Rules for solving inequalities with one variable. Solving inequalities with one variable – Knowledge Hypermarket

Now you can understand how linear inequalities a x + b are solved<0 (они могут быть записаны и с помощью любого другого знака неравенства).

The main way to solve them is to use equivalent transformations that allow one to arrive at a≠0 to elementary inequalities type x

, ≥), p - a certain number, which are the desired solution, and for a=0 - to numerical inequalities of the form a

, ≥), from which a conclusion is drawn about the solution of the original inequality. We will analyze it first.

It also doesn’t hurt to look at solving linear inequalities in one variable from other perspectives. Therefore, we will also show how linear inequality can be solved graphically and using the interval method.

Using equivalent transformations

Let us need to solve the linear inequality a x+b<0 (≤, >, ≥). Let's show how to do this using equivalent inequality transformations.

The approaches differ depending on whether the coefficient a of the variable x is equal or not equal to zero. Let's look at them one by one. Moreover, when considering, we will adhere to a three-point scheme: first we will give the essence of the process, then we will give an algorithm for solving a linear inequality, and finally, we will give solutions to typical examples.

Let's start with algorithm for solving linear inequality a x+b<0 (≤, >, ≥) for a≠0.

Firstly, the number b is transferred to the right side of the inequality with the opposite sign. This allows us to pass to the equivalent inequality a x<−b (≤, >, ≥).
Secondly, both sides of the resulting inequality are divided by a non-zero number a. Moreover, if a is a positive number, then the inequality sign is preserved, and if a is a negative number, then the inequality sign is reversed. The result is an elementary inequality equivalent to the original linear inequality, and this is the answer.

It remains to understand the application of the announced algorithm using examples. Let's consider how it can be used to solve linear inequalities for a≠0.

Example.

Solve the inequality 3·x+12≤0.

Solution.

For a given linear inequality we have a=3 and b=12. Obviously, the coefficient a for the variable x is different from zero. Let's use the corresponding solution algorithm given above.

First, we move the term 12 to the right side of the inequality, not forgetting to change its sign, that is, −12 will appear on the right side. As a result, we arrive at the equivalent inequality 3·x≤−12.

And, secondly, we divide both sides of the resulting inequality by 3, since 3 is a positive number, we do not change the sign of the inequality. We have (3 x):3≤(−12):3, which is the same as x≤−4.

The resulting elementary inequality x≤−4 is equivalent to the original linear inequality and is its desired solution.

So, the solution to the linear inequality 3 x + 12≤0 is any real number less than or equal to minus four. The answer can also be written in the form of a numerical interval corresponding to the inequality x≤−4, that is, as (−∞, −4] .

Having acquired skill in working with linear inequalities, their solutions can be written down briefly without explanation. In this case, first write down the original linear inequality, and below - equivalent inequalities obtained at each step of the solution:
3 x+12≤0 ;
3 x≤−12 ;
x≤−4 .

Answer:

x≤−4 or (−∞, −4] .

Example.

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

Solution.

Here the coefficient a for the variable z is equal to −2.7. And the coefficient b is absent in 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 moving a zero from the left side to the right will not change the form of the original inequality.

It remains to divide both sides of the inequality by −2.7, not forgetting to change the sign of the inequality to the opposite one, since −2.7 is a negative number. We have (−2.7 z):(−2.7)<0:(−2,7) , and then z<0 .

And now briefly:
−2.7·z>0 ;
z<0 .

Answer:

z<0 или (−∞, 0) .

Example.

Solve the inequality .

Solution.

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

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 this based on? Very simple: on determining the solution to the inequality. How? Yes, here’s how: no matter what value of the variable x we substitute into the original linear inequality, we will 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 arguments 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 understand this with examples.

Example.

Solve the inequality 0·x+7>0.

Solution.

For any value of the variable x, the linear inequality 0 x+7>0 will turn into the 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 0·x−12.7≥0 have solutions?

Solution.

If you substitute any number instead of the variable x, then the original inequality turns into a numerical inequality −12.7≥0, which is incorrect. This means that not a single number is a solution to the linear inequality 0·x−12.7≥0.

Answer:

no, it doesn't.

To conclude this section, we will analyze the solutions to 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?

Solution.

If you 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 of them is incorrect, and the second is correct. Consequently, 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 method of intervals is studied in a school algebra course later than the topic of solving linear inequalities in one variable. But the interval method allows you to solve a variety of inequalities, including linear ones. Therefore, let's dwell on it.

Let us immediately note that it is advisable to use the interval method to solve linear inequalities with a non-zero coefficient for the variable x. Otherwise, it is faster and more convenient to draw a conclusion about the solution of the inequality using the method discussed at the end of the previous paragraph.

The interval method implies

introducing 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 function values 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 (≤, >, ≥) for a≠0 using the interval method:

The zeros of the function y=a·x+b are found, for which a·x+b=0 is solved. As is known, for a≠0 it has a single root, which we denote as x 0 .
It is constructed, 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 punctuated (with an empty center), and if it is not strict (with a sign ≤ or ≥), then a regular point is placed. 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 in 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 in this interval. But you can do without these calculations, and draw conclusions about the signs based on 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 inequalities with signs > or ≥ are being solved, then a hatch is placed over the gap with a plus sign, and if inequalities with signs are being solved< или ≤, то – со знаком минус. В результате получается , которое и является искомым решением линейного неравенства.

Let's consider an example of solving a linear inequality using the interval method.

Example.

Solve the inequality −3·x+12>0.

Solution.

Since we are analyzing the interval method, 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 draw a coordinate line and mark a point on it with coordinate 4, and we make this point punctured, since we are solving a strict inequality:

Now we determine 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, at x=3. We have −3·3+12=3>0, which means there is a + sign 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 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 shading 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 understanding of the geometric interpretation of solving linear inequalities in one variable. To get it, let's consider four linear inequalities with the same left-hand 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 x<2 , x≤2 , x>2 and x≥2, and also draw a graph of the linear function y=0.5 x−1.

It's easy to notice that

solution to 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 represents the interval in 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 in 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 in which the graph of the function is higher or coincides with the abscissa axis.

Graphical method for solving inequalities, in particular linear, and implies finding intervals in which the graph of the function corresponding to the left side of the inequality is located above, below, not below or not above 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 information given, it is easy to formulate algorithm for solving linear inequalities graphically:

A graph of the function y=a x+b is constructed (schematically possible) 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 in which the graph is lower or coincides with the Ox axis,
when solving the inequality a x+b>0, the interval is determined in which the graph is above the Ox axis,
when solving the inequality a·x+b≥0, the interval in which the graph is higher or coincides with the Ox axis is determined.

Example.

Solve the inequality graphically.

Solution.

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

Since we are solving an inequality with a > sign, we are interested in the interval in which the graph of the function is above the Ox axis. For clarity, let’s highlight this part of the graph in red, and in order to easily determine the interval corresponding to this part, let’s 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 gap we are interested in is the part of the Ox axis that is highlighted in red. Obviously this is an open number beam . This is the solution we are looking for. Note that if we were solving the inequality not with the sign >, but with the sign of the non-strict inequality ≥, 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 Ox axis. Therefore, the solution to the inequality 0·x+0≥0 is the set of all real numbers.

Answer:

second inequality, its solution is any real number.

Inequalities that reduce to linear

A huge number of inequalities can be replaced by equivalent linear inequalities using equivalent transformations, in other words, reduced to a linear inequality. Such inequalities are called inequalities that reduce to linear.

At school, almost simultaneously with solving linear inequalities, simple inequalities that reduce to linear ones are also considered. They are special cases entire inequalities, namely in their left and right parts there are whole 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 parentheses, bringing similar terms, rearranging terms, and moving terms from one side of the inequality to another with the opposite sign.

For example, to reduce the inequality 5−2 x>0 to linear, it is enough 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 linear, you need a little more steps: on the left side we open the brackets 7·x−7+3≤4·x−2+x , after To do this, we present similar terms in both sides 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 present similar terms in the left side 2 ·x−2≤0 . Similarly, the third inequality can be reduced to a linear inequality.

Due to the fact that such inequalities can always be reduced to linear ones, some authors even call them linear as well. But we will still consider them reducible to 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 that represent the desired solutions.

To solve an inequality of this type, you can first reduce it to a linear one, and then solve this linear inequality. But it’s more rational and 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,
then bring similar terms,
and then divide both sides of the resulting inequality by the coefficient of x (if it is, of course, different from zero). This will give the answer.

Example.

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

Solution.

First, let's open the brackets, as a result we come to the inequality 5 x + 15 + x ≤ 6 x − 18 + 1 . Now let’s give similar terms: 6 x+15≤6 x−17 . Then we move the terms from the left side, we get 6 x+15−6 x+17≤0, and again we bring similar terms (which leads us to the linear inequality 0 x+32≤0) and we have 32≤0. This is how we came to an incorrect numerical inequality, from which we conclude that the original inequality has no solutions.

Answer:

no solutions.

In conclusion, we note that there are a lot of other inequalities that can be reduced to linear inequalities, or to inequalities of the type 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 analyzing solutions to inequalities of the corresponding type.

Bibliography.

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

Romanishina Dina Solomonovna, mathematics teacher at gymnasium No. 2 in Khabarovsk

1. Equations with one variable.

An equation containing a variable is called an equation with one variable, or an equation with one unknown. For example, an equation with one variable is 3(2x+7)=4x-1.

The root or solution of an equation is the value of a variable at which the equation becomes a true numerical equality. For example, the number 1 is a solution to the equation 2x+5=8x-1. The equation x2+1=0 has no solution, because the left side of the equation is always greater than zero. The equation (x+3)(x-4) =0 has two roots: x1= -3, x2=4.

Solving an equation means finding all its roots or proving that there are no roots.

Equations are called equivalent if all the roots of the first equation are roots of the second equation and vice versa, all the roots of the second equation are roots of the first equation or if both equations have no roots. For example, the equations x-8=2 and x+10=20 are equivalent, because the root of the first equation x=10 is also the root of the second equation, and both equations have the same root.

When solving equations, the following properties are used:

If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one.

If both sides of an equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given one.

The equation ax=b, where x is a variable and a and b are some numbers, is called a linear equation with one variable.

If a¹0, then the equation has a unique solution

If a=0, b=0, then the equation is satisfied by any value of x.

If a=0, b¹0, then the equation has no solutions, because 0x=b is not executed for any value of the variable.

Example 1. Solve the equation: -8(11-2x)+40=3(5x-4)

Let's open the brackets on both sides of the equation, move all terms with x to the left side of the equation, and terms that do not contain x to the right side, we get:

16x-15x=88-40-12

Example 2. Solve the equations:

x3-2x2-98x+18=0;

These equations are not linear, but we will show how such equations can be solved.

3x2-5x=0; x(3x-5)=0. The product is equal to zero, if one of the factors is equal to zero, we get x1=0; x2=

. .

Factor the left side of the equation:

x2(x-2)-9(x-2)=(x-2)(x2-9)=(x-2)(x-3)(x-3), i.e. (x-2)(x-3)(x+3)=0. This shows that the solutions to this equation are the numbers x1=2, x2=3, x3=-3.

c) Imagine 7x as 3x+4x, then we have: x2+3x+4x+12=0, x(x+3)+4(x+3)=0, (x+3)(x+4)= 0, hence x1=-3, x2=- 4.

Answer: -3; - 4.

Example 3. Solve the equation: ½x+1ç+½x-1ç=3.

Let us recall the definition of the modulus of a number:

For example: ½3½=3, ½0½=0, ½- 4½= 4.

In this equation, under the modulus sign are the numbers x-1 and x+1. If x is less than –1, then the number x+1 is negative, then ½x+1½=-x-1. And if x>-1, then ½x+1½=x+1. At x=-1 ½x+1½=0.

Thus,

Likewise

a) Consider this equation½x+1½+½x-1½=3 for x £-1, it is equivalent to the equation -x-1-x+1=3, -2x=3, x=

, this number belongs to the set x £ -1.

b) Let -1< х £ 1, тогда данное уравнение равносильно уравнению х+1-х+1=3, 2¹3 уравнение не имеет решения на данном множестве.

c) Consider the case x>1.

x+1+x-1=3, 2x=3, x=

. This number belongs to the set x>1.

Answer: x1=-1.5; x2=1.5.

Example 4. Solve the equation:½x+2½+3½x½=2½x-1½.

Let us show a short record of the solution to the equation, revealing the sign of the modulus “over intervals”.

x £-2, -(x+2)-3x=-2(x-1), - 4x=4, x=-2О(-¥; -2]

–2<х£0, х+2-3х=-2(х-1), 0=0, хÎ(-2; 0]

0<х£1, х+2+3х=-2(х-1), 6х=0, х=0Ï(0; 1]

x>1, x+2+3x=2(x-1), 2x=- 4, x=-2П(1; +¥)

Answer: [-2; 0]

Example 5. Solve the equation: (a-1)(a+1)x=(a-1)(a+2), for all values of parameter a.

There are actually two variables in this equation, but consider x to be the unknown and a to be the parameter. It is required to solve the equation for the variable x for any value of the parameter a.

If a=1, then the equation has the form 0×x=0; any number satisfies this equation.

If a=-1, then the equation looks like 0×x=-2; not a single number satisfies this equation.

If a¹1, a¹-1, then the equation has a unique solution

Answer: if a=1, then x is any number;

if a=-1, then there are no solutions;

if a¹±1, then

2. Systems of equations with two variables.

A solution to a system of equations with two variables is a pair of values of variables that turns each equation of the system into a true equality. Solving a system means finding all its solutions or proving that they do not exist. Two systems of equations are said to be equivalent if each solution of the first system is a solution of the second system and each solution of the second system is a solution of the first system, or they both have no solutions.

When solving linear systems, the substitution method and the addition method are used.

Example 1. Solve the system of equations:

To solve this system, we apply the substitution method. Let's express x from the first equation and substitute this value

into the second equation of the system, we get

Answer: (2; 3).

Example 2. Solve the system of equations:

To solve this system, we apply the method of adding equations. 8x=16, x=2. Substitute the value x=2 into the first equation, we get 10-y=9, y=1.

Answer: (2; 1).

Example 3. Solve the system of equations:

This system is equivalent to one equation 2x+y=5, because the second equation is obtained from the first by multiplying by 3. Therefore, any pair of numbers (x; 5-2x) satisfies it. The system has an infinite number of solutions.

Answer: (x; 5-2x), x – any.

Example 4. Solve the system of equations:

Let's multiply the first equation by –2 and add it with the second equation, we get 0×x+0×y=-6. No pair of numbers satisfies this equation. Therefore, this system has no solutions.

Answer: the system has no solutions.

Example 5. Solve the system:

From the second equation we express x = y + 2a + 1 and substitute this value of x into the first equation of the system, we get

. When a=-2, the equation does not have solutions, but if a¹-2, then

Answer: when a=-2 the system has no solution,

at a¹-2 the system has a solution

Example 6. Solve the system of equations:

We are given a system of three equations with three unknowns. Let us apply the Gaussian method, which consists of equivalent transformations bringing the given system to a triangular shape. Let's add the second equation to the first equation, multiplied by –2.

2х-2у-2z=-12

3х-3у-3z=-18

Finally, we add to this equation the equation y-z=-1, multiplied by 2, we get - 4z=-12, z=3. So we get a system of equations:

x+y+z=6

z=3, which is equivalent to this one.

A system of this type is called triangular.

Answer: (1; 2; 3).

3. Solving problems using equations and systems of equations.

We will show with examples how you can solve problems using equations and systems of equations.

Example 1. An alloy of tin and copper weighing 32 kg contains 55% tin. How much pure tin must be added to the alloy so that the new alloy contains 60% tin?

Solution. Let the mass of tin added to the original alloy be x kg. Then an alloy weighing (32+x) kg will contain 60% tin and 40% copper. The original alloy contained 55% tin and 45% copper, i.e. there was 32·0.45 kg of copper in it. Since the mass of copper in the original and new alloys is the same, we obtain the equation 0.45·32=0.4(32+x).

Having solved it, we find x=4, i.e. 4 kg of tin must be added to the alloy.

Example 2. A two-digit number is conceived whose tens digit is 2 less than the units digit. If this number is divided by the sum of its digits, then the quotient will be 4 and the remainder 6. What number is intended?

Solution. Let the units digit be x, then the tens digit is x-2 (x>2), the intended number is 10(x-2)+x=11x-20. The sum of the digits of the number x-2+x=2x-2. Therefore, dividing 11x-20 by 2x-2, we get 4 as a quotient and 6 as a remainder. We create the equation: 11x-20=4(2x-2)+6, because The dividend is equal to the divisor times the quotient plus the remainder. Solving this equation, we get x=6. So, the number 46 was conceived.

X and scope X. Then an inequality of the form f(x) > g(x) or f(x) < g(x) is called inequality with one variable . A bunch of X called the scope of its definition.

Variable value X from many X, at which the inequality turns into a true numerical inequality, it is called decision. Solving an inequality means finding many solutions to it.

The basis for solving inequalities with one variable is the concept of equivalence.

The two inequalities are called equivalent, if their solution sets are equal.

Theorems on the equivalence of inequalities and the consequences from them are similar to the corresponding theorems on the equivalence of equations. Their proof uses the properties of true numerical inequalities.

Theorem 1. Let inequality f(x) > g(x) defined on the set X And h(x) is an expression defined on the same set. Then the inequalities f(x) > g(x) And f(x) + h(x) > g(x)+h(x) are equivalent on the set X.

From this theorem it follows consequences, which are often used when solving inequalities:

1) If to both sides of the inequality f(x) > g(x) add the same number d, then we get the inequality f(x) + d > g(x)+ d, equivalent to the original one.

2) If any term (or expression with a variable) is transferred from one part of the inequality to another, changing the sign of the term to the opposite, then we obtain an inequality equivalent to the given one.

Theorem 2. Let inequality f(x) > g(x) defined on the set X And h(x X from many X expression h(x) takes positive values. Then the inequalities f(x) > g(x) And f(x) × h(x) > g(x) × h(x) are equivalent on the set X.

A corollary follows from this theorem: if both sides of the inequality f(x) > g(x) multiplied by the same positive number d, then we get the inequality f(x) × d > g(x) × d, equivalent to this one.

Theorem 3. Let inequality f(x) > g(x) defined on the set X And h(x) - an expression defined on the same set, and for all X from many X expression h(x) takes negative values. Then the inequalities f(x) > g(x) And f(x) × h(x) < g(x) × h(x) are equivalent on the set X.

From this theorem it follows consequence: if both sides of the inequality f(x) > g(x) multiplied by the same negative number d and change the inequality sign to the opposite one, we get the inequality f(x) × d < g(x) × d, equivalent to this one.

Task. Is the number X= 5 solution of inequality 2 X+ 7 > 10 - x, xО R? Find many solutions to this inequality.

Solution. Number X= 5 is a solution to the inequality
2X + 7 > 10 - X, since 2×5 + 7 > 10 - 5 is a true numerical inequality. And the set of its solutions is the interval (1; ¥), which is found by performing the transformation of inequality 2 X+ 7 > 10 - XÞ 3X> 3 Þ X > 1.

Task. Solve inequality 5 X- 5 < 2X+ 16 and justify all the transformations that will be performed during the solution process.

Solution.

Transformations	Rationale for transformations
1. Let's move expression 2 X to the left side, and the number -5 to the right, changing their signs to the opposite: 5 X- 2X < 16 + 5.	We used Corollary 2 from Theorem 3 and obtained an inequality equivalent to the original one.
2. Let us present similar terms on the left and right sides of the inequality: 3 X < 21.	We performed identical transformations of expressions on the left and right sides of the inequality - they did not violate the equivalence of the inequalities: the given and the original.
3. Divide both sides of the inequality by 3: X < 7.	We used the corollary of Theorem 4 and obtained an inequality equivalent to the original one.

How to solve linear inequalities with one variable of the form ax+b>cx+d?

To do this, we use only two rules.

1) The terms can be transferred from one part of the inequality to another with the opposite sign. The sign of inequality does not change.

2) Both sides of the inequality can be (or another variable). When divided by a positive number, the inequality sign does not change. When dividing by a negative number, the inequality sign is reversed.

In general, the solution to a linear inequality in one variable

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can be depicted like this:

1) We move the unknowns to one side, the known ones to the other with opposite signs:

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2) If the number in front of X is not equal to zero (a-c≠0), divide both sides of the inequality by a-c.

If a-c>0, the inequality sign does not change:

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If a-c<0, знак неравенства изменяется на противоположный:

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If a-c=0, then this is a special case. We will consider special cases of solving linear inequalities separately.

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This is a linear inequality. We move the unknowns in one direction, the knowns in the other with opposite signs:

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We divide both sides of the inequality by the number in front of X. Since -2<0, знак неравенства изменяется на противоположный:

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Since , 10 on the number line is marked with a punctured dot. , to minus infinity.

Since the inequality is strict and the point is missing, we write 10 in the answer with a parenthesis.

This is a linear inequality. Unknowns - in one direction, knowns - in the other with opposite signs:

We divide both sides of the inequality by the number in front of X. Since 10>

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Since the inequality is not strict, we mark -2.3 on the number line with a filled dot. The shading from -2,3 goes to the right, to plus infinity.

Since the inequality is strict and the point is shaded, we write -2.3 in the answer with a square bracket.

This is a linear inequality. Unknowns go in one direction, knowns go in the other direction with the opposite sign.

We divide both sides of the inequality by the number in front of X. Since 3>0, the inequality sign does not change:

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Since the inequality is strict, we represent x=2/3 on the number line as a punctured point.

Since the inequality is strict and the point is missing, we write 2/3 in the answer with a parenthesis.

1. The concept of inequality with one variable

2. Equivalent inequalities. Theorems on the equivalence of inequalities

3. Solving inequalities with one variable

4. Graphical solution of inequalities with one variable

5. Inequalities containing a variable under the modulus sign

6. Main conclusions

Inequalities with one variable

Offers 2 X + 7 > 10's, x 2 +7x< 2,(х + 2)(2х-3)> 0 are called inequalities with one variable.

In general, this concept is defined as follows:

Definition. Let f(x) and g(x) be two expressions with variable x and domain X. Then an inequality of the form f(x) > g(x) or f(x)< g(х) называется неравенством с одной переменной. Множество X называется областью его определения.

Variable value x from many X, in which the inequality turns into a true numerical inequality is called decision. Solving an inequality means finding many solutions to it.

Thus, by solving inequality 2 x + 7 > 10 -x, x? R is the number x= 5, since 2 5 + 7 > 10 - 5 is a true numerical inequality. And the set of its solutions is the interval (1, ∞), which is found by performing the transformation of the inequality: 2 x + 7 > 10-x => 3x >3 => x >1.

Equivalent inequalities. Theorems on the equivalence of inequalities

The basis for solving inequalities with one variable is the concept of equivalence.

Definition. Two inequalities are said to be equivalent if their solution sets are equal.

For example, inequalities 2 x+ 7 > 10 and 2 x> 3 are equivalent, since their solution sets are equal and represent the interval (2/3, ∞).

Theorem 3. Let inequality f(x) > g(x) defined on the set X And h(x) is an expression defined on the same set. Then the inequalities f(x) > g(x) and f(x)+ h(x) > g(x) + h(x) are equivalent on the set X.

Corollaries follow from this theorem, which are often used when solving inequalities:

1) If to both sides of the inequality f(x) > g(x) add the same number d, then we get the inequality f(x) + d > g(x)+ d, equivalent to the original one.

2) If any term (numerical expression or expression with a variable) is transferred from one part of the inequality to another, changing the sign of the term to the opposite, then we obtain an inequality equivalent to the given one.

Theorem 4. Let inequality f(x) > g(x) defined on the set X And h(X X from many X expression h(x) takes positive values. Then the inequalities f(x) > g(x) and f(x) h(x) > g(x) h(x) are equivalent on the set X.

f(x) > g(x) multiply by the same positive number d, then we get the inequality f(x) d > g(x) d, equivalent to this.

Theorem 5. Let inequality f(x) > g(x) defined on the set X And h(X) - an expression defined on the same set, and for all X there are many of them X expression h(X) takes negative values. Then the inequalities f(x) > g(x) and f(x) h(x) > g(x) h(x) are equivalent on the set X.

A corollary follows from this theorem: if both sides of the inequality f(x) > g(x) multiply by the same negative number d and change the inequality sign to the opposite one, we get the inequality f(x) d > g(x) d, equivalent to this.

Solving inequalities with one variable

Let's solve inequality 5 X - 5 < 2х - 16, X? R, and we will justify all the transformations that we will perform in the solution process.

Solving the inequality X < 7 является промежуток (-∞, 7) и, следовательно, множеством решений неравенства 5X - 5 < 2x + 16 is the interval (-∞, 7).

Exercises

1. Determine which of the following entries are inequalities with one variable:

a) -12 - 7 X< 3x+ 8; d) 12 x + 3(X- 2);

b) 15( x+ 2)>4; e) 17-12·8;

c) 17-(13 + 8)< 14-9; е) 2x 2+ 3x-4> 0.

2. Is the number 3 a solution to the inequality 6(2x + 7) < 15(X + 2), X? R? And the number 4.25?

3. Are the following pairs of inequalities equivalent on the set of real numbers:

a) -17 X< -51 и X > 3;

b) (3 x-1)/4 >0 and 3 X-1>0;

c) 6-5 x>-4 and X<2?

4. Which of the following statements are true:

a) -7 X < -28 => x>4;

b) x < 6 => x < 5;

V) X< 6 => X< 20?

5. Solve inequality 3( x - 2) - 4(X + 1) < 2(х - 3) - 2 and justify all the transformations that you will perform.

6. Prove that by solving the inequality 2(x+ 1) + 5 > 3 - (1 - 2X) is any real number.

7. Prove that there is no real number that would be a solution to the inequality 3(2 - X) - 2 > 5 - 3X.

8. One side of the triangle is 5 cm, and the other is 8 cm. What can be the length of the third side if the perimeter of the triangle is:

a) less than 22 cm;

b) more than 17 cm?

GRAPHICAL SOLUTION OF INEQUALITIES WITH ONE VARIABLE. To solve the inequality graphically f (x) > g (x) need to build graphs of functions

y = f (x) = g (x) and select those intervals of the abscissa axis on which the graph of the function y = f(x) located above the graph of the function y = g(x).

Example 17.8. Solve graphically the inequality x 2- 4 > 3X.

Y - x* - 4

Solution. Let's construct graphs of functions in one coordinate system

y = x 2 - 4 and y = Zx (Fig. 17.5). The figure shows that the graphs of functions at= x 2- 4 is located above the graph of the function y = 3 X at X< -1 and x > 4, i.e. the set of solutions to the original inequality is the set

(- ¥; -1) È (4; + oo) .

Answer: x О(- oo; -1) and ( 4; + oo).

Graph of a quadratic function at= ax 2 + bx + c is a parabola with branches pointing upward if a > 0, and down if A< 0. In this case, three cases are possible: the parabola intersects the axis Oh(i.e. equation ah 2+ bx+ c = 0 has two different roots); parabola touches axis X(i.e. equation ax 2 + bx+ c = 0 has one root); the parabola does not intersect the axis Oh(i.e. equation ah 2+ bx+ c = 0 has no roots). Thus, there are six possible positions of the parabola, which serves as a graph of the function y = ah 2+b x + c(Fig. 17.6). Using these illustrations, you can solve quadratic inequalities.

Example 17.9. Solve the inequality: a) 2 x g+ 5x - 3 > 0; b) -Zx 2 - 2x- 6 < 0.

Solution, a) The equation 2x 2 + 5x -3 = 0 has two roots: x, = -3, x 2 = 0.5. Parabola serving as a graph of a function at= 2x 2+ 5x -3, shown in Fig. A. Inequality 2x 2+ 5x -3 > 0 is satisfied for those values X, for which the points of the parabola lie above the axis Oh: it will be at X< х х or when X> x g> those. at X< -3 or at x > 0.5. This means that the set of solutions to the original inequality is the set of (- ¥; -3) and (0.5; + ¥).

b) Equation -Зх 2 + 2x- 6 = 0 has no real roots. Parabola serving as a graph of a function at= - 3x 2 - 2x - 6, shown in Fig. 17.6 Inequality -3x 2 - 2x - 6 < О выполняется при тех значениях X, for which the points of the parabola lie below the axis Oh. Since the entire parabola lies below the axis Oh, then the set of solutions to the original inequality is the set R .

INEQUALITIES CONTAINING A VARIABLE UNDER THE MODULE SIGN. When solving these inequalities, it should be kept in mind that:

|f(x) | =

f(x), If f(x) ³ 0,

- f(x), If f(x) < 0,

In this case, the range of permissible values of the inequality should be divided into intervals, at each of which the expressions under the modulus sign retain their sign. Then, expanding the modules (taking into account the signs of the expressions), you need to solve the inequality on each interval and combine the resulting solutions into a set of solutions to the original inequality.

Example 17.10. Solve the inequality:

|x -1| + |2- x| > 3+x.

Solution. Points x = 1 and x = 2 divide the numerical axis (ODZ of inequality (17.9) into three intervals: x< 1, 1 £ х £.2, х >2. Let’s solve this inequality for each of them. If x< 1, то х - 1 < 0 и 2 – х >0; therefore |x -1| = - (x - I), |2 - x | = 2 - x. This means that inequality (17.9) takes the form: 1- x + 2 - x > 3 + x, i.e. X< 0. Таким образом, в этом случае решениями неравенства (17.9) являются все отрицательные числа.

If 1 £ x £.2, then x - 1 ³ 0 and 2 – x ³ 0; therefore | x- 1| = x - 1, |2 - x| = 2 – x. This means that the system holds:

x – 1 + 2 – x > 3 + x,

The resulting system of inequalities has no solutions. Therefore, on the interval [ 1; 2] the set of solutions to inequality (17.9) is empty.

If x > 2, then x - 1 >0 and 2 – x<0; поэтому | х - 1| = х- 1, |2-х| = -(2- х). Значит, имеет место система:

x -1 + x – 2 > 3+x,

x > 6 or

Combining the solutions found on all parts of the ODZ inequality (17.9), we obtain its solution - the set (-¥; 0) È (6; +oo).

Sometimes it is useful to use the geometric interpretation of the modulus of a real number, according to which | a | means the distance of point a of the coordinate line from the origin O, and | a - b | means the distance between points a and b on the coordinate line. Alternatively, you can use the method of squaring both sides of the inequality.

Theorem 17.5. If expressions f(x) and g(x) for any x take only non-negative values, then the inequalities f (x) > g (x) And f (x) ² > g (x) ² are equivalent.

58. Main conclusions § 12

In this section we have defined the following concepts:

Numeric expression;

The value of a numeric expression;

An expression that has no meaning;

Expression with variable(s);

Expression definition area;

Identically equal expressions;

Identity;

Identical transformation of an expression;

Numerical equality;

Numerical inequality;

Equation with one variable;

Root of the equation;

What does it mean to solve an equation;

Equivalent equations;

Inequality with one variable;

Solving Inequalities;

What does it mean to solve inequality;

Equivalent inequalities.

In addition, we examined theorems on the equivalence of equations and inequalities, which are the basis for their solution.

Knowledge of the definitions of all the above concepts and theorems on the equivalence of equations and inequalities is a necessary condition for methodologically competent study of algebraic material with primary school students.