Theory:

When solving inequalities, the following rules are used:

1. Any term of the inequality can be transferred from one part
inequality to another with the opposite sign, while the inequality sign does not change.

2. Both parts of the inequality can be multiplied or divided by one
and the same positive number without changing the inequality sign.

3. Both parts of the inequality can be multiplied or divided by one
and the same negative number, while changing the inequality sign to
opposite.

Solve the inequality − 8 x + 11< − 3 x − 4
Decision.

1. Move the member − 3 x to the left side of the inequality, and the term 11 - to the right side of the inequality, while changing the signs to opposite y − 3 x and at 11 .
Then we get

− 8 x + 3 x< − 4 − 11

− 5 x< − 15

2. Divide both sides of the inequality − 5 x< − 15 to a negative number − 5 , while the inequality sign < , will change to > , i.e. we will move on to an inequality of the opposite meaning.
We get:

− 5 x< − 15 | : (− 5 )

x > −15 : (−5)

x > 3

x > 3 is the solution of the given inequality.

Pay attention!

There are two options for writing a solution: x > 3 or as a numeric range.

We mark the set of solutions of the inequality on the real line and write the answer as a numerical interval.

x ∈ (3 ; + ∞ )

Answer: x > 3 or x ∈ (3 ; + ∞ )

Algebraic inequalities.

Square inequalities. Rational inequalities of higher degrees.

Methods for solving inequalities depend mainly on which class the functions that make up the inequality belong to.

1. I. Square inequalities, that is, inequalities of the form

ax 2 + bx + c > 0 (< 0), a ≠ 0.

To solve the inequality, you can:

1. Factorize the square trinomial, that is, write the inequality as

a (x - x 1) (x - x 2) > 0 (< 0).

1. Put the roots of the polynomial on the number line. The roots divide the set of real numbers into intervals, in each of which the corresponding quadratic function will be of constant sign.
2. Determine the sign of a (x - x 1) (x - x 2) in each gap and write down the answer.

If a square trinomial has no roots, then for D<0 и a>0 is a square trinomial for any x is positive.

• Solve the inequality. x 2 + x - 6 > 0.

Factoring the square trinomial (x + 3) (x - 2) > 0

Answer: x (-∞; -3) (2; +∞).

2) (x - 6) 2 > 0

This inequality is true for any x except x = 6.

Answer: (-∞; 6) (6; +∞).

3) x² + 4x + 15< 0.

Here D< 0, a = 1 >0. The square trinomial is positive for all x.

Answer: x О Ø.

Solve inequalities:

1. 1 + x - 2x²< 0. Ответ:
2. 3x² - 12x + 12 ≤ 0. Answer:
3. 3x² - 7x + 5 ≤ 0. Answer:
4. 2x² - 12x + 18 > 0. Answer:
5. For what values ​​of a does the inequality

x² - ax > holds for any x? Answer:

1. II. Rational inequalities of higher degrees, that is, inequalities of the form

a n x n + a n-1 x n-1 + … + a 1 x + a 0 > 0 (<0), n>2.

The polynomial of the highest degree should be factored, that is, the inequality should be written in the form

a n (x - x 1) (x - x 2) ... (x - x n) > 0 (<0).

Mark on the number line the points where the polynomial vanishes.

Determine the signs of the polynomial on each interval.

1) Solve the inequality x 4 - 6x 3 + 11x 2 - 6x< 0.

x 4 - 6x 3 + 11x 2 - 6x = x (x 3 - 6x 2 + 11x -6) = x (x 3 - x 2 - 5x 2 + 5x +6x - 6) = x (x - 1)(x 2-5x + 6) =

x (x - 1) (x - 2) (x - 3). So x (x - 1) (x - 2) (x - 3)<0

Answer: (0; 1) (2; 3).

2) Solve the inequality (x -1) 5 (x + 2) (x - ½) 7 (2x + 1) 4<0.

On the real axis, mark the points where the polynomial vanishes. This is x \u003d 1, x \u003d -2, x \u003d ½, x \u003d - ½.

At the point x \u003d - ½, there is no sign change, because the binomial (2x + 1) is raised to an even power, that is, the expression (2x + 1) 4 does not change sign when passing through the point x \u003d - ½.

Answer: (-∞; -2) (½; 1).

3) Solve the inequality: x 2 (x + 2) (x - 3) ≥ 0.

This inequality is equivalent to the following set

The solution to (1) is x (-∞; -2) (3; +∞). Solution (2) is x = 0, x = -2, x = 3. Combining the obtained solutions, we get x н (-∞; -2] (0) (0) .

By gaining a knack for working with linear inequalities, their solutions can be written briefly without explanation. In this case, the initial linear inequality is first written, and below are equivalent inequalities obtained at each step of the solution:
3x+12≤0 ;
3 x≤−12 ;
x≤−4 .

Answer:

x≤−4 or (−∞, −4] .

Example.

List all solutions of the linear inequality −2.7 z>0 .

Decision.

Here the coefficient a with the variable z is −2.7. And the coefficient b is absent in an explicit form, that is, it is equal to zero. Therefore, the first step of the algorithm for solving a linear inequality with one variable does not need to be performed, since the transfer of zero from the left side to the right does not change the form of the original inequality.

It remains to divide both parts of the inequality by −2.7, remembering to reverse the sign of the inequality, since −2.7 is a negative number. We have (−2.7 z):(−2.7)<0:(−2,7) , and further z<0 .

And now briefly:
−2.7 z>0 ;
z<0 .

Answer:

z<0 или (−∞, 0) .

Example.

Solve the inequality .

Decision.

We need to solve a linear inequality with coefficient a for variable x equal to −5 and with coefficient b to which the fraction corresponds −15/22. We act according to a well-known scheme: first we transfer −15/22 to the right side with the opposite sign, after which we divide both parts of the inequality by a negative number −5, while changing the inequality sign:

The last transition on the right side uses , then executed .

Answer:

Now let's move on to the case when a=0 . The principle of solving the linear inequality a x+b<0 (знак, естественно, может быть и другим) при a=0 , то есть, неравенства 0·x+b<0 , заключается в рассмотрении числового неравенства b<0 и выяснении, верное оно или нет.

What is it based on? Very simply: on the definition of a solution to an inequality. How? Yes, here it is: no matter what value of the variable x we ​​substitute into the original linear inequality, we get a numerical inequality of the form b<0 (так как при подстановке любого значения t вместо переменной x мы имеем 0·t+b<0 , откуда b<0 ). Если оно верное, то это означает, что любое число является решением исходного неравенства. Если же числовое неравенство b<0 оказывается неверным, то это говорит о том, что исходное линейное неравенство не имеет решений, так как не существует ни одного значения переменной, которое обращало бы его в верное числовое равенство.

Let us formulate the above reasoning in the form algorithm for solving linear inequalities 0 x+b<0 (≤, >, ≥) :

• Consider the numerical inequality b<0 (≤, >, ≥) and
• if it is true, then the solution to the original inequality is any number;
• if it is false, then the original linear inequality has no solutions.

Now let's look at this with examples.

Example.

Solve the inequality 0 x+7>0 .

Decision.

For any value of the variable x, the linear inequality 0 x+7>0 turns into a numerical inequality 7>0 . The last inequality is true, therefore, any number is a solution to the original inequality.

Answer:

the solution is any number or (−∞, +∞) .

Example.

Does the linear inequality have solutions 0 x−12.7≥0 .

Decision.

If we substitute any number instead of the variable x, then the original inequality turns into a numerical inequality −12.7≥0, which is incorrect. And this means that no number is a solution to the linear inequality 0 x−12.7≥0 .

Answer:

no, it doesn't.

To conclude this subsection, we will analyze the solutions of two linear inequalities, both of whose coefficients are equal to zero.

Example.

Which of the linear inequalities 0 x+0>0 and 0 x+0≥0 has no solutions, and which has infinitely many solutions?

Decision.

If we substitute any number instead of the variable x, then the first inequality will take the form 0>0 , and the second - 0≥0 . The first one is incorrect, and the second one is correct. Therefore, the linear inequality 0 x+0>0 has no solutions, and the inequality 0 x+0≥0 has infinitely many solutions, namely, its solution is any number.

Answer:

the inequality 0 x+0>0 has no solutions, and the inequality 0 x+0≥0 has infinitely many solutions.

interval method

In general, the interval method is studied in the school algebra course later than the topic of solving linear inequalities with one variable is covered. But the interval method allows solving a variety of inequalities, including linear ones. Therefore, let's dwell on it.

We note right away that it is advisable to use the interval method for solving linear inequalities with a non-zero coefficient for the variable x. Otherwise, the conclusion about the solution of the inequality is faster and more convenient to make in the way discussed at the end of the previous paragraph.

The interval method implies

• introduction of a function corresponding to the left side of the inequality, in our case - linear function y=a x+b ,
• finding its zeros, which divide the domain of definition into intervals,
• determination of the signs that have the values ​​of the function on these intervals, on the basis of which a conclusion is made about the solution of a linear inequality.

Let's collect these moments in algorithm, revealing how to solve linear inequalities a x+b<0 (≤, >, ≥) at a≠0 by the interval method:

• The zeros of the function y=a x+b are found, for which a x+b=0 is solved. As you know, for a≠0 it has a single root, which we denote x 0 .
• It is built, and a point with coordinate x 0 is depicted on it. Moreover, if a strict inequality is solved (with the sign< или >), then this point is made punctured (with an empty center), and if it is not strict (with the sign ≤ or ≥), then a regular point is put. This point divides the coordinate line into two intervals (−∞, x 0) and (x 0 , +∞) .
• The signs of the function y=a·x+b on these intervals are determined. To do this, the value of this function is calculated at any point of the interval (−∞, x 0) , and the sign of this value will be the desired sign on the interval (−∞, x 0) . Similarly, the sign on the interval (x 0 , +∞) coincides with the sign of the value of the function y=a·x+b at any point of this interval. But you can do without these calculations, and draw conclusions about the signs from the value of the coefficient a: if a>0, then on the intervals (−∞, x 0) and (x 0, +∞) there will be signs - and +, respectively, and if a >0 , then + and −.
• If an inequality with > or ≥ signs is solved, then hatching is placed over the gap with a plus sign, and if inequalities with signs are solved< или ≤, то – со знаком минус. В результате получается , которое и является искомым решением линейного неравенства.

Consider an example of solving a linear inequality by the interval method.

Example.

Solve the inequality −3 x+12>0 .

Decision.

As soon as we analyze the method of intervals, then we will use it. According to the algorithm, first we find the root of the equation −3 x+12=0 , −3 x=−12 , x=4 . Next, we depict the coordinate line and mark on it a point with coordinate 4, and we make this point punched out, since we solve a strict inequality:

Now we define the signs on the intervals. To determine the sign on the interval (−∞, 4), you can calculate the value of the function y=−3 x+12 , for example, for x=3 . We have −3 3+12=3>0 , which means that the + sign is on this interval. To determine the sign on another interval (4, +∞), you can calculate the value of the function y=−3 x+12 , for example, at the point x=5 . We have −3 5+12=−3<0 , значит, на этом промежутке знак −. Эти же выводы можно было сделать на основании значения коэффициента при x : так как он равен −3 , то есть, он отрицательный, то на промежутке (−∞, 4) будет знак +, а на промежутке (4, +∞) знак −. Проставляем определенные знаки над соответствующими промежутками:

Since we are solving the inequality with the > sign, we draw a hatch over the gap with the + sign, the drawing takes the form

Based on the resulting image, we conclude that the desired solution is (−∞, 4) or in another notation x<4 .

Answer:

(−∞, 4) or x<4 .

Graphically

It is useful to have an idea of ​​the geometric interpretation of solving linear inequalities in one variable. To get it, let's consider four linear inequalities with the same left side: 0.5 x−1<0 , 0,5·x−1≤0 , 0,5·x−1>0 and 0.5 x−1≥0 , their solutions are respectively x<2 , x≤2 , x>2 and x≥2 , and also draw a graph of a linear function y=0.5 x−1 .

It is easy to see that

• solution of the inequality 0.5 x−1<0 представляет собой промежуток, на котором график функции y=0,5·x−1 располагается ниже оси абсцисс (эта часть графика изображена синим цветом),
• the solution to the inequality 0.5 x−1≤0 is the interval on which the graph of the function y=0.5 x−1 is below the Ox axis or coincides with it (in other words, not above the abscissa axis),
• similarly, the solution to the inequality 0.5 x−1>0 is the interval on which the graph of the function is above the Ox axis (this part of the graph is shown in red),
• and the solution to the inequality 0.5 x−1≥0 is the interval on which the graph of the function is higher or coincides with the x-axis.

Graphical way to solve inequalities, in particular linear ones, and implies finding the intervals on which the graph of the function corresponding to the left side of the inequality is located above, below, not lower or not higher than the graph of the function corresponding to the right side of the inequality. In our case of linear inequality, the function corresponding to the left side is y=a x+b , and the right side is y=0 , coinciding with the Ox axis.

Given the above information, it is easy to formulate algorithm for solving linear inequalities graphically:

• A graph of the function y=a x+b is constructed (you can schematically) and
• when solving the inequality a x+b<0 определяется промежуток, на котором график ниже оси Ox ,
• when solving the inequality a x+b≤0, the interval is determined on which the graph is lower or coincides with the axis Ox ,
• when solving the inequality a x+b>0, the interval is determined on which the graph is above the Ox axis,
• when solving the inequality a x+b≥0, the interval is determined on which the graph is higher or coincides with the axis Ox .

Example.

Solve the inequality graphically.

Decision.

Let's build a sketch of a graph of a linear function . This is a straight line that decreases since the coefficient at x is negative. We also need the coordinate of the point of its intersection with the abscissa axis, it is the root of the equation , which is equal to . For our purposes, we don't even need to draw the Oy axis. So our schematic drawing will look like this

Since we are solving the inequality with the > sign, we are interested in the interval on which the graph of the function is above the Ox axis. For clarity, we will highlight this part of the graph in red, and in order to easily determine the interval corresponding to this part, we will highlight in red the part of the coordinate plane in which the selected part of the graph is located, as in the figure below:

The interval of interest to us is a part of the Ox axis, which turned out to be highlighted in red. Obviously this is an open number beam . This is the desired solution. Note that if we were solving the inequality not with the > sign, but with the non-strict inequality sign ≥, then we would have to add in the answer, since at this point the graph of the function coincides with the Ox axis .y=0·x+7 , which is the same as y=7 , defines a straight line on the coordinate plane parallel to the Ox axis and lying above it. Therefore, the inequality 0 x+7<=0 не имеет решений, так как нет промежутков, на которых график функции y=0·x+7 ниже оси абсцисс.

And the graph of the function y=0 x+0 , which is the same as y=0 , is a straight line coinciding with the axis Ox . Therefore, the solution to the inequality 0 x+0≥0 is the set of all real numbers.

Answer:

the second inequality, its solution is any real number.

Linear inequalities

A huge number of inequalities with the help of equivalent transformations can be replaced by an equivalent linear inequality, in other words, reduced to a linear inequality. Such inequalities are called inequalities reducing to linear.

At school, almost simultaneously with the solution of linear inequalities, they also consider simple inequalities that reduce to linear ones. They are special cases. integer inequalities, namely, in their left and right parts there are integer expressions that represent or linear binomials, or are converted to them by and . For clarity, we give several examples of such inequalities: 5−2 x>0 , 7 (x−1)+3≤4 x−2+x , .

Inequalities that are similar in form to those indicated above can always be reduced to linear ones. This can be done by opening brackets, bringing like terms, rearranging terms and moving terms from one part of the inequality to another with the opposite sign.

For example, to reduce the inequality 5−2 x>0 to a linear one, it suffices to rearrange the terms on its left side, we have −2 x+5>0 . To reduce the second inequality 7 (x−1)+3≤4 x−2+x to a linear one, we need a little more work: on the left side we open the brackets 7 x−7+3≤4 x−2+x , after then we bring like terms in both parts 7 x−4≤5 x−2 , then we transfer the terms from the right side to the left 7 x−4−5 x+2≤0 , finally we give like terms on the left side 2 ·x−2≤0 . Similarly, the third inequality can be reduced to a linear inequality.

Because such inequalities can always be reduced to linear ones, some authors even call them linear as well. However, we will consider them to be linear.

Now it becomes clear why such inequalities are considered together with linear inequalities. And the principle of their solution is absolutely the same: by performing equivalent transformations, they can be reduced to elementary inequalities, which are the desired solutions.

To solve an inequality of this kind, you can first reduce it to a linear one, and then solve this linear inequality. But it is more rational and more convenient to do this:

• after opening the brackets, collect all the terms with the variable on the left side of the inequality, and all the numbers on the right,
• and then add like terms,
• and then, divide both parts of the obtained inequality by the coefficient at x (if, of course, it is different from zero). This will give the answer.

Example.

Solve the inequality 5 (x+3)+x≤6 (x−3)+1 .

Decision.

First, we open the brackets, as a result we arrive at the inequality 5 x+15+x≤6 x−18+1 . Now we present similar terms: 6 x+15≤6 x−17 . Then we transfer the terms from the left side, we get 6 x+15−6 x+17≤0 , and again bring similar terms (which leads us to the linear inequality 0 x+32≤0 ) and we have 32≤0 . So we came to an incorrect numerical inequality, from which we conclude that the original inequality has no solutions.

Answer:

there are no solutions.

In conclusion, we note that there are many other inequalities that reduce to linear inequalities, or to inequalities of the form considered above. For example, the solution exponential inequality 5 2 x−1 ≥1 reduces to solving the linear inequality 2 x−1≥0 . But we will talk about this when we analyze solutions of inequalities of the corresponding form.

Bibliography.

• Algebra: textbook for 8 cells. general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2008. - 271 p. : ill. - ISBN 978-5-09-019243-9.
• Algebra: Grade 9: textbook. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; ed. S. A. Telyakovsky. - 16th ed. - M. : Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
• Mordkovich A. G. Algebra. 8th grade. At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich. - 11th ed., erased. - M.: Mnemozina, 2009. - 215 p.: ill. ISBN 978-5-346-01155-2.
• Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., Sr. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
• Mordkovich A. G. Algebra and beginning of mathematical analysis. Grade 11. At 2 pm Part 1. Textbook for students of educational institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.

First, some lyrics to get a feel for the problem that the interval method solves. Suppose we need to solve the following inequality:

(x − 5)(x + 3) > 0

What are the options? The first thing that comes to mind for most students is the rules "plus times plus makes plus" and "minus times minus makes plus." Therefore, it suffices to consider the case when both brackets are positive: x − 5 > 0 and x + 3 > 0. Then we also consider the case when both brackets are negative: x − 5< 0 и x + 3 < 0. Таким образом, наше неравенство свелось к совокупности двух систем, которая, впрочем, легко решается:

More advanced students will remember (perhaps) that on the left is a quadratic function whose graph is a parabola. Moreover, this parabola intersects the OX axis at the points x = 5 and x = −3. For further work, you need to open the brackets. We have:

x 2 − 2x − 15 > 0

Now it is clear that the branches of the parabola are directed upwards, because coefficient a = 1 > 0. Let's try to draw a diagram of this parabola:

The function is greater than zero where it passes above the OX axis. In our case, these are the intervals (−∞ −3) and (5; +∞) - this is the answer.

Please note that the picture shows exactly function diagram, not her schedule. Because for a real graph, you need to calculate coordinates, calculate offsets and other crap, which we don’t need at all now.

Why are these methods ineffective?

So, we have considered two solutions to the same inequality. Both of them turned out to be very cumbersome. The first decision arises - just think about it! is a set of systems of inequalities. The second solution is also not very easy: you need to remember the parabola graph and a bunch of other small facts.

It was a very simple inequality. It has only 2 multipliers. Now imagine that there will be not 2 multipliers, but at least 4. For example:

(x − 7)(x − 1)(x + 4)(x + 9)< 0

How to solve such inequality? Go through all possible combinations of pros and cons? Yes, we will fall asleep faster than we find a solution. Drawing a graph is also not an option, since it is not clear how such a function behaves on the coordinate plane.

For such inequalities, a special solution algorithm is needed, which we will consider today.

What is the interval method

The interval method is a special algorithm designed to solve complex inequalities of the form f (x) > 0 and f (x)< 0. Алгоритм состоит из 4 шагов:

1. Solve the equation f (x) \u003d 0. Thus, instead of an inequality, we get an equation that is much easier to solve;
2. Mark all the obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find out the sign (plus or minus) of the function f (x) on the rightmost interval. To do this, it is enough to substitute in f (x) any number that will be to the right of all the marked roots;
4. Mark marks on other intervals. To do this, it is enough to remember that when passing through each root, the sign changes.

That's all! After that, it remains only to write out the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f (x) > 0, or a “−” sign if the inequality was of the form f (x)< 0.

At first glance, it may seem that the interval method is some kind of tin. But in practice, everything will be very simple. It takes a little practice - and everything will become clear. Take a look at the examples and see for yourself:

Task. Solve the inequality:

(x − 2)(x + 7)< 0

We work on the method of intervals. Step 1: Replace the inequality with an equation and solve it:

(x − 2)(x + 7) = 0

The product is equal to zero if and only if at least one of the factors is equal to zero:

x − 2 = 0 ⇒ x = 2;
x + 7 = 0 ⇒ x = −7.

Got two roots. Go to step 2: mark these roots on the coordinate line. We have:

Now step 3: we find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that is greater than the number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10, and even x = 10,000). We get:

f(x) = (x − 2)(x + 7);
x=3;
f (3) = (3 − 2)(3 + 7) = 1 10 = 10;

We get that f (3) = 10 > 0, so we put a plus sign in the rightmost interval.

We pass to the last point - it is necessary to note the signs on the remaining intervals. Remember that when passing through each root, the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus on the left.

This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, there is a plus to the left of the root x = −7. It remains to mark these signs on the coordinate axis. We have:

Let's return to the original inequality, which looked like:

(x − 2)(x + 7)< 0

So the function must be less than zero. This means that we are interested in the minus sign, which occurs only on one interval: (−7; 2). This will be the answer.

Task. Solve the inequality:

(x + 9)(x − 3)(1 − x )< 0

Step 1: Equate the left side to zero:

(x + 9)(x − 3)(1 − x ) = 0;
x + 9 = 0 ⇒ x = −9;
x − 3 = 0 ⇒ x = 3;
1 − x = 0 ⇒ x = 1.

Remember: the product is zero when at least one of the factors is zero. That is why we have the right to equate to zero each individual bracket.

Step 2: mark all the roots on the coordinate line:

Step 3: find out the sign of the rightmost gap. We take any number that is greater than x = 1. For example, we can take x = 10. We have:

f (x) \u003d (x + 9) (x - 3) (1 - x);
x=10;
f (10) = (10 + 9)(10 − 3)(1 − 10) = 19 7 (−9) = − 1197;
f(10) = -1197< 0.

Step 4: Place the rest of the signs. Remember that when passing through each root, the sign changes. As a result, our picture will look like this:

That's all. It remains only to write the answer. Take another look at the original inequality:

(x + 9)(x − 3)(1 − x )< 0

This is an inequality of the form f (x)< 0, т.е. нас интересуют интервалы, отмеченные знаком минус. А именно:

x ∈ (−9; 1) ∪ (3; +∞)

This is the answer.

A note about function signs

Practice shows that the greatest difficulties in the interval method arise at the last two steps, i.e. when placing signs. Many students begin to get confused: what numbers to take and where to put signs.

To finally understand the interval method, consider two remarks on which it is built:

1. A continuous function changes sign only at the points where it is equal to zero. Such points break the coordinate axis into pieces, within which the sign of the function never changes. That's why we solve the equation f (x) \u003d 0 and mark the found roots on a straight line. The numbers found are the "boundary" points separating the pluses from the minuses.
2. To find out the sign of a function on any interval, it is enough to substitute any number from this interval into the function. For example, for the interval (−5; 6) we can take x = −4, x = 0, x = 4 and even x = 1.29374 if we want. Why is it important? Yes, because many students begin to gnaw doubts. Like, what if for x = −4 we get a plus, and for x = 0 we get a minus? Nothing like that will ever happen. All points in the same interval give the same sign. Remember this.

That's all you need to know about the interval method. Of course, we have dismantled it in its simplest form. There are more complex inequalities - non-strict, fractional and with repeated roots. For them, you can also apply the interval method, but this is a topic for a separate large lesson.

Now I would like to analyze an advanced trick that drastically simplifies the interval method. More precisely, the simplification affects only the third step - the calculation of the sign on the rightmost piece of the line. For some reason, this technique is not held in schools (at least no one explained this to me). But in vain - in fact, this algorithm is very simple.

So, the sign of the function is on the right piece of the numerical axis. This piece has the form (a; +∞), where a is the largest root of the equation f (x) = 0. In order not to blow our brains, consider a specific example:

(x − 1)(2 + x )(7 − x )< 0;
f (x) \u003d (x - 1) (2 + x) (7 - x);
(x − 1)(2 + x )(7 − x ) = 0;
x − 1 = 0 ⇒ x = 1;
2 + x = 0 ⇒ x = −2;
7 − x = 0 ⇒ x = 7;

We got 3 roots. We list them in ascending order: x = −2, x = 1 and x = 7. Obviously, the largest root is x = 7.

For those who find it easier to reason graphically, I will mark these roots on the coordinate line. Let's see what happens:

It is required to find the sign of the function f (x) on the rightmost interval, i.e. on (7; +∞). But as we have already noted, to determine the sign, you can take any number from this interval. For example, you can take x = 8, x = 150, etc. And now - the same technique that is not taught in schools: let's take infinity as a number. More precisely, plus infinity, i.e. +∞.

"Are you stoned? How can you substitute infinity into a function? perhaps, you ask. But think about it: we do not need the value of the function itself, we only need the sign. Therefore, for example, the values ​​f (x) = −1 and f (x) = −938 740 576 215 mean the same thing: the function is negative on this interval. Therefore, all that is required of you is to find the sign that occurs at infinity, and not the value of the function.

In fact, substituting infinity is very simple. Let's go back to our function:

f(x) = (x − 1)(2 + x)(7 − x)

Imagine that x is a very large number. A billion or even a trillion. Now let's see what happens in each parenthesis.

First bracket: (x − 1). What happens if you subtract one from a billion? The result will be a number not much different from a billion, and this number will be positive. Similarly with the second bracket: (2 + x). If we add a billion to two, we get a billion with kopecks - this is a positive number. Finally, the third bracket: (7 − x ). Here there will be minus a billion, from which a miserable piece in the form of a seven has been “gnawed off”. Those. the resulting number will not differ much from minus a billion - it will be negative.

It remains to find the sign of the whole work. Since we had a plus in the first brackets, and a minus in the last bracket, we get the following construction:

(+) · (+) · (−) = (−)

The final sign is minus! It doesn't matter what the value of the function itself is. The main thing is that this value is negative, i.e. on the rightmost interval there is a minus sign. It remains to complete the fourth step of the interval method: arrange all the signs. We have:

The original inequality looked like:

(x − 1)(2 + x )(7 − x )< 0

Therefore, we are interested in the intervals marked with a minus sign. We write out the answer:

x ∈ (−2; 1) ∪ (7; +∞)

That's the whole trick that I wanted to tell. In conclusion, there is one more inequality, which is solved by the interval method using infinity. To visually shorten the solution, I will not write step numbers and detailed comments. I will write only what really needs to be written when solving real problems:

Task. Solve the inequality:

x (2x + 8)(x − 3) > 0

We replace the inequality with an equation and solve it:

x (2x + 8)(x − 3) = 0;
x = 0;
2x + 8 = 0 ⇒ x = −4;
x − 3 = 0 ⇒ x = 3.

We mark all three roots on the coordinate line (immediately with signs):

There is a plus on the right side of the coordinate axis, because the function looks like:

f(x) = x(2x + 8)(x − 3)

And if we substitute infinity (for example, a billion), we get three positive brackets. Since the original expression must be greater than zero, we are only interested in pluses. It remains to write the answer:

x ∈ (−4; 0) ∪ (3; +∞)

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

What "square inequality"? Not a question!) If you take any quadratic equation and change the sign in it "=" (equal) to any inequality icon ( > ≥ < ≤ ≠ ), we get a quadratic inequality. For example:

1. x2 -8x+12 ≥ 0

2. -x 2 +3x > 0

3. x2 ≤ 4

Well, you get the idea...)

I knowingly linked equations and inequalities here. The fact is that the first step in solving any square inequality - solve the equation from which this inequality is made. For this reason - the inability to solve quadratic equations automatically leads to a complete failure in inequalities. Is the hint clear?) If anything, look at how to solve any quadratic equations. Everything is detailed there. And in this lesson we will deal with inequalities.

The inequality ready for solution has the form: left - square trinomial ax 2 +bx+c, on the right - zero. The inequality sign can be absolutely anything. The first two examples are here are ready for a decision. The third example still needs to be prepared.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

After receiving the initial information about inequalities with variables, we turn to the question of their solution. Let's analyze the solution of linear inequalities with one variable and all methods for their resolution with algorithms and examples. Only linear equations with one variable will be considered.

Yandex.RTB R-A-339285-1

What is a linear inequality?

First you need to define a linear equation and find out its standard form and how it will differ from others. From the school course we have that inequalities do not have a fundamental difference, so several definitions must be used.

Definition 1

Linear inequality with one variable x is an inequality of the form a x + b > 0 when any inequality sign is used instead of >< , ≤ , ≥ , а и b являются действительными числами, где a ≠ 0 .

Definition 2

Inequalities a x< c или a · x >c , with x being a variable and a and c some numbers, is called linear inequalities with one variable.

Since nothing is said about whether the coefficient can be equal to 0 , then a strict inequality of the form 0 x > c and 0 x< c может быть записано в виде нестрогого, а именно, a · x ≤ c , a · x ≥ c . Такое уравнение считается линейным.

Their differences are:

• notation a · x + b > 0 in the first, and a · x > c – in the second;
• admissibility of zero coefficient a , a ≠ 0 - in the first, and a = 0 - in the second.

It is believed that the inequalities a x + b > 0 and a x > c are equivalent, because they are obtained by transferring the term from one part to another. Solving the inequality 0 · x + 5 > 0 will lead to the fact that it will need to be solved, and the case a = 0 will not work.

Definition 3

It is considered that linear inequalities in one variable x are inequalities of the form a x + b< 0 , a · x + b >0 , a x + b ≤ 0 and a x + b ≥ 0, where a and b are real numbers. Instead of x, there can be an ordinary number.

Based on the rule, we have that 4 x − 1 > 0 , 0 z + 2 , 3 ≤ 0 , - 2 3 x - 2< 0 являются примерами линейных неравенств. А неравенства такого плана, как 5 · x >7 , − 0 , 5 · y ≤ − 1 , 2 are called linear.

How to solve a linear inequality

The main way to solve such inequalities is to use equivalent transformations to find the elementary inequalities x< p (≤ , >, ≥) , p being some number, for a ≠ 0 , and of the form a< p (≤ , >, ≥) for a = 0 .

To solve an inequality with one variable, you can apply the interval method or represent it graphically. Any of them can be used in isolation.

Using equivalent transformations

To solve a linear inequality of the form a x + b< 0 (≤ , >, ≥) , it is necessary to apply equivalent transformations of the inequality. The coefficient may or may not be zero. Let's consider both cases. To clarify, it is necessary to adhere to a scheme consisting of 3 points: the essence of the process, the algorithm, the solution itself.

Definition 4

Algorithm for solving a linear inequality a x + b< 0 (≤ , >, ≥) for a ≠ 0

• the number b will be transferred to the right side of the inequality with the opposite sign, which will allow us to come to the equivalent a x< − b (≤ , > , ≥) ;
• both parts of the inequality will be divided by a number not equal to 0. Moreover, when a is positive, the sign remains, when a is negative, it changes to the opposite.

Consider the application of this algorithm to solving examples.

Example 1

Solve an inequality of the form 3 · x + 12 ≤ 0 .

Decision

This linear inequality has a = 3 and b = 12 . Hence, the coefficient a of x is not equal to zero. Let's apply the above algorithms and solve.

It is necessary to transfer the term 12 to another part of the inequality with a sign change in front of it. Then we obtain an inequality of the form 3 · x ≤ − 12 . It is necessary to divide both parts by 3. The sign will not change because 3 is a positive number. We get that (3 x) : 3 ≤ (− 12) : 3 , which will give the result x ≤ − 4 .

An inequality of the form x ≤ − 4 is equivalent. That is, the solution for 3 x + 12 ≤ 0 is any real number that is less than or equal to 4 . The answer is written as an inequality x ≤ − 4 , or a numerical interval of the form (− ∞ , − 4 ] .

The entire algorithm described above is written as follows:

3 x + 12 ≤ 0; 3 x ≤ − 12 ; x ≤ − 4 .

Answer: x ≤ − 4 or (− ∞ , − 4 ] .

Example 2

Indicate all available solutions of the inequality − 2 , 7 · z > 0 .

Decision

From the condition we see that the coefficient a at z is equal to - 2, 7, and b is explicitly absent or equal to zero. You can not use the first step of the algorithm, but immediately go to the second.

We divide both parts of the equation by the number - 2, 7. Since the number is negative, it is necessary to change the inequality sign to the opposite. That is, we get that (− 2 , 7 z) : (− 2 , 7)< 0: (− 2 , 7) , и дальше z < 0 .

We write the whole algorithm in a short form:

− 2 , 7 z > 0 ; z< 0 .

Answer: z< 0 или (− ∞ , 0) .

Example 3

Solve the inequality - 5 · x - 15 22 ≤ 0 .

Decision

According to the condition, we see that it is necessary to solve the inequality with the coefficient a for the variable x, which is equal to - 5, with the coefficient b, which corresponds to the fraction - 15 22 . It is necessary to solve the inequality following the algorithm, that is: transfer - 15 22 to another part with the opposite sign, divide both parts by - 5, change the inequality sign:

5 x ≤ 15 22 ; - 5 x: - 5 ≥ 15 22: - 5 x ≥ - 3 22

At the last transition, for the right side, the rule for dividing a number with different signs is used 15 22: - 5 \u003d - 15 22: 5, after which we divide the ordinary fraction by a natural number - 15 22: 5 \u003d - 15 22 1 5 \u003d - 15 1 22 5 = - 3 22 .

Answer: x ≥ - 3 22 and [ - 3 22 + ∞) .

Consider the case when a = 0. Linear expression of the form a x + b< 0 является неравенством 0 · x + b < 0 , где на рассмотрение берется неравенство вида b < 0 , после чего выясняется, оно верное или нет.

Everything is based on the definition of the solution of the inequality. For any value of x, we obtain a numerical inequality of the form b< 0 , потому что при подстановке любого t вместо переменной x , тогда получаем 0 · t + b < 0 , где b < 0 . В случае, если оно верно, то для его решения подходит любое значение. Когда b < 0 неверно, тогда линейное уравнение не имеет решений, потому как не имеется ни одного значения переменной, которое привело бы верному числовому равенству.

We consider all judgments in the form of an algorithm for solving linear inequalities 0 x + b< 0 (≤ , > , ≥) :

Definition 5

Numerical inequality of the form b< 0 (≤ , >, ≥) is true, then the original inequality has a solution for any value, and false when the original inequality has no solutions.

Example 4

Solve the inequality 0 · x + 7 > 0 .

Decision

This linear inequality 0 · x + 7 > 0 can take any value x . Then we get an inequality of the form 7 > 0 . The last inequality is considered true, so any number can be its solution.

Answer: interval (− ∞ , + ∞) .

Example 5

Find a solution to the inequality 0 · x − 12 , 7 ≥ 0 .

Decision

Substituting the variable x for any number, we get that the inequality will take the form − 12 , 7 ≥ 0 . It is incorrect. That is, 0 · x − 12 , 7 ≥ 0 has no solutions.

Answer: there are no solutions.

Consider the solution of linear inequalities, where both coefficients are equal to zero.

Example 6

Determine an unsolvable inequality from 0 · x + 0 > 0 and 0 · x + 0 ≥ 0 .

Decision

When substituting any number instead of x, we get two inequalities of the form 0 > 0 and 0 ≥ 0 . The first is incorrect. This means that 0 x + 0 > 0 has no solutions, and 0 x + 0 ≥ 0 has an infinite number of solutions, that is, any number.

Answer: the inequality 0 x + 0 > 0 has no solutions, and 0 x + 0 ≥ 0 has solutions.

This method is considered in the school course of mathematics. The interval method is capable of resolving various kinds of inequalities, including linear ones.

The interval method is used for linear inequalities when the value of the coefficient x is not equal to 0 . Otherwise, you will have to calculate using another method.

Definition 6

The spacing method is:

• introduction of the function y = a x + b ;
• search for zeros to split the domain of definition into intervals;
• determination of signs for the concept of them on intervals.

Let's assemble an algorithm for solving linear equations a x + b< 0 (≤ , >, ≥) for a ≠ 0 using the interval method:

• finding the zeros of the function y = a · x + b to solve an equation of the form a · x + b = 0 . If a ≠ 0, then the solution will be the only root that will take the designation x 0;
• construction of a coordinate line with the image of a point with a coordinate x 0, with a strict inequality, the point is denoted by a punched out, with a non-strict inequality, it is shaded;
• determination of the signs of the function y = a x + b on the intervals, for this it is necessary to find the values ​​of the function at points on the interval;
• the solution of the inequality with the > or ≥ signs on the coordinate line, hatching is added above the positive gap,< или ≤ над отрицательным промежутком.

Consider several examples of solving a linear inequality using the interval method.

Example 6

Solve the inequality − 3 · x + 12 > 0 .

Decision

It follows from the algorithm that first you need to find the root of the equation − 3 · x + 12 = 0 . We get that − 3 · x = − 12 , x = 4 . It is necessary to depict the coordinate line, where we mark the point 4. It will be punctured since the inequality is strict. Consider the drawing below.

It is necessary to determine the signs on the intervals. To determine it on the interval (− ∞ , 4) , it is necessary to calculate the function y = − 3 · x + 12 for x = 3 . From here we get that − 3 3 + 12 = 3 > 0 . The sign on the interval is positive.

We determine the sign from the interval (4, + ∞), then we substitute the value x \u003d 5. We have − 3 5 + 12 = − 3< 0 . Знак на промежутке является отрицательным. Изобразим на числовой прямой, приведенной ниже.

We perform the solution of the inequality with the sign > , and the hatching is performed over the positive gap. Consider the drawing below.

It can be seen from the drawing that the desired solution has the form (− ∞ , 4) or x< 4 .

Answer: (− ∞ , 4) or x< 4 .

To understand how to represent graphically, it is necessary to consider 4 linear inequalities as an example: 0, 5 x − 1< 0 , 0 , 5 · x − 1 ≤ 0 , 0 , 5 · x − 1 >0 and 0 , 5 x − 1 ≥ 0 . Their solutions will be x< 2 , x ≤ 2 , x >2 and x ≥ 2 . To do this, draw a graph of the linear function y = 0 , 5 · x − 1 below.

It's clear that

Definition 7

• solution of the inequality 0 , 5 x − 1< 0 считается промежуток, где график функции y = 0 , 5 · x − 1 располагается ниже О х;
• the solution 0 , 5 x − 1 ≤ 0 is the interval where the function y = 0 , 5 x − 1 is below 0 x or coincides;
• the solution 0 , 5 x − 1 > 0 is considered to be the interval, where the function is located above O x;
• the solution 0 , 5 x − 1 ≥ 0 is the interval where the graph is higher than O x or coincides.

The meaning of the graphical solution of inequalities is to find the gaps, which must be depicted on the graph. In this case, we get that the left side has y \u003d a x + b, and the right side has y \u003d 0, and it coincides with About x.

Definition 8

The plotting of the function y = a x + b is performed:

• while solving the inequality a x + b< 0 определяется промежуток, где график изображен ниже О х;
• while solving the inequality a x + b ≤ 0, the interval is determined where the graph is displayed below the O x axis or coincides;
• while solving the inequality a x + b > 0, the interval is determined, where the graph is displayed above O x;
• while solving the inequality a x + b ≥ 0, the interval is determined where the graph is above O x or coincides.

Example 7

Solve the inequality - 5 · x - 3 > 0 using the graph.

Decision

It is necessary to build a graph of a linear function - 5 · x - 3 > 0 . This line is decreasing because the coefficient of x is negative. To determine the coordinates of the point of its intersection with O x - 5 · x - 3 > 0, we obtain the value - 3 5 . Let's graph it.

The solution of the inequality with the sign >, then you need to pay attention to the interval above O x. We highlight the necessary part of the plane in red and get that

The required gap is the O x part of the red color. Hence, the open number ray - ∞ , - 3 5 will be the solution of the inequality. If, by condition, they had a non-strict inequality, then the value of the point - 3 5 would also be a solution to the inequality. And would coincide with O x.

Answer: - ∞ , - 3 5 or x< - 3 5 .

The graphical solution is used when the left side will correspond to the function y = 0 x + b , that is, y = b . Then the line will be parallel to O x or coinciding at b \u003d 0. These cases show that an inequality may have no solutions, or any number can be a solution.

Example 8

Determine from inequalities 0 x + 7< = 0 , 0 · x + 0 ≥ 0 то, которое имеет хотя бы одно решение.

Decision

The representation y = 0 x + 7 is y = 7 , then a coordinate plane with a straight line parallel to O x and above O x will be given. So 0 x + 7< = 0 решений не имеет, потому как нет промежутков.

The graph of the function y \u003d 0 x + 0 is considered y \u003d 0, that is, the line coincides with O x. Hence, the inequality 0 · x + 0 ≥ 0 has many solutions.

Answer: the second inequality has a solution for any value of x .

Linear inequalities

The solution of inequalities can be reduced to the solution of a linear equation, which are called linear inequalities.

These inequalities were considered in the school course, since they were a special case of solving inequalities, which led to the opening of brackets and the reduction of similar terms. For example, consider that 5 − 2 x > 0 , 7 (x − 1) + 3 ≤ 4 x − 2 + x , x - 3 5 - 2 x + 1 > 2 7 x .

The inequalities given above are always reduced to the form of a linear equation. After that, the brackets are opened and similar terms are given, transferred from different parts, changing the sign to the opposite.

When reducing the inequality 5 − 2 x > 0 to a linear one, we represent it in such a way that it has the form − 2 x + 5 > 0 , and to reduce the second we get that 7 (x − 1) + 3 ≤ 4 x − 2 + x . It is necessary to open the brackets, bring like terms, move all terms to the left side and bring like terms. It looks like this:

7 x − 7 + 3 ≤ 4 x − 2 + x 7 x − 4 ≤ ​​5 x − 2 7 x − 4 − 5 x + 2 ≤ 0 2 x − 2 ≤ 0

This brings the solution to a linear inequality.

These inequalities are considered as linear, since they have the same principle of solution, after which it is possible to reduce them to elementary inequalities.

To solve this kind of inequality of this kind, it is necessary to reduce it to a linear one. It should be done like this:

Definition 9

• open brackets;
• collect variables on the left, and numbers on the right;
• bring like terms;
• divide both parts by the coefficient of x .

Example 9

Solve the inequality 5 · (x + 3) + x ≤ 6 · (x − 3) + 1 .

Decision

We expand the brackets, then we get an inequality of the form 5 · x + 15 + x ≤ 6 · x − 18 + 1 . After reducing similar terms, we have that 6 · x + 15 ≤ 6 · x − 17 . After moving the terms from the left to the right, we get that 6 x + 15 − 6 x + 17 ≤ 0 . Hence, it has an inequality of the form 32 ≤ 0 from the result obtained in the calculation 0 · x + 32 ≤ 0 . It can be seen that the inequality is false, which means that the inequality given by the condition has no solutions.

Answer: no solutions.

It is worth noting that there are many inequalities of another kind, which can be reduced to a linear one or an inequality of the kind shown above. For example, 5 2 x − 1 ≥ 1 is an exponential equation that reduces to a linear solution 2 · x − 1 ≥ 0 . These cases will be considered when solving inequalities of this type.

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