What is the solution to the inequality. Solving Linear Inequalities

First, a little lyrics to get a feel for the problem that the interval method solves. Let's say we need to solve the following inequality:

(x − 5)(x + 3) > 0

What are the options? The first thing that comes to mind for most students is the rules “plus on plus gives plus” and “minus on minus gives plus.” Therefore, it is enough to consider the case when both brackets are positive: x − 5 > 0 and x + 3 > 0. Then we also consider the case when both brackets are negative: x − 5< 0 и x + 3 < 0. Таким образом, наше неравенство свелось к совокупности двух систем, которая, впрочем, легко решается:

More advanced students will (maybe) remember that on the left is quadratic function, whose graph is a parabola. Moreover, this parabola intersects the OX axis at points x = 5 and x = −3. For further work you need to open the brackets. We have:

x 2 − 2x − 15 > 0

Now it is clear that the branches of the parabola are directed upward, because coefficient a = 1 > 0. Let's try to draw a diagram of this parabola:

The function is greater than zero where it passes above the OX axis. In our case, these are the intervals (−∞ −3) and (5; +∞) - this is the answer.

Please note: the picture shows exactly function diagram, not her schedule. Because for a real graph you need to count coordinates, calculate displacements and other crap that we have absolutely no use for now.

Why are these methods ineffective?

So, we have considered two solutions to the same inequality. Both of them turned out to be quite cumbersome. The first decision arises - just think about it! — a set of systems of inequalities. The second solution is also not particularly easy: you need to remember the graph of the parabola and a bunch of other small facts.

It was a very simple inequality. It only has 2 multipliers. Now imagine that there will be not 2, but at least 4 multipliers. For example:

(x − 7)(x − 1)(x + 4)(x + 9)< 0

How to solve such inequality? Go through all possible combinations of pros and cons? Yes, we will fall asleep faster than we find a solution. Drawing a graph is also not an option, since it is not clear how such a function behaves on the coordinate plane.

For such inequalities, a special solution algorithm is needed, which we will consider today.

What is the interval method

The interval method is a special algorithm designed to solve complex inequalities of the form f (x) > 0 and f (x)< 0. Алгоритм состоит из 4 шагов:

1. Solve the equation f (x) = 0. Thus, instead of an inequality, we get an equation that is much simpler to solve;
2. Mark all obtained roots on the coordinate line. Thus, the straight line will be divided into several intervals;
3. Find out the sign (plus or minus) of the function f (x) on the rightmost interval. To do this, it is enough to substitute into f (x) any number that will be to the right of all marked roots;
4. Mark the signs at the remaining intervals. To do this, just remember that when passing through each root, the sign changes.

That's all! After this, all that remains is to write down the intervals that interest us. They are marked with a “+” sign if the inequality was of the form f (x) > 0, or with a “−” sign if the inequality was of the form f (x)< 0.

At first glance, it may seem that the interval method is some kind of tinny thing. But in practice everything will be very simple. Just practice a little and everything will become clear. Take a look at the examples and see for yourself:

Task. Solve the inequality:

(x − 2)(x + 7)< 0

We work using the interval method. Step 1: replace the inequality with an equation and solve it:

(x − 2)(x + 7) = 0

The product is zero if and only if at least one of the factors is zero:

x − 2 = 0 ⇒ x = 2;
x + 7 = 0 ⇒ x = −7.

We got two roots. Let's move on to step 2: mark these roots on the coordinate line. We have:

Now step 3: find the sign of the function on the rightmost interval (to the right of the marked point x = 2). To do this, you need to take any number that more number x = 2. For example, let's take x = 3 (but no one forbids taking x = 4, x = 10 and even x = 10,000). We get:

f (x) = (x − 2)(x + 7);
x = 3;
f (3) = (3 − 2)(3 + 7) = 1 10 = 10;

We find that f (3) = 10 > 0, so we put a plus sign in the rightmost interval.

Let's move on to the last point - we need to note the signs on the remaining intervals. We remember that when passing through each root the sign must change. For example, to the right of the root x = 2 there is a plus (we made sure of this in the previous step), so there must be a minus to the left.

This minus extends to the entire interval (−7; 2), so there is a minus to the right of the root x = −7. Therefore, to the left of the root x = −7 there is a plus. It remains to mark these signs on the coordinate axis. We have:

Let's return to the original inequality, which had the form:

(x − 2)(x + 7)< 0

So the function must be less than zero. This means that we are interested in the minus sign, which appears only on one interval: (−7; 2). This will be the answer.

Task. Solve the inequality:

(x + 9)(x − 3)(1 − x )< 0

Step 1: set the left side to zero:

(x + 9)(x − 3)(1 − x ) = 0;
x + 9 = 0 ⇒ x = −9;
x − 3 = 0 ⇒ x = 3;
1 − x = 0 ⇒ x = 1.

Remember: the product is equal to zero when at least one of the factors is equal to zero. That is why we have the right to equate each individual bracket to zero.

Step 2: mark all the roots on the coordinate line:

Step 3: find out the sign of the rightmost gap. We take any number that is greater than x = 1. For example, we can take x = 10. We have:

f (x) = (x + 9)(x − 3)(1 − x);
x = 10;
f (10) = (10 + 9)(10 − 3)(1 − 10) = 19 · 7 · (−9) = − 1197;
f (10) = −1197< 0.

Step 4: placing the remaining signs. We remember that when passing through each root the sign changes. As a result, our picture will look like this:

That's all. All that remains is to write out the answer. Take another look at the original inequality:

(x + 9)(x − 3)(1 − x )< 0

This is an inequality of the form f(x)< 0, т.е. нас интересуют интервалы, отмеченные знаком минус. А именно:

x ∈ (−9; 1) ∪ (3; +∞)

This is the answer.

A note about function signs

Practice shows that the greatest difficulties in the interval method arise in the last two steps, i.e. when placing signs. Many students begin to get confused: which numbers to take and where to put the signs.

To finally understand the interval method, consider two observations on which it is based:

1. A continuous function changes sign only at those points where it is equal to zero. Such points split the coordinate axis into pieces, within which the sign of the function never changes. That's why we solve the equation f (x) = 0 and mark the found roots on the straight line. The numbers found are “borderline” points separating the pros and cons.
2. To find out the sign of a function on any interval, it is enough to substitute any number from this interval into the function. For example, for the interval (−5; 6) we have the right to take x = −4, x = 0, x = 4 and even x = 1.29374 if we want. Why is it important? Yes, because doubts begin to gnaw at many students. Like, what if for x = −4 we get a plus, and for x = 0 we get a minus? But nothing like this will ever happen. All points on the same interval give the same sign. Remember this.

That's all you need to know about the interval method. Of course, we took it apart simple version. There are more complex inequalities - non-strict, fractional and with repeated roots. You can also use the interval method for them, but this is a topic for a separate large lesson.

Now I would like to look at an advanced technique that dramatically simplifies the interval method. More precisely, the simplification affects only the third step - calculating the sign on the rightmost piece of the line. For some reason, this technique is not taught in schools (at least no one explained this to me). But in vain - because in fact this algorithm is very simple.

So, the sign of the function is on the right piece of the number line. This piece has the form (a ; +∞), where a is the largest root of the equation f (x) = 0. In order not to blow your mind, let’s consider a specific example:

(x − 1)(2 + x )(7 − x )< 0;
f (x) = (x − 1)(2 + x)(7 − x);
(x − 1)(2 + x)(7 − x) = 0;
x − 1 = 0 ⇒ x = 1;
2 + x = 0 ⇒ x = −2;
7 − x = 0 ⇒ x = 7;

We got 3 roots. Let's list them in ascending order: x = −2, x = 1 and x = 7. Obviously, the largest root is x = 7.

For those who find it easier to reason graphically, I will mark these roots on the coordinate line. Let's see what happens:

It is required to find the sign of the function f (x) on the rightmost interval, i.e. to (7; +∞). But as we have already noted, to determine the sign you can take any number from this interval. For example, you can take x = 8, x = 150, etc. And now - the same technique that is not taught in schools: let's take infinity as a number. More precisely, plus infinity, i.e. +∞.

“Are you stoned? How can you substitute infinity into a function?” - you might ask. But think about it: we don’t need the value of the function itself, we only need the sign. Therefore, for example, the values ​​f (x) = −1 and f (x) = −938 740 576 215 mean the same thing: the function on this interval is negative. Therefore, all that is required of you is to find the sign that appears at infinity, and not the value of the function.

In fact, substituting infinity is very simple. Let's return to our function:

f (x) = (x − 1)(2 + x)(7 − x)

Imagine that x is very big number. Billion or even trillion. Now let's see what happens in each bracket.

First bracket: (x − 1). What happens if you subtract one from a billion? The result will be a number not much different from a billion, and this number will be positive. Similarly with the second bracket: (2 + x). If we add a billion to two, we get a billion and kopecks - this is positive number. Finally, the third bracket: (7 − x). Here there will be a minus billion, from which a pathetic piece in the form of a seven was “gnawed off”. Those. the resulting number will not differ much from minus billion - it will be negative.

All that remains is to find the sign of the entire work. Since we had a plus in the first brackets and a minus in the last, we get the following construction:

(+) · (+) · (−) = (−)

The final sign is minus! And it doesn’t matter what the value of the function itself is. The main thing is that this value is negative, i.e. the rightmost interval has a minus sign. All that remains is to complete the fourth step of the interval method: arrange all the signs. We have:

The original inequality was:

(x − 1)(2 + x )(7 − x )< 0

Therefore, we are interested in the intervals marked with a minus sign. We write out the answer:

x ∈ (−2; 1) ∪ (7; +∞)

That's the whole trick I wanted to tell you. In conclusion, here is another inequality that can be solved by the interval method using infinity. To visually shorten the solution, I will not write step numbers and detailed comments. I will only write what you really need to write when solving real problems:

Task. Solve the inequality:

x (2x + 8)(x − 3) > 0

We replace the inequality with an equation and solve it:

x (2x + 8)(x − 3) = 0;
x = 0;
2x + 8 = 0 ⇒ x = −4;
x − 3 = 0 ⇒ x = 3.

We mark all three roots on the coordinate line (with signs at once):

There is a plus on the right side of the coordinate axis, because the function looks like:

f (x) = x (2x + 8)(x − 3)

And if we substitute infinity (for example, a billion), we get three positive brackets. Since the original expression must be greater than zero, we are only interested in the positives. All that remains is to write out the answer:

x ∈ (−4; 0) ∪ (3; +∞)

For example, the inequality is the expression \(x>5\).

Types of inequalities:

If \(a\) and \(b\) are numbers or , then the inequality is called numerical. It's actually just comparing two numbers. Such inequalities are divided into faithful And unfaithful.

For example:
\(-5<2\) - верное числовое неравенство, ведь \(-5\) действительно меньше \(2\);

\(17+3\geq 115\) is an incorrect numerical inequality, since \(17+3=20\), and \(20\) is less than \(115\) (and not greater than or equal to).

If \(a\) and \(b\) are expressions containing a variable, then we have inequality with variable. Such inequalities are divided into types depending on the content:

\(2x+1\geq4(5-x)\)				Variable only to the first power
\(3x^2-x+5>0\)				There is a variable in the second power (square), but there are no higher powers (third, fourth, etc.)
\(\log_(4)((x+1))<3\)
\(2^(x)\leq8^(5x-2)\)

... and so on.

What is the solution to an inequality?

If you substitute a number instead of a variable into an inequality, it will turn into a numeric one.

If a given value for x turns the original inequality into a true numerical one, then it is called solution to inequality. If not, then this value is not a solution. And to solve inequality– you need to find all its solutions (or show that there are none).

For example, if we substitute the number \(7\) into the linear inequality \(x+6>10\), we get the correct numerical inequality: \(13>10\). And if we substitute \(2\), there will be an incorrect numerical inequality \(8>10\). That is, \(7\) is a solution to the original inequality, but \(2\) is not.

However, the inequality \(x+6>10\) has other solutions. Indeed, we will get the correct numerical inequalities when substituting \(5\), and \(12\), and \(138\)... And how can we find all possible solutions? For this they use For our case we have:

\(x+6>10\) \(|-6\)
\(x>4\)

That is, any number greater than four will suit us. Now you need to write down the answer. Solutions to inequalities are usually written numerically, additionally marking them on the number axis with shading. For our case we have:

Answer: \(x\in(4;+\infty)\)

When does the sign of an inequality change?

There is one big trap in inequalities that students really “love” to fall into:

When multiplying (or dividing) an inequality by a negative number, it is reversed (“more” by “less”, “more or equal” by “less than or equal”, and so on)

Why is this happening? To understand this, let's look at the transformations of the numerical inequality \(3>1\). It is correct, three is indeed greater than one. First, let's try to multiply it by any positive number, for example, two:

\(3>1\) \(|\cdot2\)
\(6>2\)

As we can see, after multiplication the inequality remains true. And no matter what positive number we multiply by, we will always get the correct inequality. Now let's try to multiply by a negative number, for example, minus three:

\(3>1\) \(|\cdot(-3)\)
\(-9>-3\)

The result is an incorrect inequality, because minus nine is less than minus three! That is, in order for the inequality to become true (and therefore, the transformation of multiplication by negative was “legal”), you need to reverse the comparison sign, like this: \(−9<− 3\).
With division it will work out the same way, you can check it yourself.

The rule written above applies to all types of inequalities, not just numerical ones.

Example: Solve the inequality \(2(x+1)-1<7+8x\)
Solution:

\(2x+2-1<7+8x\)	Let's move \(8x\) to the left, and \(2\) and \(-1\) to the right, not forgetting to change the signs
\(2x-8x<7-2+1\)
\(-6x<6\) \(|:(-6)\)	Let's divide both sides of the inequality by \(-6\), not forgetting to change from “less” to “more”
	Let's mark a numerical interval on the axis. Inequality, therefore we “prick out” the value \(-1\) itself and do not take it as an answer
	Let's write the answer as an interval

Answer: \(x\in(-1;\infty)\)

Inequalities and disability

Inequalities, just like equations, can have restrictions on , that is, on the values ​​of x. Accordingly, those values ​​that are unacceptable according to the DZ should be excluded from the range of solutions.

Example: Solve the inequality \(\sqrt(x+1)<3\)

Solution: It is clear that in order for the left side to be less than \(3\), the radical expression must be less than \(9\) (after all, from \(9\) just \(3\)). We get:

\(x+1<9\) \(|-1\)
\(x<8\)

All? Any value of x smaller than \(8\) will suit us? No! Because if we take, for example, the value \(-5\) that seems to fit the requirement, it will not be a solution to the original inequality, since it will lead us to calculating the root of a negative number.

\(\sqrt(-5+1)<3\)
\(\sqrt(-4)<3\)

Therefore, we must also take into account the restrictions on the value of X - it cannot be such that there is a negative number under the root. Thus, we have the second requirement for x:

\(x+1\geq0\)
\(x\geq-1\)

And for x to be the final solution, it must satisfy both requirements at once: it must be less than \(8\) (to be a solution) and greater than \(-1\) (to be admissible in principle). Plotting it on the number line, we have the final answer:

Answer: \(\left[-1;8\right)\)

What you need to know about inequality icons? Inequalities with icon more (> ), or less (< ) are called strict. With icons more or equal (≥ ), less or equal (≤ ) are called not strict. Icon not equal (≠ ) stands apart, but you also have to solve examples with this icon all the time. And we will decide.)

The icon itself does not have much influence on the solution process. But at the end of the decision, when choosing the final answer, the meaning of the icon appears in full force! This is what we will see below in examples. There are some jokes there...

Inequalities, like equalities, exist faithful and unfaithful. Everything is simple here, no tricks. Let's say 5 > 2 is a true inequality. 5 < 2 - incorrect.

This preparation works for inequalities any kind and simple to the point of horror.) You just need to correctly perform two (only two!) elementary actions. These actions are familiar to everyone. But, characteristically, mistakes in these actions are the main mistake in solving inequalities, yes... Therefore, these actions must be repeated. These actions are called as follows:

Identical transformations of inequalities.

Identical transformations of inequalities are very similar to identical transformations of equations. Actually, this is the main problem. The differences go over your head and... here you are.) Therefore, I will especially highlight these differences. So, the first identical transformation of inequalities:

1. The same number or expression can be added (subtracted) to both sides of the inequality. Any. This will not change the inequality sign.

In practice, this rule is used as a transfer of terms from the left side of the inequality to the right (and vice versa) with a change of sign. With a change in the sign of the term, not the inequality! The one-to-one rule is the same as the rule for equations. But the following identical transformations in inequalities differ significantly from those in equations. So I highlight them in red:

2. Both sides of the inequality can be multiplied (divided) by the same thingpositivenumber. For anypositive Will not change.

3. Both sides of the inequality can be multiplied (divided) by the same thingnegative number. For anynegativenumber. The inequality sign from thiswill change to the opposite.

You remember (I hope...) that the equation can be multiplied/divided by anything. And for any number, and for an expression with an X. If only it wasn't zero. This makes him, the equation, neither hot nor cold.) It does not change. But inequalities are more sensitive to multiplication/division.

A clear example for a long memory. Let us write an inequality that does not raise doubts:

5 > 2

Multiply both sides by +3, we get:

15 > 6

Any objections? There are no objections.) And if we multiply both sides of the original inequality by -3, we get:

15 > -6

And this is an outright lie.) A complete lie! Deception of the people! But as soon as you change the inequality sign to the opposite one, everything falls into place:

15 < -6

I’m not just swearing about lies and deception.) "Forgot to change the equal sign..."- This home error in solving inequalities. This trivial and simple rule has hurt so many people! Which they forgot...) So I’m swearing. Maybe I'll remember...)

Particularly attentive people will notice that inequality cannot be multiplied by an expression with an X. Respect to those who are attentive!) Why not? The answer is simple. We don’t know the sign of this expression with an X. It can be positive, negative... Therefore, we do not know which inequality sign to put after multiplication. Should I change it or not? Unknown. Of course, this restriction (the prohibition of multiplying/dividing an inequality by an expression with an x) can be circumvented. If you really need it. But this is a topic for other lessons.

That's all the identical transformations of inequalities. Let me remind you once again that they work for any inequalities Now you can move on to specific types.

Linear inequalities. Solution, examples.

Linear inequalities are inequalities in which x is in the first power and there is no division by x. Type:

x+3 > 5x-5

How are such inequalities resolved? They are very easy to solve! Namely: with the help of we reduce the most confusing linear inequality straight to the answer. That's the solution. I will highlight the main points of the decision. To avoid stupid mistakes.)

Let's solve this inequality:

x+3 > 5x-5

We solve it in exactly the same way as a linear equation. With the only difference:

We carefully monitor the inequality sign!

The first step is the most common. With X's - to the left, without X's - to the right... This is the first identical transformation, simple and trouble-free.) Just don't forget to change the signs of the transferred terms.

The inequality sign remains:

x-5x > -5-3

Here are similar ones.

The inequality sign remains:

4x > -8

It remains to apply the last identical transformation: divide both sides by -4.

Divide by negative number.

The inequality sign will change to the opposite:

X < 2

This is the answer.

This is how all linear inequalities are solved.

Attention! Point 2 is drawn white, i.e. unpainted. Empty inside. This means that she is not included in the answer! I drew her so healthy on purpose. Such a point (empty, not healthy!)) in mathematics is called punctured point.

The remaining numbers on the axis can be marked, but not necessary. Extraneous numbers that are not related to our inequality can be confusing, yes... You just need to remember that the numbers increase in the direction of the arrow, i.e. numbers 3, 4, 5, etc. are to the right are twos, and numbers are 1, 0, -1, etc. - to the left.

Inequality x < 2 - strict. X is strictly less than two. If in doubt, checking is simple. We substitute the dubious number into the inequality and think: “Two is less than two? No, of course!” Exactly. Inequality 2 < 2 incorrect. A two in return is not appropriate.

Is one okay? Certainly. Less... And zero is good, and -17, and 0.34... Yes, all numbers that are less than two are good! And even 1.9999.... At least a little bit, but less!

So let's mark all these numbers on the number axis. How? There are options here. Option one - shading. We move the mouse over the picture (or touch the picture on the tablet) and see that the area of ​​​​all x's that meet the condition x is shaded < 2 . That's all.

Let's look at the second option using the second example:

X ≥ -0,5

Draw an axis and mark the number -0.5. Like this:

Notice the difference?) Well, yes, it’s hard not to notice... This dot is black! Painted over. This means -0.5 is included in the answer. Here, by the way, the verification may confuse someone. Let's substitute:

-0,5 ≥ -0,5

How so? -0.5 is no more than -0.5! And there is more icon...

It's OK. In a non-strict inequality, everything that fits the icon is suitable. AND equals good, and more good. Therefore, -0.5 is included in the response.

So, we marked -0.5 on the axis; it remains to mark all the numbers that are greater than -0.5. This time I mark the area of ​​suitable x values bow(from the word arc), rather than shading. We hover the cursor over the drawing and see this bow.

There is no particular difference between the shading and the arms. Do as the teacher says. If there is no teacher, draw arches. In more complex tasks, shading is less obvious. You can get confused.

This is how linear inequalities are drawn on an axis. Let us move on to the next feature of the inequalities.

Writing the answer for inequalities.

The equations were good.) We found x and wrote down the answer, for example: x=3. There are two forms of writing answers in inequalities. One is in the form of final inequality. Good for simple cases. For example:

X< 2.

This is a complete answer.

Sometimes you need to write the same thing, but in a different form, at numerical intervals. Then the recording starts to look very scientific):

x ∈ (-∞; 2)

Under the icon ∈ the word is hidden "belongs".

The entry reads like this: x belongs to the interval from minus infinity to two not including. Quite logical. X can be any number from all possible numbers from minus infinity to two. There cannot be a double X, which is what the word tells us "not including".

And where in the answer is it clear that "not including"? This fact is noted in the answer round bracket immediately after the two. If the two were included, the bracket would be square. Here it is:]. The following example uses such a parenthesis.

Let's write down the answer: x ≥ -0,5 at intervals:

x ∈ [-0.5; +∞)

Read: x belongs to the interval from minus 0.5, including, to plus infinity.

Infinity can never be turned on. It's not a number, it's a symbol. Therefore, in such notations, infinity is always adjacent to a parenthesis.

This form of recording is convenient for complex answers consisting of several spaces. But - just for final answers. In intermediate results, where a further solution is expected, it is better to use the usual form, in the form of a simple inequality. We will deal with this in the relevant topics.

Popular tasks with inequalities.

The linear inequalities themselves are simple. Therefore, tasks often become more difficult. So it was necessary to think. This, if you’re not used to it, is not very pleasant.) But it’s useful. I will show examples of such tasks. Not for you to learn them, it's unnecessary. And in order not to be afraid when meeting such examples. Just think a little - and it’s simple!)

1. Find any two solutions to the inequality 3x - 3< 0

If it’s not very clear what to do, remember the main rule of mathematics:

If you don’t know what you need, do what you can!)

X < 1

And what? Nothing special. What are they asking us? We are asked to find two specific numbers that are the solution to an inequality. Those. fit the answer. Two any numbers. Actually, this is confusing.) A couple of 0 and 0.5 are suitable. A couple -3 and -8. There are an infinite number of these couples! Which answer is correct?!

I answer: everything! Any pair of numbers, each of which is less than one, will be the correct answer. Write which one you want. Let's move on.

2. Solve the inequality:

4x - 3 ≠ 0

Tasks in this form are rare. But, as auxiliary inequalities, when finding ODZ, for example, or when finding the domain of definition of a function, they occur all the time. Such a linear inequality can be solved as an ordinary linear equation. Only everywhere except the "=" sign ( equals) put a sign " ≠ " (not equal). This is how you approach the answer, with an inequality sign:

X ≠ 0,75

In more complex examples, it is better to do things differently. Make inequality out of equality. Like this:

4x - 3 = 0

Calmly solve it as taught and get the answer:

x = 0.75

The main thing is, at the very end, when writing down the final answer, do not forget that we found x, which gives equality. And we need - inequality. Therefore, we don’t really need this X.) And we need to write it down with the correct symbol:

X ≠ 0,75

This approach results in fewer errors. Those who solve equations automatically. And for those who don’t solve equations, inequalities are, in fact, of no use...) Another example of a popular task:

3. Find the smallest integer solution to the inequality:

3(x - 1) < 5x + 9

First we simply solve the inequality. We open the brackets, move them, bring similar ones... We get:

X > - 6

Didn't it work out that way!? Did you follow the signs!? And behind the signs of members, and behind the sign of inequality...

Let's think again. We need to find a specific number that matches both the answer and the condition "smallest integer". If it doesn’t dawn on you right away, you can just take any number and figure it out. Two over minus six? Certainly! Is there suitable number slightly less? Of course. For example, zero is greater than -6. And even less? We need the smallest thing possible! Minus three is more than minus six! You can already catch the pattern and stop stupidly going through numbers, right?)

Let's take a number closer to -6. For example, -5. The answer is fulfilled, -5 > - 6. Is it possible to find another number less than -5 but greater than -6? You can, for example, -5.5... Stop! We are told whole solution! Doesn't roll -5.5! What about minus six? Uh-uh! The inequality is strict, minus 6 is in no way less than minus 6!

Therefore, the correct answer is -5.

I hope everything is clear with the choice of value from the general solution. Another example:

4. Solve inequality:

7 < 3x+1 < 13

Wow! This expression is called triple inequality. Strictly speaking, this is an abbreviated form of a system of inequalities. But such triple inequalities still have to be solved in some tasks... It can be solved without any systems. According to the same identical transformations.

We need to simplify, bring this inequality to pure X. But... What should be transferred where?! This is where it’s time to remember that moving left and right is abbreviated form first identity transformation.

And the full form sounds like this: Any number or expression can be added/subtracted to both sides of the equation (inequality).

There are three parts here. So we will apply identical transformations to all three parts!

So, let's get rid of the one in the middle part of the inequality. Let's subtract one from the entire middle part. So that the inequality does not change, we subtract one from the remaining two parts. Like this:

7 -1< 3x+1-1 < 13-1

6 < 3x < 12

That’s better, right?) All that remains is to divide all three parts into three:

2 < X < 4

That's all. This is the answer. X can be any number from two (not including) to four (not including). This answer is also written at intervals; such entries will be in quadratic inequalities. There they are the most common thing.

At the end of the lesson I will repeat the most important thing. Success in solving linear inequalities depends on the ability to transform and simplify linear equations. If at the same time watch for the inequality sign, there won't be any problems. That's what I wish for you. No problems.)

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You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

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 solution set, so the bracket is square and the point on the line is indicated by a filled circle.
The answer will be: x)