Equations

How to solve equations?

In this section we will recall (or study, depending on who you choose) the most elementary equations. So what is the equation? In human language, this is some kind of mathematical expression where there is an equal sign and an unknown. Which is usually denoted by the letter "X". Solve the equation- this is to find such values ​​of x that, when substituted into original expression will give us the correct identity. Let me remind you that identity is an expression that is beyond doubt even for a person who is absolutely not burdened with mathematical knowledge. Like 2=2, 0=0, ab=ab, etc. So how to solve equations? Let's figure it out.

There are all sorts of equations (I’m surprised, right?). But all their infinite variety can be divided into only four types.

4. Other.)

All the rest, of course, most of all, yes...) This includes cubic, exponential, logarithmic, trigonometric and all sorts of others. We will work closely with them in the relevant sections.

I’ll say right away that sometimes the equations of the first three types are so screwed up that you won’t even recognize them... Nothing. We will learn how to unwind them.

And why do we need these four types? And then what linear equations solved in one way square others, fractional rationals - third, A rest They don’t dare at all! Well, it’s not that they can’t decide at all, it’s that I was wrong with mathematics.) It’s just that they have their own special techniques and methods.

But for any (I repeat - for any!) equations provide a reliable and fail-safe basis for solving. Works everywhere and always. This foundation - Sounds scary, but it's very simple. And very (Very!) important.

Actually, the solution to the equation consists of these very transformations. 99% Answer to the question: " How to solve equations?" lies precisely in these transformations. Is the hint clear?)

Identical transformations of equations.

IN any equations To find the unknown, you need to transform and simplify the original example. And so that when the appearance changes the essence of the equation has not changed. Such transformations are called identical or equivalent.

Note that these transformations apply specifically to the equations. There are also identity transformations in mathematics expressions. This is another topic.

Now we will repeat all, all, all basic identical transformations of equations.

Basic because they can be applied to any equations - linear, quadratic, fractional, trigonometric, exponential, logarithmic, etc. and so on.

First identity transformation: you can add (subtract) to both sides of any equation any(but one and the same!) number or expression (including an expression with an unknown!). This does not change the essence of the equation.

By the way, you constantly used this transformation, you just thought that you were transferring some terms from one part of the equation to another with a change of sign. Type:

The case is familiar, we move the two to the right, and we get:

Actually you taken away from both sides of the equation is two. The result is the same:

x+2 - 2 = 3 - 2

Moving terms left and right with a change of sign is simply a shortened version of the first identity transformation. And why do we need such deep knowledge? - you ask. Nothing in the equations. For God's sake, bear it. Just don’t forget to change the sign. But in inequalities, the habit of transference can lead to a dead end...

Second identity transformation: both sides of the equation can be multiplied (divided) by the same thing non-zero number or expression. Here an understandable limitation already appears: multiplying by zero is stupid, and dividing is completely impossible. This is the transformation you use when you solve something cool like

It's clear X= 2. How did you find it? By selection? Or did it just dawn on you? In order not to select and not wait for insight, you need to understand that you are just divided both sides of the equation by 5. When dividing the left side (5x), the five was reduced, leaving pure X. Which is exactly what we needed. And when dividing the right side of (10) by five, we get, you know, two.

That's all.

It's funny, but these two (only two!) identical transformations are the basis of the solution all equations of mathematics. Wow! It makes sense to look at examples of what and how, right?)

Examples of identical transformations of equations. Main problems.

Let's start with first identity transformation. Transfer left-right.

An example for the younger ones.)

Let's say we need to solve the following equation:

3-2x=5-3x

Let's remember the spell: "with X's - to the left, without X's - to the right!" This spell is instructions for using the first identity transformation.) What expression with an X is on the right? 3x? The answer is incorrect! On our right - 3x! Minus three x! Therefore, when moving to the left, the sign will change to plus. It will turn out:

3-2x+3x=5

So, the X’s were collected in a pile. Let's get into the numbers. There is a three on the left. With what sign? The answer “with none” is not accepted!) In front of the three, indeed, nothing is drawn. And this means that before the three there is plus. So the mathematicians agreed. Nothing is written, which means plus. Therefore, the triple will be transferred to the right side with a minus. We get:

-2x+3x=5-3

There are mere trifles left. On the left - bring similar ones, on the right - count. The answer comes straight away:

In this example, one identity transformation was enough. The second one was not needed. Well, okay.)

An example for older children.)

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.

Replacing a polynomial or. Here is a polynomial of degree, for example, the expression is a polynomial of degree.

Let's say we have an example:

Let's use the variable replacement method. What do you think should be taken for? Right, .

The equation becomes:

We perform a reverse change of variables:

Let's solve the first equation:

Let's decide second the equation:

… What does this mean? Right! That there are no solutions.

Thus, we received two answers - ; .

Do you understand how to use the variable replacement method for a polynomial? Practice doing this yourself:

Decided? Now let's check the main points with you.

You need to take it.

We get the expression:

Solving a quadratic equation, we find that it has two roots: and.

The solution to the first quadratic equation is the numbers and

Solving the second quadratic equation - numbers and.

Answer: ; ; ;

Let's sum it up

The variable replacement method has the main types of variable replacements in equations and inequalities:

1. Power substitution, when we take for some unknown, raised to a power.

2. Replacement of a polynomial, when we take for an entire expression containing an unknown.

3. Fractional-rational replacement, when we take any relation containing an unknown variable.

Important adviсe when introducing a new variable:

1. Replacement of variables must be done immediately, at the first opportunity.

2. The equation for a new variable must be solved to the end and only then returned to the old unknown.

3. When returning to the original unknown (and indeed throughout the entire solution), do not forget to check the roots for ODZ.

A new variable is introduced in a similar way, both in equations and in inequalities.

Let's look at 3 problems

Answers to 3 problems

1. Let, then the expression takes the form.

Since, it can be both positive and negative.

Answer:

2. Let, then the expression takes the form.

there is no solution because...

Answer:

3. By grouping we get:

Let then the expression take the form
.

Answer:

REPLACEMENT OF VARIABLES. AVERAGE LEVEL.

Replacing variables- this is the introduction of a new unknown, with respect to which the equation or inequality has a simpler form.

I will list the main types of replacements.

Power substitution

Power substitution.

For example, using a substitution, a biquadratic equation is reduced to a quadratic one: .

In inequalities everything is similar.

For example, we make a replacement in the inequality and get a quadratic inequality: .

Example (decide for yourself):

Solution:

This is a fractional-rational equation (repeat), but solving it using the usual method (reduction to a common denominator) is inconvenient, since we will obtain an equation of degree, so a change of variables is used.

Everything will become much easier after replacing: . Then:

Now let's do it reverse replacement:

Answer: ; .

Replacing a polynomial

Replacing a polynomial or.

Here is a polynomial of degree, i.e. expression of the form

(for example, the expression is a polynomial of degree, that is).

The most commonly used substitution for the quadratic trinomial is: or.

Example:

Solve the equation.

Solution:

And again, substitution of variables is used.

Then the equation will take the form:

The roots of this quadratic equation are: and.

We have two cases. Let's make a reverse substitution for each of them:

This means that this equation has no roots.

The roots of this equation are: i.

Answer. .

Fractional-rational substitution

Fractional-rational replacement.

and are polynomials of degrees and, respectively.

For example, when solving reciprocal equations, that is, equations of the form

replacement is usually used.

Now I'll show you how it works.

It is easy to check what is not the root of this equation: after all, if we substitute it into the equation, we get what contradicts the condition.

Let's divide the equation into:

Let's regroup:

Now we make a replacement: .

The beauty of it is that when squaring the double product of the terms, x is reduced:

It follows that.

Let's return to our equation:

Now it is enough to solve the quadratic equation and make the reverse substitution.

Example:

Solve the equation: .

Solution:

When equality does not hold, therefore. Let's divide the equation into:

The equation will take the form:

Its roots:

Let's make a reverse replacement:

Let's solve the resulting equations:

Answer: ; .

Another example:

Solve the inequality.

Solution:

By direct substitution we are convinced that it is not included in the solution of this inequality. Divide the numerator and denominator of each fraction by:

Now the replacement of the variable is obvious: .

Then the inequality will take the form:

We use the interval method to find y:

in front of everyone, because

in front of everyone, because

So the inequality is equivalent to the following:

in front of everyone, because...

This means that the inequality is equivalent to the following: .

So, inequality turns out to be equivalent to aggregate:

Answer: .

Replacing variables- one of the most important methods for solving equations and inequalities.

Finally, I’ll give you a couple of important tips:

REPLACEMENT OF VARIABLES. SUMMARY AND BASIC FORMULAS.

Replacing variables- a method for solving complex equations and inequalities, which allows you to simplify the original expression and bring it to a standard form.

Types of variable replacement:

1. Power substitution: is taken to be some unknown, raised to a power - .
2. Fractional-rational replacement: is taken to be any relation containing an unknown variable - , where and are polynomials of degrees n and m, respectively.
3. Replacing a polynomial: the whole expression containing the unknown is taken as - or, where is a polynomial of degree.

After solving a simplified equation/inequality, it is necessary to make a reverse substitution.

In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems in which certain conditions are introduced on the coefficients of the equation that limit them fall out of sight. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although problems of this kind are found more and more often in the Unified State Examination materials and in entrance exams.

Which equation will be called an equation with two variables?

So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.

Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation in question.

Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values ​​of the variables that turn this equation into a true numerical equality.

An equation with two unknowns can:

A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);

b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;

G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.

The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy – 2 = 2x – y.

Solution.

We group the terms for the purpose of factorization:

(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y – 2) = 0. We have:

y = 2, x – any real number or x = -1, y – any real number.

Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality of non-negative numbers to zero

Example 2.

Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).

Solution.

Grouping:

(9x 2 – 12x + 4) + (4y 2 – 12y + 9) = 0. Now each bracket can be folded using the squared difference formula.

(3x – 2) 2 + (2y – 3) 2 = 0.

The sum of two non-negative expressions is zero only if 3x – 2 = 0 and 2y – 3 = 0.

This means x = 2/3 and y = 3/2.

Answer: (2/3; 3/2).

Estimation method

Example 3.

Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.

Solution.

In each bracket we select a complete square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate the meaning of the expressions in parentheses.

(x + 1) 2 + 1 ≥ 1 and (y – 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:

(x + 1) 2 + 1 = 1 and (y – 2) 2 + 2 = 2, which means x = -1, y = 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method consists of treating the equation as square with respect to some variable.

Example 4.

Solve the equation: x 2 – 6x + y – 4√y + 13 = 0.

Solution.

Let's solve the equation as a quadratic equation for x. Let's find the discriminant:

D = 36 – 4(y – 4√y + 13) = -4y + 16√y – 16 = -4(√y – 2) 2 . The equation will have a solution only when D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.

Answer: (3; 4).

Often in equations with two unknowns they indicate restrictions on variables.

Example 5.

Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.

Solution.

Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number not divisible by 5 gives a remainder of 1 or 4. Thus, equality is impossible and there are no solutions.

Answer: no roots.

Example 6.

Solve the equation: (x 2 – 4|x| + 5)(y 2 + 6y + 12) = 3.

Solution.

Let's highlight the complete squares in each bracket:

((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.

Answer: (2; -3) and (-2; -3).

Example 7.

For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.

Solution.

Let's select complete squares:

(x 2 – 2xy + y 2) + (y 2 + 4y + 4) = 37;

(x – y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. We get the sum of the squares of two integers equal to 37 if we add 1 + 36. Therefore:

(x – y) 2 = 36 and (y + 2) 2 = 1

(x – y) 2 = 1 and (y + 2) 2 = 36.

Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).

Answer: -17.

Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.

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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 bring 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 give 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 simplest 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 this step. In the second step we need to isolate the variables. Please note: we are talking 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 using standard algorithms: we will get what we saw in the calculations above.

Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things 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 necessarily 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 of which simply do not have 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 of elementary transformations, where the inability to clearly and competently perform simple actions 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 a simple construction: 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:

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

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 have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
• 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!

The author's approach to this topic is not accidental. Equations with two variables are first encountered in the 7th grade course. One equation with two variables has an infinite number of solutions. This is clearly demonstrated by the graph of a linear function, given as ax + by=c. In the school course, students study systems of two equations with two variables. As a result, a whole series of problems with limited conditions on the coefficient of the equation, as well as methods for solving them, fall out of the sight of the teacher and, therefore, the student.

We are talking about solving an equation with two unknowns in integers or natural numbers.

At school, natural numbers and integers are studied in grades 4-6. By the time they graduate from school, not all students remember the differences between the sets of these numbers.

However, a problem like “solve an equation of the form ax + by=c in integers” is increasingly found on entrance exams to universities and in Unified State Examination materials.

Solving uncertain equations develops logical thinking, intelligence, and attention to analysis.

I propose developing several lessons on this topic. I do not have clear recommendations on the timing of these lessons. Some elements can also be used in 7th grade (for a strong class). These lessons can be taken as a basis and developed a small elective course on pre-vocational training in the 9th grade. And, of course, this material can be used in grades 10-11 to prepare for exams.

The purpose of the lesson:

• repetition and generalization of knowledge on the topic “First and second order equations”
• nurturing cognitive interest in the subject
• developing the ability to analyze, make generalizations, transfer knowledge to a new situation

Lesson 1.

During the classes.

1) Org. moment.

2) Updating basic knowledge.

Definition. A linear equation in two variables is an equation of the form

mx + ny = k, where m, n, k are numbers, x, y are variables.

Example: 5x+2y=10

Definition. A solution to an equation with two variables is a pair of values ​​of variables that turns the equation into a true equality.

Equations with two variables that have the same solutions are called equivalent.

1. 5x+2y=12 (2)y = -2.5x+6

This equation can have any number of solutions. To do this, it is enough to take any x value and find the corresponding y value.

Let x = 2, y = -2.5 2+6 = 1

x = 4, y = -2.5 4+6 =- 4

Pairs of numbers (2;1); (4;-4) – solutions to equation (1).

This equation has infinitely many solutions.

3) Historical background

Indefinite (Diophantine) equations are equations containing more than one variable.

In the 3rd century. AD – Diophantus of Alexandria wrote “Arithmetic”, in which he expanded the set of numbers to rational ones and introduced algebraic symbolism.

Diophantus also considered the problems of solving indefinite equations and he gave methods for solving indefinite equations of the second and third degree.

4) Studying new material.

Definition: A first-order inhomogeneous Diophantine equation with two unknowns x, y is an equation of the form mx + ny = k, where m, n, k, x, y Z k0

Statement 1.

If the free term k in equation (1) is not divisible by the greatest common divisor (GCD) of the numbers m and n, then equation (1) has no integer solutions.

Example: 34x – 17y = 3.

GCD (34; 17) = 17, 3 is not evenly divisible by 17, there is no solution in integers.

Let k be divided by gcd (m, n). By dividing all coefficients, we can ensure that m and n become relatively prime.

Statement 2.

If m and n of equation (1) are relatively prime numbers, then this equation has at least one solution.

Statement 3.

If the coefficients m and n of equation (1) are coprime numbers, then this equation has infinitely many solutions:

Where (; ) is any solution to equation (1), t Z

Definition. A first-order homogeneous Diophantine equation with two unknowns x, y is an equation of the form mx + ny = 0, where (2)

Statement 4.

If m and n are coprime numbers, then any solution to equation (2) has the form

5) Homework. Solve the equation in whole numbers:

1. 9x – 18y = 5
2. x + y= xy
3. Several children were picking apples. Each boy collected 21 kg, and the girl collected 15 kg. In total they collected 174 kg. How many boys and how many girls picked apples?

Comment. This lesson does not provide examples of solving equations in integers. Therefore, children solve homework based on statement 1 and selection.

Lesson 2.

1) Organizational moment

2) Checking homework

1) 9x – 18y = 5

5 is not divisible by 9; there are no solutions in whole numbers.

Using the selection method you can find a solution

Answer: (0;0), (2;2)

3) Let's make an equation:

Let the boys be x, x Z, and the girls y, y Z, then we can create the equation 21x + 15y = 174

Many students, having written an equation, will not be able to solve it.

Answer: 4 boys, 6 girls.

3) Learning new material

Having encountered difficulties in completing homework, students were convinced of the need to learn their methods for solving uncertain equations. Let's look at some of them.

I. Method for considering division remainders.

Example. Solve the equation in whole numbers 3x – 4y = 1.

The left side of the equation is divisible by 3, therefore the right side must be divisible. Let's consider three cases.

Answer: where m Z.

The described method is convenient to use if the numbers m and n are not small, but can be decomposed into simple factors.

Example: Solve equations in whole numbers.

Let y = 4n, then 16 - 7y = 16 – 7 4n = 16 – 28n = 4*(4-7n) is divided by 4.

y = 4n+1, then 16 – 7y = 16 – 7 (4n + 1) = 16 – 28n – 7 = 9 – 28n is not divisible by 4.

y = 4n+2, then 16 – 7y = 16 – 7 (4n + 2) = 16 – 28n – 14 = 2 – 28n is not divisible by 4.

y = 4n+3, then 16 – 7y = 16 – 7 (4n + 3) = 16 – 28n – 21 = -5 – 28n is not divisible by 4.

Therefore y = 4n, then

4x = 16 – 7 4n = 16 – 28n, x = 4 – 7n

Answer: , where n Z.

II. Uncertain equations of 2nd degree

Today in the lesson we will only touch on the solution of second-order Diophantine equations.

And of all types of equations, we will consider the case when we can apply the difference of squares formula or another method of factorization.

Example: Solve an equation in whole numbers.

13 is a prime number, so it can only be factored in four ways: 13 = 13 1 = 1 13 = (-1)(-13) = (-13)(-1)

Let's consider these cases

Answer: (7;-3), (7;3), (-7;3), (-7;-3).

4) Homework.

Examples. Solve the equation in whole numbers:

(x - y)(x + y)=4

2x = 4	2x = 5	2x = 5
x = 2	x = 5/2	x = 5/2
y = 0	doesn't fit	doesn't fit
2x = -4	doesn't fit	doesn't fit
x = -2
y = 0

Answer: (-2;0), (2;0).

Answers: (-10;9), (-5;3), (-2;-3), (-1;-9), (1;9), (2;3), (5;-3) , (10;-9).

V)

Answer: (2;-3), (-1;-1), (-4;0), (2;2), (-1;3), (-4;5).

Results. What does it mean to solve an equation in whole numbers?

What methods for solving uncertain equations do you know?

Application:

Exercises for training.

1) Solve in whole numbers.

a) 8x + 12y = 32	x = 1 + 3n, y = 2 - 2n, n Z
b) 7x + 5y = 29	x = 2 + 5n, y = 3 – 7n, n Z
c) 4x + 7y = 75	x = 3 + 7n, y = 9 – 4n, n Z
d) 9x – 2y = 1	x = 1 – 2m, y = 4 + 9m, m Z
e) 9x – 11y = 36	x = 4 + 11n, y = 9n, n Z
e) 7x – 4y = 29	x = 3 + 4n, y = -2 + 7n, n Z
g) 19x – 5y = 119	x = 1 + 5p, y = -20 + 19p, p Z
h) 28x – 40y = 60	x = 45 + 10t, y = 30 + 7t, t Z

2) Find integer non-negative solutions to the equation:

Solution:Z (2; -1)

Literature.

1. Children's encyclopedia “Pedagogy”, Moscow, 1972.
2. Algebra-8, N.Ya. Vilenkin, VO “Science”, Novosibirsk, 1992
3. Competition problems based on number theory.
4. V.Ya. Galkin, D.Yu. Sychugov. MSU, VMK, Moscow, 2005.
5. Algebra 7, Makarychev Yu.N., “Enlightenment”.