding roots. We mark four numbers on a straight line:

We place the signs:

We write down the answer.

Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5,5;+\infty \ right)$.

All! Like that, I read up to this line. :)

Inequality is a numerical ratio that illustrates the magnitude of numbers relative to each other. Inequalities are widely used in the search for quantities in applied sciences. Our calculator will help you deal with such a difficult topic as solving linear inequalities.

What is inequality

Unequal ratios in real life correspond to the constant comparison of different objects: higher or lower, farther or closer, heavier or lighter. Intuitively or visually, we can understand that one object is larger, higher or heavier than another, but in fact it is always a matter of comparing numbers that characterize the corresponding quantities. You can compare objects on any basis, and in any case, we can make a numerical inequality.

If the unknown quantities under specific conditions are equal, then for their numerical determination we make an equation. If not, then instead of the "equal" sign, we can indicate any other ratio between these quantities. Two numbers or mathematical objects can be greater than ">", less than "<» или равны «=» относительно друг друга. В этом случае речь идет о строгих неравенствах. Если же в неравных соотношениях присутствует знак равно и числовые элементы больше или равны (a ≥ b) или меньше или равны (a ≤ b), то такие неравенства называются нестрогими.

Inequality signs in their modern form were invented by the British mathematician Thomas Harriot, who in 1631 published a book on unequal ratios. Greater than ">" and less than "<» представляли собой положенные на бок буквы V, поэтому пришлись по вкусу не только математикам, но и типографам.

Solving inequalities

Inequalities, like equations, come in different types. Linear, square, logarithmic or exponential unequal ratios are unleashed by various methods. However, regardless of the method, any inequality must first be reduced to a standard form. For this, identical transformations are used, which are identical to the modifications of equalities.

Identity transformations of inequalities

Such transformations of expressions are very similar to the ghost of equations, but they have nuances that are important to consider when untying inequalities.

The first identity transformation is identical to the analogous operation with equalities. To both sides of the unequal ratio, you can add or subtract the same number or expression with an unknown x, while the inequality sign remains the same. Most often, this method is used in a simplified form as the transfer of the terms of the expression through the inequality sign with the change of the sign of the number to the opposite. This refers to the change of the sign of the term itself, that is, + R when transferred through any inequality sign will change to - R and vice versa.

The second transformation has two points:

1. Both sides of an unequal ratio are allowed to be multiplied or divided by the same positive number. The sign of the inequality itself will not change.
2. Both sides of the inequality are allowed to be divided or multiplied by the same negative number. The sign of the inequality itself will change to the opposite.

The second identical transformation of inequalities has serious differences with the modification of equations. First, when multiplying/dividing by a negative number, the sign of an unequal expression always reverses. Secondly, dividing or multiplying parts of a relation is allowed only by a number, and not by any expression containing an unknown. The fact is that we cannot know for sure whether a number greater or less than zero is hidden behind the unknown, so the second identical transformation is applied to inequalities exclusively with numbers. Let's look at these rules with examples.

Examples of Untying Inequalities

In algebra assignments, there are a variety of assignments on the topic of inequalities. Let's give us an expression:

6x − 3(4x + 1) > 6.

First, open the brackets and move all unknowns to the left, and all numbers to the right.

6x − 12x > 6 + 3

We need to divide both parts of the expression by −6, so when finding an unknown x, the inequality sign will change to the opposite.

When solving this inequality, we used both identical transformations: we moved all the numbers to the right of the sign and divided both sides of the ratio by a negative number.

Our program is a calculator for solving numerical inequalities that do not contain unknowns. The program contains the following theorems for the ratios of three numbers:

• if A< B то A–C< B–C;
• if A > B, then A–C > B–C.

Instead of subtracting terms A-C, you can specify any arithmetic operation: addition, multiplication, or division. Thus, the calculator will automatically present the inequalities of sums, differences, products or fractions.

Conclusion

In real life, inequalities are as common as equations. Naturally, in everyday life, knowledge about the resolution of inequalities may not be needed. However, in applied sciences, inequalities and their systems are widely used. For example, various studies of the problems of the global economy are reduced to the compilation and unleashing of systems of linear or square inequalities, and some unequal relationships serve as an unambiguous way of proving the existence of certain objects. Use our programs to solve linear inequalities or check your own calculations.