The solution to an inequality in one variable is the value. Equations and inequalities with one variable

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 you can solve linear inequality 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 c 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 negative number, then the sign of inequality changes to the opposite. 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 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 get numerical inequality type 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 we 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 it is decided strict inequality(with 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 to easily determine the interval corresponding to this part, let's highlight the part in red 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 action: on the left side we open the brackets 7 x−7+3≤4 x−2+x , after which we give similar terms in both sides 7 x−4≤5 x−2 , then we transfer the terms from the right side to the left side 7 x−4−5 x+2≤0 , finally, we present similar terms on 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 . Next we move the terms from 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 form.

References.

• 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. At 2 p.m. Part 1. Textbook for students educational 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 beginnings mathematical analysis. 11th grade. At 2 p.m. 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.

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 view solving 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 - 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.

With one variable: what are equivalent inequalities; which transformations of inequalities are equivalent and which are not. We discussed these questions in the algebra course, starting from the 8th grade, and in this textbook they were already discussed, for example, when solving exponential and logarithmic inequalities. We return to these questions again because, after completing the study school course algebra, it is advisable to rethink general ideas and methods.

1. Equivalence of inequalities

Recall that the solution to the inequality a(x) > n(x) is any value variable x, which turns a given inequality with a variable into a true numerical inequality. Sometimes the term private solution is used. The set of all particular solutions to an inequality is called a general solution, but the term solution is more often used. Thus, the term solution is used in three senses: and how general solution, both as a private decision and as a process, but usually in terms of meaning it is clear what we are talking about.

Definition 1. Two inequalities with one variable f(x)>g(x) and p(x)>h(x) are called equivalent if their solutions (i.e. sets of partial solutions) coincide.

You, of course, understand that using the > sign in the definition is unimportant. You can use any other inequality sign, both strict and non-strict, in this definition and in all statements in this paragraph.

Definition 2. If the solution to the inequality

contained in the solution of the inequality

then inequality (2) is called a consequence of inequality (1)

For example, the inequality x 2 >9 is a consequence of the inequality 2x>6. In fact, transforming the first inequality to the form x 2 -9 >0 and further to the form (x-3)(x+3) >0 and applying the method of intervals (Fig. 245), we find that the solution to the inequality is the union of two open rays : The solution to the second inequality 2x>6 has the form x>3, i.e. represents open beam The solution to the second inequality is part of the solution to the first inequality, and therefore the first inequality is a consequence of the second.
It is curious that the situation will change radically if the sign of inequality in both inequalities is changed. 2x inequality< 6 будет следствием неравенства x 2 < 9. В самом деле, решением первого неравенства служит открытый луч . Преобразовав второе неравенство к виду х r - 9 <0 и далее к виду (х-3)(х+3) <06 применив метод интервалов (см. рис. 245), получаем, что решением неравенства служит интервал (-3, 3). Решение второго неравенства является частью решения первого неравенства, а потому первое неравенство - следствие второго.

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Offers 2x+7>10x, 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 q(x) be two expressions with variable x and domain X. Then an inequality of the form f(x)< q(х) или f(х) >q(x) is called an inequality in one variable. The set X is called its domain of definition.

The value of the variable x from the set X, at which the inequality turns into a true numerical inequality, is called its solution. Solving an inequality means finding many solutions to it.

Thus, by solving inequality 2 X+7>10-X, XÎ 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 transforming the inequality: 2x+7>10's Þ 3x> Þ x>1.

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, the inequalities 2x+7>10 and 2x>3 are equivalent, since their solution sets are equal and represent an interval

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

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

1) If we add the same number d to both sides of the inequality f(x) > q(x), we obtain the inequality f(x)+ d > q(x)+ d, which is equivalent to the original one.

2) If any term ( numeric expression or an expression with a variable) transfer 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 the inequality f(x) > q(x) be defined on the set X and h(x) be an expression defined on the same set, and for all x from the set X the expression h(x) takes positive values. Then the inequalities f(x)× h(x) > q(x)× h(x) are equivalent on the set X.

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

Theorem 5. Let the inequality f(x) > q(x) be defined on the set X and let h(x) be an expression defined on the same set, and for all x of the set X the expression h(x) takes negative values. Then the inequalities f(x) > q(x) b f(x)× h(x)< q(х)× h(х) равносильны на множестве X.

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

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