Quadratic equations and how to solve them. Linear equations. Solution, examples

52. More complex examples equations.
Example 1.

5/(x – 1) – 3/(x + 1) = 15/(x 2 – 1)

The common denominator is x 2 – 1, since x 2 – 1 = (x + 1)(x – 1). Let's multiply both sides of this equation by x 2 – 1. We get:

or, after reduction,

5(x + 1) – 3(x – 1) = 15

5x + 5 – 3x + 3 = 15

2x = 7 and x = 3½

Let's consider another equation:

5/(x-1) – 3/(x+1) = 4(x 2 – 1)

Solving as above, we get:

5(x + 1) – 3(x – 1) = 4
5x + 5 – 3x – 3 = 4 or 2x = 2 and x = 1.

Let's see whether our equalities are justified if we replace x in each of the considered equations with the found number.

For the first example we get:

We see that there is no room for any doubts: we have found a number for x such that the required equality is justified.

For the second example we get:

5/(1-1) – 3/2 = 15/(1-1) or 5/0 – 3/2 = 15/0

Here doubts arise: we are faced with division by zero, which is impossible. If in the future we manage to give a certain, albeit indirect, meaning to this division, then we can agree that the found solution x – 1 satisfies our equation. Until then, we must admit that our equation does not have a solution that has a direct meaning.

Such cases can occur when the unknown is somehow included in the denominators of the fractions present in the equation, and some of these denominators, when the solution is found, turn to zero.

Example 2.

You can immediately see that this equation has the form of a proportion: the ratio of the number x + 3 to the number x – 1 is equal to the ratio of the number 2x + 3 to the number 2x – 2. Let someone, in view of this circumstance, decide to apply here to free the equation from fractions, the main property of proportion (the product of the extreme terms is equal to the product of the middle terms). Then he will get:

(x + 3) (2x – 2) = (2x + 3) (x – 1)

2x 2 + 6x – 2x – 6 = 2x 2 + 3x – 2x – 3.

Here, fears that we will not cope with this equation may be raised by the fact that the equation includes terms with x 2. However, we can subtract 2x 2 from both sides of the equation - this will not break the equation; then the terms with x 2 are destroyed and we get:

6x – 2x – 6 = 3x – 2x – 3

Let's move the unknown terms to the left and the known ones to the right - we get:

3x = 3 or x = 1

Remembering this equation

(x + 3)/(x – 1) = (2x + 3)/(2x – 2)

We will immediately notice that the found value for x (x = 1) makes the denominators of each fraction vanish; We must abandon such a solution until we have considered the question of division by zero.

If we also note that the application of the property of proportion has complicated matters and that a simpler equation could be obtained by multiplying both sides of the given by common denominator, namely on 2(x – 1) - after all, 2x – 2 = 2 (x – 1), then we get:

2(x + 3) = 2x – 3 or 2x + 6 = 2x – 3 or 6 = –3,

which is impossible.

This circumstance indicates that this equation does not have any solutions that have a direct meaning that would not turn the denominators of this equation to zero.
Let us now solve the equation:

(3x + 5)/(x – 1) = (2x + 18)/(2x – 2)

Let's multiply both sides of the equation 2(x – 1), i.e. by a common denominator, we get:

6x + 10 = 2x + 18

The found solution does not make the denominator vanish and has a direct meaning:

or 11 = 11

If someone, instead of multiplying both parts by 2(x – 1), were to use the property of proportion, they would get:

(3x + 5)(2x – 2) = (2x + 18)(x – 1) or
6x 2 + 4x – 10 = 2x 2 + 16x – 18.

Here the terms with x 2 would not be destroyed. By transferring all unknown members to left side, and those known to the right would get

4x 2 – 12x = –8

x 2 – 3x = –2

Now we will not be able to solve this equation. In the future, we will learn how to solve such equations and find two solutions for it: 1) you can take x = 2 and 2) you can take x = 1. It’s easy to check both solutions:

1) 2 2 – 3 2 = –2 and 2) 1 2 – 3 1 = –2

If we remember the initial equation

(3x + 5) / (x – 1) = (2x + 18) / (2x – 2),

then we will see that now we get both of its solutions: 1) x = 2 is the solution that has a direct meaning and does not turn the denominator to zero, 2) x = 1 is the solution that turns the denominator to zero and does not have a direct meaning .

Example 3.

Let’s find the common denominator of the fractions included in this equation by factoring each of the denominators:

1) x 2 – 5x + 6 = x 2 – 3x – 2x + 6 = x(x – 3) – 2(x – 3) = (x – 3)(x – 2),

2) x 2 – x – 2 = x 2 – 2x + x – 2 = x (x – 2) + (x – 2) = (x – 2)(x + 1),

3) x 2 – 2x – 3 = x 2 – 3x + x – 3 = x (x – 3) + (x – 3) = (x – 3) (x + 1).

The common denominator is (x – 3)(x – 2)(x + 1).

Let's multiply both sides of this equation (and we can now rewrite it as:

by a common denominator (x – 3) (x – 2) (x + 1). Then, after reducing each fraction we get:

3(x + 1) – 2(x – 3) = 2(x – 2) or
3x + 3 – 2x + 6 = 2x – 4.

From here we get:

–x = –13 and x = 13.

This solution has a direct meaning: it does not make any of the denominators vanish.

If we took the equation:

then, doing exactly the same as above, we would get

3(x + 1) – 2(x – 3) = x – 2

3x + 3 – 2x + 6 = x – 2

3x – 2x – x = –3 – 6 – 2,

where would you get it from?

which is impossible. This circumstance shows that it is impossible to find a solution for the last equation that has a direct meaning.

In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

1. First of all, you need to expand the parentheses, if there are any (as in our last example);
2. Then combine similar
3. Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.

Then, as a rule, you need to bring similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.

In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the very simple tasks.

Scheme for solving simple linear equations

First, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the brackets, if any.
2. We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
3. We present similar terms.
4. We divide everything by the coefficient of “x”.

Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip them this stage. In the second step we need to isolate the variables. Note: we're talking about only about individual terms. Let's write it down:

We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:

\[\frac(6x)(6)=-\frac(72)(6)\]

So we got the answer.

Task No. 2

We can see the parentheses in this problem, so let's expand them:

Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:

Here are some similar ones:

At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.

Task No. 3

The third linear equation is more interesting:

\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]

There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:

We perform the second step already known to us:

\[-x+x+2x=15-6-12+3\]

Let's do the math:

We carry out the last step - divide everything by the coefficient of “x”:

\[\frac(2x)(x)=\frac(0)(2)\]

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, I would like to say the following:

• As I said above, not every linear equation has a solution - sometimes there are simply no roots;
• Even if there are roots, there may be zero among them - there is nothing wrong with that.

Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.

Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it by standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will allow you to avoid making stupid and offensive mistakes in high school, when doing such actions is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.

Example No. 1

Obviously, the first step is to open the brackets. Let's do this very carefully:

Now let's take a look at privacy:

\[-x+6((x)^(2))-6((x)^(2))+x=-12\]

Here are some similar ones:

Obviously, this equation has no solutions, so we’ll write this in the answer:

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

Let's move everything with a variable to the left, and without it - to the right:

Here are some similar ones:

Obviously, this linear equation has no solution, so we’ll write it this way:

\[\varnothing\],

or there are no roots.

Nuances of the solution

Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.

But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:

Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.

We do the same with the second equation:

It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence elementary transformations, where the inability to clearly and competently perform simple steps leads to the fact that high school students come to me and again learn to solve such simple equations.

Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.

Solving even more complex linear equations

What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.

Task No. 1

\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]

Let's multiply all the elements in the first part:

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

\[\frac(-4x)(4)=\frac(4)(-4)\]

Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.

Task No. 2

\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]

Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:

Now let’s carefully perform the multiplication in each term:

Let’s move the terms with “X” to the left, and those without - to the right:

\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.

About the algebraic sum

With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean simple design: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.

Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.

Solving equations with fractions

To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:

1. Open the brackets.
2. Separate variables.
3. Bring similar ones.
4. Divide by the ratio.

Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.

How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:

1. Get rid of fractions.
2. Open the brackets.
3. Separate variables.
4. Bring similar ones.
5. Divide by the ratio.

What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.

Example No. 1

\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]

Let's get rid of the fractions in this equation:

\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]

Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:

\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

\[-4x=-1\left| :\left(-4 \right) \right.\]

\[\frac(-4x)(-4)=\frac(-1)(-4)\]

We have received the final solution, let's move on to the second equation.

Example No. 2

\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]

Here we perform all the same actions:

\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]

\[\frac(4x)(4)=\frac(4)(4)\]

The problem is solved.

That, in fact, is all I wanted to tell you today.

Key points

Key findings are:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Don't worry if you see quadratic functions, most likely, in the process of further transformations they will decrease.
• There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.

I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!

Usually, equations appear in problems in which you need to find a certain quantity. The equation allows you to formulate the problem in the language of algebra. Having solved the equation, we obtain the value of the desired quantity, which is called the unknown. “Andrey has several rubles in his wallet. If you multiply this number by 2 and then subtract 5, you get 10. How much money does Andrey have?” Let's designate the unknown amount of money as x and write the equation: 2x-5=10.

To talk about ways to solve equations, first you need to define the basic concepts and become familiar with the generally accepted notations. For different types equations, there are various algorithms for solving them. The easiest way to solve equations is of the first degree with one unknown. Many people are familiar with the formula for solving quadratic equations from school. Techniques higher mathematics will help solve higher order equations. The set of numbers on which an equation is defined is closely related to its solutions. The relationship between equations and function graphs is also interesting, since representing equations graphically is a great help in solving them.

Description. An equation is a mathematical equality with one or more unknown quantities, for example 2x+3y=0.

Expressions on both sides of the equal sign are called left and right sides of the equation. The letters of the Latin alphabet indicate unknowns. Although there can be any number of unknowns, below we will only talk about equations with one unknown, which we will denote by x.

Degree of equation is the maximum power to which the unknown can be raised. For example,
$3x^4+6x-1=0$ is an equation of the fourth degree, $x-4x^2+6x=8$ is an equation of the second degree.

The numbers by which the unknown is multiplied are called coefficients. In the previous example, the unknown to the fourth power has a coefficient of 3. If, when replacing x with this number, the given equality is satisfied, then this number is said to satisfy the equation. It's called solving the equation, or its root. For example, 3 is the root, or solution, of the equation 2x+8=14, since 2*3+8=6+8=14.

Solving equations. Let's say we want to solve the equation 2x+5=11.

You can substitute some value x into it, for example x=2. Replace x with 2 and get: 2*2+5=4+5=9.

There's something wrong here because on the right side of the equation we should have gotten 11. Let's try x=3: 2*3+5=6+5=11.

The answer is correct. It turns out that if the unknown takes the value 3, then equality is satisfied. Therefore, we have shown that the number 3 is a solution to the equation.

The method we used to solve this equation is called selection method. Obviously it is inconvenient to use. Moreover, it cannot even be called a method. To verify this, just try to apply it to an equation of the form $x^4-5x^2+16=2365$.

Solution methods. There are so-called “rules of the game” that will be useful to familiarize yourself with. Our goal is to determine the value of the unknown that satisfies the equation. Therefore, it is necessary to identify the unknown in some way. To do this, it is necessary to transfer the terms of the equation from one part to another. The first rule for solving equations is...

1. When moving a member of an equation from one part to another, its sign changes to the opposite: plus changes to minus and vice versa. Consider as an example the equation 2x+5=11. Let's move 5 from the left side to the right: 2x=11-5. The equation will become 2x=6.

Let's move on to the second rule.
2. Both sides of the equation can be multiplied and divided by a number that is not equal to zero. Let's apply this rule to our equation: $x=\frac62=3$. On the left side of the equality, only the unknown x remained, therefore, we found its value and solved the equation.

We have just looked at the simplest problem - linear equation with one unknown. Equations of this type always have a solution, moreover, they can always be solved using the simplest operations: addition, subtraction, multiplication and division. Alas, not all equations are so simple. Moreover, their degree of complexity increases very quickly. For example, equations of the second degree can be easily solved by any student high school, but methods for solving systems of linear equations or equations of higher degrees are studied only in high school.

Linear equations. Solution, examples.

Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)

Linear equations.

Linear equations- not the most difficult topic in school mathematics. But there are some tricks there that can puzzle even a trained student. Let's figure it out?)

Typically a linear equation is defined as an equation of the form:

ax + b = 0 Where a and b– any numbers.

2x + 7 = 0. Here a=2, b=7

0.1x - 2.3 = 0 Here a=0.1, b=-2.3

12x + 1/2 = 0 Here a=12, b=1/2

Nothing complicated, right? Especially if you don’t notice the words: "where a and b are any numbers"... And if you notice and carelessly think about it?) After all, if a=0, b=0(any numbers are possible?), then we get a funny expression:

But that's not all! If, say, a=0, A b=5, This turns out to be something completely out of the ordinary:

Which is annoying and undermines confidence in mathematics, yes...) Especially during exams. But out of these strange expressions you also need to find X! Which doesn't exist at all. And, surprisingly, this X is very easy to find. We will learn to do this. In this lesson.

How to recognize a linear equation by its appearance? It depends what appearance.) The trick is that not only equations of the form are called linear equations ax + b = 0 , but also any equations that can be reduced to this form by transformations and simplifications. And who knows whether it comes down or not?)

A linear equation can be clearly recognized in some cases. Let's say, if we have an equation in which there are only unknowns to the first degree and numbers. And in the equation there is no fractions divided by unknown , it is important! And division by number, or a numerical fraction - that's welcome! For example:

This is a linear equation. There are fractions here, but there are no x's in the square, cube, etc., and no x's in the denominators, i.e. No division by x. And here is the equation

cannot be called linear. Here the X's are all in the first degree, but there are division by expression with x. After simplifications and transformations, you can get a linear equation, a quadratic equation, or anything you want.

It turns out that it is impossible to recognize the linear equation in some complicated example until you almost solve it. This is upsetting. But in assignments, as a rule, they don’t ask about the form of the equation, right? The assignments ask for equations decide. This makes me happy.)

Solving linear equations. Examples.

The entire solution of linear equations consists of identical transformations of the equations. By the way, these transformations (two of them!) are the basis of the solutions all equations of mathematics. In other words, the solution any the equation begins with these very transformations. In the case of linear equations, it (the solution) is based on these transformations and ends with a full answer. It makes sense to follow the link, right?) Moreover, there are also examples of solving linear equations there.

First, let's look at the simplest example. Without any pitfalls. Suppose we need to solve this equation.

x - 3 = 2 - 4x

This is a linear equation. The X's are all in the first power, there is no division by X's. But, in fact, it doesn’t matter to us what kind of equation it is. We need to solve it. The scheme here is simple. Collect everything with X's on the left side of the equation, everything without X's (numbers) on the right.

To do this you need to transfer - 4x to the left side, with a change of sign, of course, and - 3 - to the right. By the way, this is the first identical transformation of equations. Surprised? This means that you didn’t follow the link, but in vain...) We get:

x + 4x = 2 + 3

Here are similar ones, we consider:

What do we need for complete happiness? Yes, so that there is a pure X on the left! Five is in the way. Getting rid of the five with the help the second identical transformation of equations. Namely, we divide both sides of the equation by 5. We get a ready answer:

An elementary example, of course. This is for warming up.) It’s not very clear why I remembered identical transformations here? OK. Let's take the bull by the horns.) Let's decide something more solid.

For example, here's the equation:

Where do we start? With X's - to the left, without X's - to the right? Could be so. In small steps long road. Or you can immediately, universally and in a powerful way. If, of course, you have identical transformations of equations in your arsenal.

I ask you a key question: What do you dislike most about this equation?

95 out of 100 people will answer: fractions ! The answer is correct. So let's get rid of them. Therefore, we start immediately with second identity transformation. What do you need to multiply the fraction on the left by so that the denominator is completely reduced? That's right, at 3. And on the right? By 4. But mathematics allows us to multiply both sides by the same number. How can we get out? Let's multiply both sides by 12! Those. to a common denominator. Then both the three and the four will be reduced. Don't forget that you need to multiply each part entirely. Here's what the first step looks like:

Expanding the brackets:

Note! Numerator (x+2) I put it in brackets! This is because when multiplying fractions, the entire numerator is multiplied! Now you can reduce fractions:

Expand the remaining brackets:

Not an example, but pure pleasure!) Now let’s remember the spell from junior classes: with an X - to the left, without an X - to the right! And apply this transformation:

Here are some similar ones:

And divide both parts by 25, i.e. apply the second transformation again:

That's all. Answer: X=0,16

Take note: to bring the original confusing equation to pleasant view, we used two (only two!) identity transformations– translation left-right with a change of sign and multiplication-division of an equation by the same number. This universal method! We will work in this way with any equations! Absolutely anyone. That’s why I tediously repeat about these identical transformations all the time.)

As you can see, the principle of solving linear equations is simple. We take the equation and simplify it using identical transformations until we get the answer. The main problems here are in the calculations, not in the principle of the solution.

But... There are such surprises in the process of solving the most elementary linear equations that they can drive you into a strong stupor...) Fortunately, there can only be two such surprises. Let's call them special cases.

Special cases in solving linear equations.

First surprise.

Suppose you come across a very basic equation, something like:

2x+3=5x+5 - 3x - 2

Slightly bored, we move it with an X to the left, without an X - to the right... With a change of sign, everything is perfect... We get:

2x-5x+3x=5-2-3

We count, and... oops!!! We get:

This equality in itself is not objectionable. Zero really is zero. But X is missing! And we must write down in the answer, what is x equal to? Otherwise, the solution doesn't count, right...) Deadlock?

Calm! In such doubtful cases, the most general rules will save you. How to solve equations? What does it mean to solve an equation? This means, find all the values ​​of x that, when substituted into the original equation, will give us the correct equality.

But we have true equality already happened! 0=0, how much more accurate?! It remains to figure out at what x's this happens. What values ​​of X can be substituted into original equation if these x's will they still be reduced to zero? Come on?)

Yes!!! X's can be substituted any! Which ones do you want? At least 5, at least 0.05, at least -220. They will still shrink. If you don’t believe me, you can check it.) Substitute any values ​​of X into original equation and calculate. All the time you will get the pure truth: 0=0, 2=2, -7.1=-7.1 and so on.

Here's your answer: x - any number.

The answer can be written in different mathematical symbols, the essence does not change. This is a completely correct and complete answer.

Second surprise.

Let's take the same elementary linear equation and change just one number in it. This is what we will decide:

2x+1=5x+5 - 3x - 2

After the same identical transformations, we get something intriguing:

Like this. We solved a linear equation and got a strange equality. In mathematical terms, we got false equality. And speaking in simple language, this is not true. Rave. But nevertheless, this nonsense is a very good reason for the right decision equations.)

Again we think based on general rules. What x's, when substituted into the original equation, will give us true equality? Yes, none! There are no such X's. No matter what you put in, everything will be reduced, only nonsense will remain.)

Here's your answer: there are no solutions.

This is also a completely complete answer. In mathematics, such answers are often found.

Like this. Now, I hope, the disappearance of X's in the process of solving any (not just linear) equation will not confuse you at all. This is already a familiar matter.)

Now that we have dealt with all the pitfalls in linear equations, it makes sense to solve them.

If you like this site...

By the way, I have a couple more interesting sites for you.)

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.

Ministry of General and vocational education RF

Municipal educational institution

Gymnasium No. 12

composition

on the topic: Equations and methods for solving them

Completed by: student of class 10 "A"

Krutko Evgeniy

Checked by: mathematics teacher Iskhakova Gulsum Akramovna

Tyumen 2001

Plan................................................. ........................................................ ................................ 1

Introduction........................................................ ........................................................ ........................ 2

Main part................................................ ........................................................ ............... 3

Conclusion................................................. ........................................................ .................... 25

Application................................................. ........................................................ ................ 26

List of used literature......................................................... ........................... 29

Plan.

Introduction.

Historical reference.

Equations. Algebraic equations.

a) Basic definitions.

b) Linear equation and method for solving it.

V) Quadratic equations and ways to solve it.

d) Binomial equations and how to solve them.

e) Cubic equations and methods for solving them.

f) Biquadratic equation and method for solving it.

g) Equations of the fourth degree and methods for solving them.

g) Equations high degrees and methods from the solution.

h) Rational algebraic equation and the way it is

And) Irrational equations and ways to solve it.

j) Equations containing an unknown under a sign.

absolute value and method for solving it.

Transcendental equations.

A) Exponential equations and a way to solve them.

b) Logarithmic equations and a way to solve them.

Introduction

Mathematics education received in secondary school, is the most important component general education and general culture modern man. Almost everything that surrounds modern man is all somehow connected with mathematics. And the latest achievements in physics, technology and information technology leave no doubt that in the future the state of affairs will remain the same. Therefore, solving many practical problems comes down to solving various types equations that you need to learn to solve.

This work is an attempt to summarize and systematize the studied material on the above topic. I have arranged the material in order of difficulty, starting with the simplest. It includes both the types of equations known to us from the school algebra course, and additional material. At the same time, I tried to show the types of equations that are not studied in school course, but knowledge of which may be needed when entering higher education educational institution. In my work, when solving equations, I did not limit myself only to the real solution, but also indicated the complex one, since I believe that otherwise the equation is simply unsolved. After all, if an equation has no real roots, this does not mean that it has no solutions. Unfortunately, due to lack of time, I was not able to present all the material I have, but even with the material presented here, many questions may arise. I hope that my knowledge is enough to answer most questions. So, I begin to present the material.

Mathematics... reveals order,

symmetry and certainty,

and this is most important species beautiful.

Aristotle.

Historical reference

In those distant times, when the sages first began to think about equalities containing unknown quantities, there were probably no coins or wallets. But there were heaps, as well as pots and baskets, which were perfect for the role of storage caches that could hold an unknown number of items. “We are looking for a heap that, together with two thirds, a half and one seventh, makes 37...”, taught the Egyptian scribe Ahmes in the 2nd millennium BC. In the ancient mathematical problems of Mesopotamia, India, China, Greece, unknown quantities expressed the number of peacocks in the garden, the number of bulls in the herd, and the totality of things taken into account when dividing property. Scribes, officials and initiates well trained in the science of accounts secret knowledge The priests coped with such tasks quite successfully.

Sources that have reached us indicate that ancient scientists owned some general techniques solving problems with unknown quantities. However, not a single papyrus or clay tablet contains a description of these techniques. The authors only occasionally supplied their numerical calculations with skimpy comments such as: “Look!”, “Do this!”, “You found the right one.” In this sense, the exception is the “Arithmetic” of the Greek mathematician Diophantus of Alexandria (III century) - a collection of problems for composing equations with a systematic presentation of their solutions.

However, the first manual for solving problems that became widely known was the work of the Baghdad scientist of the 9th century. Muhammad bin Musa al-Khwarizmi. The word "al-jabr" from the Arabic name of this treatise - "Kitab al-jaber wal-mukabala" ("Book of restoration and opposition") - over time turned into the well-known word "algebra", and al-Khwarizmi's work itself served the starting point in the development of the science of solving equations.

equations Algebraic equations

Basic definitions

In algebra, two types of equalities are considered - identities and equations.

Identity is an equality that holds for all (admissible) values ​​of the letters included in it). To record an identity along with a sign

the sign is also used.

The equation is an equality that holds only for certain values ​​of the letters included in it. The letters included in the equation, according to the conditions of the problem, may be unequal: some can accept all of their valid values(they are called parameters or coefficients equations and are usually denoted by the first letters of the Latin alphabet:

, , ... - or the same letters provided with indices: , , ... or , , ...); others whose values ​​need to be found are called unknown(they are usually designated by the last letters of the Latin alphabet: , , , ... - or the same letters provided with indices: , , ... or , , ...).

IN general view the equation can be written like this:

(, , ..., ).

Depending on the number unknown equation called an equation with one, two, etc. unknowns.