Inequality calculator with online solution. Linear inequalities

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

What is inequality

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

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

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

Solving inequalities

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

Identity transformations of inequalities

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

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

The second transformation has two points:

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

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

Examples of Untying Inequalities

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

6x − 3(4x + 1) > 6.

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

6x − 12x > 6 + 3

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

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

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

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

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

Conclusion

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

The form ax 2 + bx + 0 0, where (instead of the > sign, there can, of course, be any other inequality sign). We have all the facts of the theory necessary for solving such inequalities, which we will now verify.

Example 1. Solve the inequality:

a) x 2 - 2x - 3 > 0; b) x 2 - 2x - 3< 0;
c) x 2 - 2x - 3 > 0; d) x 2 - 2x - 3< 0.
Solution,

a) Consider the parabola y \u003d x 2 - 2x - 3 shown in fig. 117.

To solve the inequality x 2 - 2x - 3 > 0 - this means answering the question, for which values of x the ordinates of the points of the parabola are positive.

We notice that y > 0, i.e., the graph of the function is located above the x-axis, at x< -1 или при х > 3.

Hence, the solutions of the inequality are all points of the open beam(- 00 , - 1), as well as all points of the open beam (3, +00).

Using the sign U (the sign of the union of sets), the answer can be written as follows: (-00 , - 1) U (3, +00). However, the answer can also be written like this:< - 1; х > 3.

b) Inequality x 2 - 2x - 3< 0, или у < 0, где у = х 2 - 2х - 3, также можно решить с помощью рис. 117: schedule located below the x-axis if -1< х < 3. Поэтому решениями данного неравенства служат все точки интервала (- 1, 3).

c) The inequality x 2 - 2x - 3 > 0 differs from the inequality x 2 - 2x - 3 > 0 in that the answer must also include the roots of the equation x 2 - 2x - 3 = 0, i.e. points x = -1

and x \u003d 3. Thus, the solutions of this non-strict inequality are all points of the beam (-00, - 1], as well as all points of the beam.

Practical mathematicians usually say this: why do we, solving the inequality ax 2 + bx + c > 0, carefully build a parabola graph of a quadratic function

y \u003d ax 2 + bx + c (as was done in example 1)? It is enough to make a schematic sketch of the graph, for which you only need to find roots square trinomial (the point of intersection of the parabola with the x-axis) and determine where the branches of the parabola are directed - up or down. This schematic sketch will give a visual interpretation of the solution of the inequality.

Example 2 Solve the inequality - 2x 2 + 3x + 9< 0.
Solution.

1) Find the roots of the square trinomial - 2x 2 + Zx + 9: x 1 \u003d 3; x 2 \u003d - 1.5.

2) The parabola, which serves as a graph of the function y \u003d -2x 2 + Zx + 9, intersects the x-axis at points 3 and - 1.5, and the branches of the parabola are directed downward, since the older coefficient- negative number - 2. In fig. 118 is a sketch of a graph.

3) Using fig. 118, we conclude:< 0 на тех промежутках оси х, где график расположен ниже оси х, т.е. на открытом луче (-оо, -1,5) или на открытом луче C, +оо).
Answer: x< -1,5; х > 3.

Example 3 Solve the inequality 4x 2 - 4x + 1< 0.
Solution.

1) From the equation 4x 2 - 4x + 1 = 0 we find.

2) The square trinomial has one root; this means that the parabola serving as the graph of a square trinomial does not intersect the x-axis, but touches it at the point. The branches of the parabola are directed upwards (Fig. 119.)

3) Using the geometric model shown in fig. 119, we establish that the specified inequality is satisfied only at the point, since for all other values of x, the ordinates of the graph are positive.
Answer: .
You probably noticed that in fact, in examples 1, 2, 3, a well-defined algorithm solving quadratic inequalities, we will formalize it.

The algorithm for solving the quadratic inequality ax 2 + bx + 0 0 (ax 2 + bx + c< 0)

The first step of this algorithm is to find the roots of a square trinomial. But the roots may not exist, so what to do? Then the algorithm is inapplicable, which means that it is necessary to reason differently. The key to these arguments is given by the following theorems.

In other words, if D< 0, а >0, then the inequality ax 2 + bx + c > 0 is satisfied for all x; on the contrary, the inequality ax 2 + bx + c< 0 не имеет решений.
Proof. schedule functions y \u003d ax 2 + bx + c is a parabola whose branches are directed upwards (since a > 0) and which does not intersect the x axis, since the square trinomial has no roots by condition. The graph is shown in fig. 120. We see that for all x the graph is located above the x axis, which means that for all x the inequality ax 2 + bx + c > 0 is satisfied, which was required to be proved.

In other words, if D< 0, а < 0, то неравенство ах 2 + bх + с < 0 выполняется при всех х; напротив, неравенство ах 2 + bх + с >0 has no solutions.

Proof. The graph of the function y \u003d ax 2 + bx + c is a parabola, the branches of which are directed downwards (since a< 0) и которая не пересекает ось х, так как корней у квадратного трехчлена по условию нет. График представлен на рис. 121. Видим, что при всех х график расположен ниже оси х, а это значит, что при всех х выполняется неравенство ах 2 + bх + с < 0, что и требовалось доказать.

Example 4. Solve the inequality:

a) 2x 2 - x + 4 > 0; b) -x 2 + Zx - 8 > 0.

a) Find the discriminant of the square trinomial 2x 2 - x + 4. We have D \u003d (-1) 2 - 4 2 4 \u003d - 31< 0.
The senior coefficient of the trinomial (number 2) is positive.

Hence, by Theorem 1, for all x, the inequality 2x 2 - x + 4 > 0 is satisfied, i.e., the solution to the given inequality is the whole (-00, + 00).

b) Find the discriminant of the square trinomial - x 2 + Zx - 8. We have D \u003d Z2 - 4 (- 1) (- 8) \u003d - 23< 0. Старший коэффициент трехчлена (число - 1) отрицателен. Следовательно, по теореме 2, при всех х выполняется неравенство - х 2 + Зx - 8 < 0. Это значит, что неравенство - х 2 + Зх - 8 0 не выполняется ни при каком значении х, т. е. заданное неравенство не имеет решений.

Answer: a) (-00, + 00); b) there are no solutions.

In the following example, we will get acquainted with another way of reasoning, which is used in solving quadratic inequalities.

Example 5 Solve the inequality 3x 2 - 10x + 3< 0.
Solution. Let us factorize the square trinomial 3x 2 - 10x + 3. The roots of the trinomial are the numbers 3 and, therefore, using ax 2 + bx + c \u003d a (x - x 1) (x - x 2), we get Zx 2 - 10x + 3 \u003d 3 (x - 3) (x - )
We note on the number line the roots of the trinomial: 3 and (Fig. 122).

Let x > 3; then x-3>0 and x->0, and hence the product 3(x - 3)(x - ) is positive. Next, let< х < 3; тогда x-3< 0, а х- >0. Therefore, the product 3(x-3)(x-) is negative. Finally, let x<; тогда x-3< 0 и x- < 0. Но в таком случае произведение
3(x -3)(x -) is positive.

Summing up the reasoning, we come to the conclusion: the signs of the square trinomial Zx 2 - 10x + 3 change as shown in Fig. 122. We are interested in for what x the square trinomial takes negative values. From fig. 122 we conclude: the square trinomial 3x 2 - 10x + 3 takes negative values for any value of x from the interval (, 3)
Answer (, 3), or< х < 3.

Comment. The method of reasoning that we applied in Example 5 is usually called the method of intervals (or the method of intervals). It is actively used in mathematics to solve rational inequalities. In 9th grade, we will study the interval method in more detail.

Example 6. At what values of the parameter p is the quadratic equation x 2 - 5x + p 2 \u003d 0:
a) has two different roots;

b) has one root;

c) has no -roots?

Solution. The number of roots of a quadratic equation depends on the sign of its discriminant D. In this case, we find D \u003d 25 - 4p 2.

a) A quadratic equation has two different roots, if D> 0, then the problem is reduced to solving the inequality 25 - 4p 2 > 0. We multiply both parts of this inequality by -1 (remembering to change the inequality sign). We obtain an equivalent inequality 4p 2 - 25< 0. Далее имеем 4 (р - 2,5) (р + 2,5) < 0.

The signs of the expression 4(p - 2.5) (p + 2.5) are shown in fig. 123.

We conclude that the inequality 4(p - 2.5)(p + 2.5)< 0 выполняется для всех значений р из интервала (-2,5; 2,5). Именно при этих значениях параметра р данное квадратное уравнение имеет два различных корня.

b) quadratic equation has one root if D is 0.
As we stated above, D = 0 at p = 2.5 or p = -2.5.

It is for these values of the parameter p that this quadratic equation has only one root.

c) A quadratic equation has no roots if D< 0. Решим неравенство 25 - 4р 2 < 0.

We get 4p 2 - 25 > 0; 4 (p-2.5) (p + 2.5)> 0, whence (see Fig. 123) p< -2,5; р >2.5. For these values of the parameter p, this quadratic equation has no roots.

Answer: a) at p (-2.5, 2.5);

b) at p = 2.5 or p = -2.5;
c) at r< - 2,5 или р > 2,5.

Mordkovich A. G., Algebra. Grade 8: Proc. for general education institutions. - 3rd ed., finalized. - M.: Mnemosyne, 2001. - 223 p.: ill.

Help a student online, Mathematics for grade 8 download, calendar-thematic planning

see also Solving a linear programming problem graphically, Canonical form of linear programming problems

The system of constraints for such a problem consists of inequalities in two variables:
and the objective function has the form F = C 1 x + C 2 y, which is to be maximized.

Let's answer the question: what pairs of numbers ( x; y) are solutions to the system of inequalities, i.e., do they satisfy each of the inequalities simultaneously? In other words, what does it mean to solve a system graphically?
First you need to understand what is the solution of one linear inequality with two unknowns.
To solve a linear inequality with two unknowns means to determine all pairs of values of the unknowns for which the inequality is satisfied.
For example, inequality 3 x – 5y≥ 42 satisfy the pairs ( x , y) : (100, 2); (3, –10), etc. The problem is to find all such pairs.
Consider two inequalities: ax + by≤ c, ax + by≥ c. Straight ax + by = c divides the plane into two half-planes so that the coordinates of the points of one of them satisfy the inequality ax + by >c, and the other inequality ax + +by <c.
Indeed, take a point with coordinate x = x 0; then a point lying on a straight line and having an abscissa x 0 , has an ordinate

Let for definiteness a<0, b>0, c>0. All points with abscissa x 0 above P(e.g. dot M), have y M>y 0 , and all points below the point P, with abscissa x 0 , have yN<y 0 . Insofar as x 0 is an arbitrary point, then there will always be points on one side of the line for which ax+ by > c, forming a half-plane, and on the other hand, points for which ax + by< c.

Picture 1

The inequality sign in the half-plane depends on the numbers a, b , c.
This implies the following method for graphical solution of systems of linear inequalities in two variables. To solve the system, you need:

For each inequality, write down the equation corresponding to the given inequality.
Construct lines that are graphs of functions given by equations.
For each straight line, determine the half-plane, which is given by the inequality. To do this, take an arbitrary point that does not lie on a straight line, substitute its coordinates into the inequality. if the inequality is true, then the half-plane containing the chosen point is the solution to the original inequality. If the inequality is false, then the half-plane on the other side of the line is the set of solutions to this inequality.
To solve a system of inequalities, it is necessary to find the area of intersection of all half-planes that are the solution to each inequality in the system.

This area may turn out to be empty, then the system of inequalities has no solutions, it is inconsistent. Otherwise, the system is said to be compatible.
Solutions can be a finite number and an infinite set. The area can be a closed polygon or it can be unlimited.

Let's look at three relevant examples.

Example 1. Graphically solve the system:
x + y- 1 ≤ 0;
–2x- 2y + 5 ≤ 0.

consider the equations x+y–1=0 and –2x–2y+5=0 corresponding to the inequalities;
let us construct the straight lines given by these equations.

Figure 2

Let us define the half-planes given by the inequalities. Take an arbitrary point, let (0; 0). Consider x+ y– 1 0, we substitute the point (0; 0): 0 + 0 – 1 ≤ 0. hence, in the half-plane where the point (0; 0) lies, x + y – 1 ≤ 0, i.e. the half-plane lying below the straight line is the solution to the first inequality. Substituting this point (0; 0) into the second one, we get: –2 ∙ 0 – 2 ∙ 0 + 5 ≤ 0, i.e. in the half-plane where the point (0; 0) lies, -2 x – 2y+ 5≥ 0, and we were asked where -2 x – 2y+ 5 ≤ 0, therefore, in another half-plane - in the one above the straight line.
Find the intersection of these two half-planes. The lines are parallel, so the planes do not intersect anywhere, which means that the system of these inequalities has no solutions, it is inconsistent.

Example 2. Find graphically solutions to the system of inequalities:

Figure 3
1. Write down the equations corresponding to the inequalities and construct straight lines.
x + 2y– 2 = 0

x	2	0
y	0	1

y – x – 1 = 0

x	0	2
y	1	3

y + 2 = 0;
y = –2.
2. Having chosen the point (0; 0), we determine the signs of inequalities in the half-planes:
0 + 2 ∙ 0 – 2 ≤ 0, i.e. x + 2y– 2 ≤ 0 in the half-plane below the straight line;
0 – 0 – 1 ≤ 0, i.e. y –x– 1 ≤ 0 in the half-plane below the straight line;
0 + 2 =2 ≥ 0, i.e. y+ 2 ≥ 0 in the half-plane above the line.
3. The intersection of these three half-planes will be an area that is a triangle. It is not difficult to find the vertices of the region as the points of intersection of the corresponding lines

In this way, BUT(–3; –2), IN(0; 1), FROM(6; –2).

Let us consider one more example, in which the resulting domain of the solution of the system is not limited.

Solving inequalities online

Before solving inequalities, it is necessary to understand well how equations are solved.

It doesn’t matter whether the inequality is strict () or non-strict (≤, ≥), the first step is to solve the equation by replacing the inequality sign with equality (=).

Explain what it means to solve an inequality?

After studying the equations, the student has the following picture in his head: you need to find such values of the variable for which both parts of the equation take the same values. In other words, find all points where the equality holds. Everything is correct!

When talking about inequalities, they mean finding the intervals (segments) on which the inequality holds. If there are two variables in the inequality, then the solution will no longer be intervals, but some areas on the plane. Guess what will be the solution of the inequality in three variables?

How to solve inequalities?

The method of intervals (aka the method of intervals) is considered to be a universal way to solve inequalities, which consists in determining all the intervals within which the given inequality will be fulfilled.

Without going into the type of inequality, in this case it is not the essence, it is required to solve the corresponding equation and determine its roots, followed by the designation of these solutions on the numerical axis.

What is the correct way to write the solution to an inequality?

When you have determined the intervals for solving the inequality, you need to correctly write out the solution itself. There is an important nuance - are the boundaries of the intervals included in the solution?

Everything is simple here. If the solution of the equation satisfies the ODZ and the inequality is not strict, then the boundary of the interval is included in the solution of the inequality. Otherwise, no.

Considering each interval, the solution to the inequality can be the interval itself, or a half-interval (when one of its boundaries satisfies the inequality), or a segment - an interval together with its boundaries.

Important point

Do not think that only intervals, half-intervals and segments can be the solution to an inequality. No, individual points can also be included in the solution.

For example, the inequality |x|≤0 has only one solution - point 0.

And the inequality |x|

What is the inequality calculator for?

The inequality calculator gives the correct final answer. In this case, in most cases, an illustration of a numerical axis or plane is given. You can see whether the boundaries of the intervals are included in the solution or not - the points are displayed filled or pierced.

Thanks to the online inequality calculator, you can check whether you have found the roots of the equation correctly, marked them on the number line and checked the inequality conditions on the intervals (and boundaries)?

If your answer differs from the answer of the calculator, then you definitely need to double-check your solution and identify the mistake made.

Inequality is an expression with, ≤, or ≥. For example, 3x - 5 To solve an inequality means to find all values of the variables for which this inequality is true. Each of these numbers is a solution to the inequality, and the set of all such solutions is its many solutions. Inequalities that have the same set of solutions are called equivalent inequalities.

Linear inequalities

The principles for solving inequalities are similar to the principles for solving equations.

Principles for solving inequalities
For any real numbers a, b, and c :
The principle of adding inequalities: If a Multiplication principle for inequalities: If a 0 is true, then ac If a bc is also true.
Similar statements also apply for a ≤ b.

When both sides of an inequality are multiplied by a negative number, the sign of the inequality needs to be reversed.
First-level inequalities, as in Example 1 (below), are called linear inequalities.

Example 1 Solve each of the following inequalities. Then draw a set of solutions.
a) 3x - 5 b) 13 - 7x ≥ 10x - 4
Solution
Any number less than 11/5 is a solution.
The set of solutions is (x|x
To make a check, we can plot y 1 = 3x - 5 and y 2 = 6 - 2x. Then it can be seen from here that for x
The solution set is (x|x ≤ 1), or (-∞, 1]. The graph of the solution set is shown below.

Double inequalities

When two inequalities are connected by a word And, or, then it is formed double inequality. Double inequality like
-3 And 2x + 5 ≤ 7
called connected because it uses And. Record -3 Double inequalities can be solved using the principles of addition and multiplication of inequalities.

Example 2 Solve -3 Solution We have

Set of solutions (x|x ≤ -1 or x > 3). We can also write the solution using the spacing notation and the symbol for associations or inclusions of both sets: (-∞ -1] (3, ∞). The graph of the set of solutions is shown below.

To test, draw y 1 = 2x - 5, y 2 = -7, and y 3 = 1. Note that for (x|x ≤ -1 or x > 3), y 1 ≤ y 2 or y 1 > y 3 .

Inequalities with absolute value (modulus)

Inequalities sometimes contain modules. The following properties are used to solve them.
For a > 0 and an algebraic expression x:
|x| |x| > a is equivalent to x or x > a.
Similar statements for |x| ≤ a and |x| ≥ a.

For example,
|x| |y| ≥ 1 is equivalent to y ≤ -1 or y ≥ 1;
and |2x + 3| ≤ 4 is equivalent to -4 ≤ 2x + 3 ≤ 4.

Example 4 Solve each of the following inequalities. Plot the set of solutions.
a) |3x + 2| b) |5 - 2x| ≥ 1

Solution
a) |3x + 2|

The solution set is (x|-7/3