# Some points about how inequalities are satisfied

One of the topics that requires maximum attention and perseverance from students is the solution of inequalities. These are similar to the equations and are very different from them. Because they need a special approach to their solution.

## Properties that will be required to find the answer

All of them are used to replace the existing record with the equivalent. Most of them are similar to what was in the equations. But there are differences.

• A function that is defined in the DSL, or any number can be added to both parts of the original inequality.
• Similarly, multiplication is possible, but only on a positive function or number.
• If this action is performed with a negative function or number, then the inequality sign should be replaced by the opposite one.
• Functions that are nonnegative can be raised to a positive degree. Sometimes the solution of inequalities is accompanied by actions that give extraneous answers. They need to be eliminated by comparing the area of ​​th

e DGS and the set of solutions.

## Using the interval method

Its essence is to reduce the inequality to an equation in which the right-hand side is zero. 1. Define the area where the permissible values ​​of the variables lie, that is, the DSA.
2. Convert the inequality using mathematical operations so that there is a zero in its right side.
3. Replace the inequality sign by "=" and solve the corresponding equation.
4. On the numeric axis, note all the answers that were obtained during the decision, as well as the intervals of the LDZ.With strict inequality, the points need to be punctured. If there is an equal sign, then they are supposed to be painted over.
5. Determine the sign of the initial function on each interval obtained from the points of the DSU and dividing its answers. If the sign of a function does not change when passing through a point, it enters the answer. Otherwise, it is excluded.
6. The boundary points for the LDZ point need to be checked additionally and only then included or not in response.
7. The answer that you get should be written in the form of merged sets.

## A bit about the double inequalities

They use two signs of inequality in the record at once. That is, some function is bounded by conditions at once twice. Such inequalities are solved as a system of two, when the original is divided into parts. And in the interval method, the answers from the solution of both equations are indicated.

For their solution, it is also permissible to use the properties mentioned above. With their help, it is convenient to reduce the inequality to zero. ## How are things with the inequalities in which there is a module?

In this case, the solution of the inequalities uses the following properties, and they are valid for a positive value of "a".

If "x" takes an algebraic expression, then such substitutions are valid:

• | x |& lt;a to -a & lt;x & lt;a;
• | x |& gt;a by x & lt;-a or x & gt;a.

If the inequalities are not strict, then the formulas are also true, only in them, except for the sign more or less, appears "=".

## How is the system of inequalities solved?

This knowledge is required in cases where such a task is given or there is a record of a double inequality or a module appeared in the record. In such a situation, the solution will be those values ​​of variables that satisfy all the inequalities in the notation. If there are no such numbers, then the system of solutions does not.

Plan for the solution of the system of inequalities:

• solve each of them separately;
• show on the numerical axis all intervals and determine their intersections;
• record the response of the system, which will be the union of what happened in the second paragraph. ## How to deal with fractional inequalities?

Since at the time of their decision it may be necessary to change the sign of inequality, it is necessary to carefully and carefully follow all the points of the plan. Otherwise, you might get the opposite answer.

The solution of fractional inequalities also uses the interval method. And the action plan will be as follows:

• Using the described properties, to give fractions such a kind that to the right of the sign there was only zero.
• Replace the inequality by "=" and determine the points at which the function will be zero.
• Mark them on the coordinate axis. In this case, the numbers resulting from the calculations in the denominator will always be punctured. All others are based on the inequality condition.
• Define intervals of sign-constant.
• In response, write the union of those gaps, the sign of which corresponds to the one that was in the original inequality.

## Situations when the irrationality of the

appears in the inequality In other words, the entry has a mathematical root. Since in the school course of algebra most of the assignments are for the square root, then it will be considered.

The solution of irrational inequalities reduces to obtaining a system of two or three that will be equivalent to the original one.

 Initial inequality condition equivalent system √ n( x) & lt;m( x) m( x) is less than or equal to 0 no solutions m( x) greater than 0 n( x) greater than or equal to 0 n( x) & lt;(m( x)) 2 √ n( x) & gt;m( x) m( x) is greater than or equal to 0 n( x) & gt;(m( x)) 2 or n( x) is greater than or equal to 0 m( x) is less than 0 √n( x) ≤ m( x) m( x) is less than 0 no solutions m) is greater than or equal to 0 n( x) is greater than or equal to 0 n( x) ≤( m( x)) 2 √n( x) ≥ m( x) m( x) greater than or equal to 0 nx) ≥( m( x)) 2 or n( x) is greater than or equal to 0 m( x) is less than 0 √ n( x) & lt;√ m( x) n( x) is greater than or equal to 0 n( x) is less than m( x) √n( x) * m( x) & lt;0 n( x) is greater than 0 m( x) is less than 0 √n( x) * m( x) & gt;0 n( x) greater than 0 m( x) ≤0 or n( x) is greater than 0 m( x) is greater than 0 √n( x) * m( x) ≤ 0 n0 m( x) is any √n( x) * m( x) ≥ 0 n( x) is greater than 0 m( x) ≥0 or n( x) is 0 m( x) -any

## Examples of solving different types of inequalities

In order to add clarity to the theory of solving inequalities, examples are given below.

First example.2x4 & gt;1 + x

Solution: in order to determine the LDZ, it is enough just to look closely at the inequality. It is formed from linear functions, so it is defined for all values ​​of the variable.

Now from both parts of the inequality it is necessary to subtract( 1 + x).It turns out: 2x - 4 -( 1 + x) & gt;0. After the parentheses are opened and similar terms are given, the inequality will take the form: x - 5 & gt;0.

Equating it to zero, it's easy to find its solution: x = 5.

Now this point with the digit 5, it is necessary to note on the coordinate beam. Then check the signs of the original function. On the first interval from minus infinity to 5 we can take the number 0 and substitute it in the inequality obtained after the transformations. After the calculations, -7 & gt; 0 is obtained.under the arc of the interval, you need to sign a minus sign.

In the next interval from 5 to infinity, the number 6 can be chosen. Then it turns out that 1 & gt;0. The sign "+" was signed under the arc. This second interval is the answer of inequality.

Answer: x lies in the interval( 5; ∞). The second example. It is required to solve a system of two equations: 3x + 3 ≤ 2x + 1 and 3x - 2 ≤ 4x + 2.

Solution. The ODZ of these inequalities also lies in the range of any numbers, since linear functions are given.

You need to proceed step by step. First, transform the first of the inequalities and equate it to zero.3x + 3 - 2x - 1 = 0. That is, x + 2 = 0. Thus, x is -2.

The second inequality takes the form of such an equation: 3x - 2 - 4x - 2 = 0. After the transformation: -x - 4 = 0.From it, the value for the variable is -4.

These two numbers must be marked on the axis, spacing out. Since the inequality is not strict, all points need to be painted over. The first interval is from minus infinity to -4.Let the number -5 be chosen. The first inequality gives a value of -3, and the second one. Hence, this interval does not enter into the answer.

The second interval is -4 to -2.You can choose a number -3 and substitute it in both inequalities. The first and the second result is -1.Hence, under the arc "-".

In the last interval from -2 to infinity, the best number is zero. It is necessary to substitute it and find the values ​​of inequalities. In the first of them a positive number is obtained, and the second is zero. This gap must also be excluded from the answer.

Of three intervals, there is only one solution to the inequality. The third example.| 1 - x |& gt;2 | x - 1 |.

Solution. The first thing to do is to determine the points at which the functions go to zero. For the left, this number will be 2, for the right one - 1. they should be noted on the ray and determine intervals of sign-constant.

On the first interval, from minus infinity to 1, the function on the left-hand side of the inequality takes positive values, and from the right-hand side negative. Under the arc, you need to write down two "+" and "-" signs next to each other.

The next interval is from 1 to 2. On it both functions take positive values. So under the arc there are two pluses.

The third interval from 2 to infinity will yield the following result: the left function is negative, the right one is positive.

Given the resulting signs, you need to calculate the values ​​of the inequality for all gaps.

On the first one, the following inequality is obtained: 2 - x & gt;- 2( x - 1).The minus before the two in the second inequality is due to the fact that this function is negative.

After the conversion, the inequality looks like this: x & gt;0. It immediately gives the values ​​of the variable. That is, from this interval only the interval from 0 to 1 will go in response.

On the second: 2 - x & gt;2( x - 1).The transformations will give the following inequality: -3x + 4 greater than zero. Its zero will be the value x = 4/3.Taking into account the inequality sign, it turns out that x must be less than this number. Hence, this interval is reduced to the interval from 1 to 4/3.

The latter gives the following inequality: -( 2 - x) & gt;2( x - 1).Its conversion results in the following: -x & gt;0. That is, the equation is true for x less than zero. This means that inequality does not give solutions to the required interval.

On the first two intervals, the number 1 was the boundary. It must be checked separately. That is, substitute in the original inequality. It turns out: | 2 - 1 |& gt;2 | 1 - 1 |.Counting gives that 1 is greater than 0. This is a true statement, so the unit is included in the answer.

Answer: x lies in the gap( 0; 4/3).

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