Single and double linear inequalities. Linear inequalities

It has been necessary to compare quantities and quantities when solving practical problems since ancient times. At the same time, words such as more and less, higher and lower, lighter and heavier, quieter and louder, cheaper and more expensive, etc. appeared, denoting the results of comparing homogeneous quantities.

The concepts of more and less arose in connection with counting objects, measuring and comparing quantities. For example, mathematicians of Ancient Greece knew that the side of any triangle is less than the sum of the other two sides and that the larger side lies opposite the larger angle in a triangle. Archimedes, while calculating the circumference, established that the perimeter of any circle is equal to three times the diameter with an excess that is less than a seventh of the diameter, but more than ten seventy times the diameter.

Symbolically write relationships between numbers and quantities using the signs > and b. Records in which two numbers are connected by one of the signs: > (greater than), You also encountered numerical inequalities in the lower grades. You know that inequalities can be true, or they can be false. For example, $\frac(1)(2) > \frac(1)(3)$ is a correct numerical inequality, 0.23 > 0.235 is an incorrect numerical inequality.

Inequalities involving unknowns may be true for some values of the unknowns and false for others. For example, the inequality 2x+1>5 is true for x = 3, but false for x = -3. For an inequality with one unknown, you can set the task: solve the inequality. Problems of solving inequalities in practice are posed and solved no less often than problems of solving equations. For example, many economic problems come down to the study and solution of systems of linear inequalities. In many branches of mathematics, inequalities are more common than equations.

Some inequalities serve as the only auxiliary, allowing you to prove or disprove the existence of a certain object, for example, the root of an equation.

Numerical inequalities

You can compare whole numbers and decimal fractions. Know the rules for comparing ordinary fractions with the same denominators but different numerators; with the same numerators but different denominators. Here you will learn how to compare any two numbers by finding the sign of their difference.

Comparing numbers is widely used in practice. For example, an economist compares planned indicators with actual ones, a doctor compares a patient’s temperature with normal, a turner compares the dimensions of a machined part with a standard. In all such cases, some numbers are compared. As a result of comparing numbers, numerical inequalities arise.

Definition. Number a more number b, if difference a-b positive. Number a less number b, if the difference a-b is negative.

If a is greater than b, then they write: a > b; if a is less than b, then they write: a Thus, the inequality a > b means that the difference a - b is positive, i.e. a - b > 0. Inequality a For any two numbers a and b, from the following three relations a > b, a = b, a To compare the numbers a and b means to find out which of the signs >, = or Theorem. If a > b and b > c, then a > c.

Theorem. If you add the same number to both sides of the inequality, the sign of the inequality will not change.
Consequence. Any term can be moved from one part of the inequality to another by changing the sign of this term to the opposite.

Theorem. If both sides of the inequality are multiplied by the same positive number, then the sign of the inequality does not change. If both sides of the inequality are multiplied by the same negative number, then the sign of the inequality will change to the opposite.
Consequence. If both sides of the inequality are divided by the same positive number, then the sign of the inequality will not change. If both sides of the inequality are divided by the same negative number, then the sign of the inequality will change to the opposite.

You know that numerical equalities can be added and multiplied term by term. Next, you will learn how to perform similar actions with inequalities. The ability to add and multiply inequalities term by term is often used in practice. These actions help solve problems of evaluating and comparing the meanings of expressions.

When solving various problems, it is often necessary to add or multiply the left and right sides of inequalities term by term. At the same time, it is sometimes said that inequalities add up or multiply. For example, if a tourist walked more than 20 km on the first day, and more than 25 km on the second, then we can say that in two days he walked more than 45 km. Similarly, if the length of a rectangle is less than 13 cm and the width is less than 5 cm, then we can say that the area of this rectangle is less than 65 cm2.

When considering these examples, the following were used: theorems on addition and multiplication of inequalities:

Theorem. When adding inequalities of the same sign, an inequality of the same sign is obtained: if a > b and c > d, then a + c > b + d.

Theorem. When multiplying inequalities of the same sign, whose left and right sides are positive, an inequality of the same sign is obtained: if a > b, c > d and a, b, c, d are positive numbers, then ac > bd.

Inequalities with the sign > (greater than) and 1/2, 3/4 b, c Along with the signs of strict inequalities > and In the same way, the inequality $a \geq b $ means that the number a is greater than or equal to b, i.e. .and not less b.

Inequalities containing the $\geq $ sign or the $\leq $ sign are called non-strict. For example, $18 \geq 12 , \; 11 \leq 12 $ are not strict inequalities.

All properties of strict inequalities are also valid for non-strict inequalities. Moreover, if for strict inequalities the signs > were considered opposite and you know that to solve a number of applied problems you have to create a mathematical model in the form of an equation or a system of equations. Next, you will learn that mathematical models for solving many problems are inequalities with unknowns. We will introduce the concept of solving an inequality and show how to test whether a given number is a solution to a particular inequality.

Inequalities of the form
\(ax > b, \quad ax in which a and b are given numbers, and x is unknown, is called linear inequalities with one unknown.

Definition. The solution to an inequality with one unknown is the value of the unknown at which this inequality becomes a true numerical inequality. Solving an inequality means finding all its solutions or establishing that there are none.

You solved the equations by reducing them to the simplest equations. Similarly, when solving inequalities, one tries to reduce them, using properties, to the form of simple inequalities.

Solving second degree inequalities with one variable

Inequalities of the form
$ax^2+bx+c >0 $ and $ax^2+bx+c where x is a variable, a, b and c are some numbers and \(a \neq 0 $, called inequalities of the second degree with one variable.

Solution to inequality
$ax^2+bx+c >0 $ or $ax^2+bx+c can be considered as finding intervals in which the function \(y= ax^2+bx+c $ takes positive or negative values To do this, it is enough to analyze how the graph of the function $y= ax^2+bx+c$ is located in the coordinate plane: where the branches of the parabola are directed - up or down, whether the parabola intersects the x-axis and if it does, then at what points.

Algorithm for solving second degree inequalities with one variable:
1) find the discriminant of the square trinomial $ax^2+bx+c$ and find out whether the trinomial has roots;
2) if the trinomial has roots, then mark them on the x-axis and through the marked points draw a schematic parabola, the branches of which are directed upward for a > 0 or downward for a 0 or at the bottom for a 3) find intervals on the x-axis for which the points parabolas are located above the x-axis (if they solve the inequality $ax^2+bx+c >0$) or below the x-axis (if they solve the inequality
\(ax^2+bx+c Solving inequalities using the interval method

Consider the function
f(x) = (x + 2)(x - 3)(x - 5)

The domain of this function is the set of all numbers. The zeros of the function are the numbers -2, 3, 5. They divide the domain of definition of the function into the intervals $(-\infty; -2), \; (-2; 3), \; (3; 5) $ and $ (5; +\infty)$

Let us find out what the signs of this function are in each of the indicated intervals.

The expression (x + 2)(x - 3)(x - 5) is the product of three factors. The sign of each of these factors in the intervals under consideration is indicated in the table:

In general, let the function be given by the formula
f(x) = (x-x 1)(x-x 2) ... (x-x n),
where x is a variable, and x 1, x 2, ..., x n are numbers that are not equal to each other. The numbers x 1 , x 2 , ..., x n are the zeros of the function. In each of the intervals into which the domain of definition is divided by zeros of the function, the sign of the function is preserved, and when passing through zero its sign changes.

This property is used to solve inequalities of the form
(x-x 1)(x-x 2) ... (x-x n) > 0,
(x-x 1)(x-x 2) ... (x-x n) where x 1, x 2, ..., x n are numbers not equal to each other

Considered method solving inequalities is called the interval method.

Let us give examples of solving inequalities using the interval method.

Solve inequality:

$x(0.5-x)(x+4) Obviously, the zeros of the function f(x) = x(0.5-x)(x+4) are the points \(x=0, \; x= \frac(1)(2) , \; x=-4 $

We plot the zeros of the function on the number axis and calculate the sign on each interval:

We select those intervals at which the function is less than or equal to zero and write down the answer.

Answer:
$x \in \left(-\infty; \; 1 \right) \cup \left[ 4; \; +\infty \right) $

In the article we will consider solving inequalities. We will tell you clearly about how to construct a solution to inequalities, with clear examples!

Before we look at solving inequalities using examples, let’s understand the basic concepts.

General information about inequalities

Inequality is an expression in which functions are connected by relation signs >, . Inequalities can be both numerical and literal.
Inequalities with two signs of the ratio are called double, with three - triple, etc. For example:
a(x) > b(x),
a(x) a(x) b(x),
a(x) b(x).
a(x) Inequalities containing the sign > or or - are not strict.
Solving the inequality is any value of the variable for which this inequality will be true.
"Solve inequality" means that we need to find the set of all its solutions. There are different methods for solving inequalities. For inequality solutions They use the number line, which is infinite. For example, solution to inequality x > 3 is the interval from 3 to +, and the number 3 is not included in this interval, so the point on the line is denoted empty circle, because inequality is strict.
+
The answer will be: x (3; +).
The value x=3 is not included in the solution set, so the parenthesis is round. The infinity sign is always highlighted with a parenthesis. The sign means "belonging."
Let's look at how to solve inequalities using another example with a sign:
x 2
-+
The value x=2 is included in the set of solutions, so the bracket is square and the point on the line is indicated by a filled circle.
The answer will be: x.

The entire algorithm described above is written like this:

3 x + 12 ≤ 0 ; 3 x ≤ − 12 ; x ≤ − 4 .

Answer: x ≤ − 4 or (− ∞ , − 4 ] .

Example 2

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

Solution

From the condition we see that the coefficient a for 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 move on to the second.

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

Let us write the entire algorithm in brief form:

− 2, 7 z > 0; z< 0 .

Answer: z< 0 или (− ∞ , 0) .

Example 3

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

Solution

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

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

During the last transition for the right side, the number division rule is used with different signs 15 22: - 5 = - 15 22: 5, after which we divide the ordinary fraction by the natural number - 15 22: 5 = - 15 22 · 1 5 = - 15 · 1 22 · 5 = - 3 22.

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

Let's 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 determining the solution to 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 will 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 it is false when the original inequality has no solutions.

Example 4

Solve the inequality 0 x + 7 > 0.

Solution

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, which means any number can be its solution.

Answer: interval (− ∞ , + ∞) .

Example 5

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

Solution

When substituting the variable x of any number, we obtain that the inequality takes the form − 12, 7 ≥ 0. It is incorrect. That is, 0 x − 12, 7 ≥ 0 has no solutions.

Answer: there are no solutions.

Let's consider solving linear inequalities where both coefficients are equal to zero.

Example 6

Determine the unsolvable inequality from 0 x + 0 > 0 and 0 x + 0 ≥ 0.

Solution

When substituting any number instead of x, we obtain 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, but 0 x + 0 ≥ 0 has solutions.

This method is discussed in school course mathematics. The interval method is capable of resolving different kinds inequalities, also linear.

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 a different method.

Definition 6

The interval method is:

introducing the function y = a · x + b ;
searching for zeros to split the domain of definition into intervals;
definition of signs for their concepts 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 a single root, which will take the designation x 0;
construction of a coordinate line with an image of a point with coordinate x 0, with a strict inequality the point is denoted by a punctured one, with a non-strict inequality – by a shaded one;
determining the signs of the function y = a · x + b on intervals; for this it is necessary to find the values of the function at points on the interval;
solving an inequality with signs > or ≥ on the coordinate line, adding shading over the positive interval,< или ≤ над отрицательным промежутком.

Let's look at several examples of solving linear inequalities using the interval method.

Example 6

Solve the inequality − 3 x + 12 > 0.

Solution

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 draw a coordinate line where we mark point 4. It will be punctured because the inequality is strict. Consider the drawing below.

It is necessary to determine the signs at the intervals. To determine it on the interval (− ∞, 4), it is necessary to calculate the function y = − 3 x + 12 at 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 substitute the value x = 5. We have that − 3 5 + 12 = − 3< 0 . Знак на промежутке является отрицательным. Изобразим на числовой прямой, приведенной ниже.

We solve the inequality with the > sign, and the shading is performed over the positive interval. Consider the drawing below.

From the drawing it is clear that the desired solution has the form (− ∞ , 4) or x< 4 .

Answer: (− ∞ , 4) or x< 4 .

To understand how to depict 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 the values of x< 2 , x ≤ 2 , x >2 and x ≥ 2. To do this, let's plot the linear function y = 0, 5 x − 1 shown below.

It's clear that

Definition 7

solving 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 considered to be the interval where the function y = 0, 5 x − 1 is lower than O x or coincides;
the solution 0, 5 · x − 1 > 0 is considered to be an interval, the function is located above O x;
the solution 0, 5 · x − 1 ≥ 0 is considered to be the interval where the graph above O x or coincides.

The point of graphically solving inequalities is to find the intervals that need to be depicted on the graph. IN in this case we find that the left side has y = a · x + b, and the right side has y = 0, and coincides with O x.

Definition 8

The graph of the function y = a x + b is plotted:

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

Solution

It is necessary to construct a graph of the 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 depict it graphically.

Solving the inequality with the > sign, then you need to pay attention to the interval above O x. Highlight it in red necessary part plane and we get that

The required gap is part O x red. This means that the open number ray - ∞ , - 3 5 will be a solution to the inequality. If, according to the condition, we had a non-strict inequality, then the value of the point - 3 5 would also be a solution to the inequality. And it would coincide with O x.

Answer: - ∞ , - 3 5 or x< - 3 5 .

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

Example 8

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

Solution

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

The graph of the function y = 0 x + 0 is considered to be y = 0, that is, the straight line coincides with O x. This means that the inequality 0 x + 0 ≥ 0 has many solutions.

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

Inequalities that reduce to linear

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

These inequalities were considered in the school course, since they were a special case of solving inequalities, which led to the opening of parentheses 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. Then the brackets are opened and similar terms are given and transferred from different parts, changing the sign to the opposite.

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

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

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

Definition 9

open parentheses;
collect variables on the left and numbers on the right;
give similar terms;
divide both sides by the coefficient of x.

Example 9

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

Solution

We open 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 find that 6 x + 15 − 6 x + 17 ≤ 0. Hence there is an inequality of the form 32 ≤ 0 from that obtained by calculating 0 x + 32 ≤ 0. It can be seen that the inequality is false, which means that the inequality given by condition has no solutions.

Answer: no solutions.

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

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Lesson and presentation on the topic: "Examples of linear inequalities and their solution"

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Linear Equations (Repeat)

Guys, we are moving on to studying the algebra course for 9th grade. As we study our course, we will learn to solve many new and exciting problems.

Let's repeat it a bit.
Do you remember what it is linear equation?
We call an equation of the form $ax+b=0$ linear, here the coefficients a and b are from the set real numbers, that is, almost any number. By the way, why is it called linear? That's right, if we draw a graph of the solution to our equation, we get a line.

How did we solve our equation? We left what was with x to the left of the equal sign, and without x we moved it to the right, not forgetting to change the sign, that is, we got an equation of the form: $ax=-b$.
Then we divided it by the coefficient of x and obtained a solution to the equation: $x=-\frac(b)(a)$.
Well, let's move on to the first topic of our course.

We remembered linear equations, now let's introduce the concept of linear inequality. I think you guessed that the definitions will not be much different.
Linear inequalities with one variable are inequalities of the following form: $ax+b>0$, where a and b are values from the set of real numbers $(a≠0)$. In general, you can write 4 types inequalities:
$ax+b>0\\ ax+b
The value of the variable x at which our inequality becomes true is called a solution. It is worth noting that there are two types of solutions: specific and general. General solution name the entire set of particular solutions.

Let's introduce a few rules when solving linear inequalities:
The terms of the inequality can be transferred from one part to another, just as in linear equations, without changing the sign of the inequality.
$3x inequality
An inequality can be multiplied and divided by the same number greater than zero without changing the sign of the inequality. Guys, don’t forget that you must multiply or divide both sides of the inequality!
$3x inequality
An inequality can be multiplied or divided by a negative number, without forgetting to change the sign of the inequality to the opposite one. Sign, ≤ on≥, and vice versa, respectively.
Let's multiply the inequality $3x-7 0$.

If the inequality in the variable x is divided or multiplied by the expression $p(x)$, which depends on x, and which is positive for any x, without changing the sign of the inequality, then we obtain an inequality equivalent to the original one.

If the inequality in the variable x is divided or multiplied by the expression $p(x)$, depending on x, b which is negative for any x, changing the sign of the inequality, then we get an inequality equivalent to the original one.

1. Solve inequality: $3x-6
Solution:
The solution method is similar to linear equations, move -6 to the right of the inequality sign $3x We can divide our inequality by any positive number without changing the sign. Let's divide by 3 and get the solution: $x Answer: $x
2. Solve inequality: $-3x+6
Solution:
Let's perform the initial steps: $-3x Divide the inequality by -3, not forgetting to change the sign: $x>2$.
Answer: $x>2$.

3. Solve the inequality: $\frac(x)(4)+\frac((3x-2))(8)>x-\frac(1)(16)$.

Solution:
Let's multiply our inequality by 16, we get: $4x+2(3x-2)>16x-1$.
Let's do it necessary actions: $4x+6x-4-16x>-1$.
$-6x>3$.
Divide the inequality by -6, changing its sign: $x Answer: $x
4. Solve the inequality: $|2x-2|
Solution:
Divide the inequality by 2. We get: $|x-1| The solution to our inequality can be represented as a segment of a coordinate line. The middle of the segment will be at point $x=1$, and the boundaries will be removed by 2.
Let's draw our segment:
The open interval $(-1;3)$ is the solution to our inequality.

Linear inequalities problems

1. Solve the inequality:
a) $2x+5 b) $-4x-9>11.$
c) $-5x+10
2. Solve the inequality: $\frac(2x)(9)+\frac(2x-4)(3)≤x-\frac(1)(18)$.

3. Solve inequality:
$a) |3x-5| b) $|5x|

How to solve linear inequalities? First, we need to simplify the inequality: open the brackets and add similar terms.

Let's look at examples of solving linear inequalities with one variable.

Opening the parentheses. If there is a factor in front of the brackets, multiply it by each term in the brackets. If the parentheses are preceded by a plus sign, the characters in the parentheses do not change. If there is a minus sign in front of the brackets, the signs in the brackets are reversed.

We present similar terms.

We got an inequality of the form ax+b≤cx+d. We move the unknowns to one side, the knowns to the other with opposite signs (we could first move the unknowns to one side, the knowns to the other, and only then bring similar terms).

We divide both sides of the inequality by the number in front of X. Since 8 is greater than zero, the inequality sign does not change: