Equations are a difficult topic to master, but they are a powerful tool for solving most problems.

Equations are used to describe various processes, occurring in nature. Equations are widely used in other sciences: economics, physics, biology and chemistry.

IN this lesson We will try to understand the essence of the simplest equations, learn to express unknowns and solve several equations. As you learn new materials, the equations will become more complex, so understanding the basics is very important.

Preliminary skills Lesson content

What is an equation?

An equation is an equality that contains a variable whose value you want to find. This value must be such that when substituted into the original equation, the correct numerical equality is obtained.

For example, the expression 2 + 2 = 4 is an equality. When calculating the left side, the correct numerical equality is obtained 4 = 4.

But the equality is 2 + x= 4 is an equation because it contains a variable x, the value of which can be found. The value must be such that when substituting this value into the original equation, the correct numerical equality is obtained.

In other words, we must find a value at which the equal sign would justify its location - the left side must be equal to the right side.

Equation 2 + x= 4 is elementary. Variable value x is equal to the number 2. For any other value, equality will not be observed

They say that the number 2 is root or solving the equation 2 + x = 4

Root or solution to the equation- this is the value of the variable at which the equation turns into a true numerical equality.

There may be several roots or none at all. Solve the equation means to find its roots or prove that there are no roots.

The variable included in the equation is otherwise called unknown. You have the right to call it what you prefer. These are synonyms.

Note. The phrase “solve an equation” speaks for itself. Solving an equation means “equalizing” the equation—making it balanced so that the left side equals the right side.

Express one thing through the other

The study of equations traditionally begins with learning to express one number included in an equality through a number of others. Let's not break this tradition and do the same.

Consider the following expression:

8 + 2

This expression is the sum of the numbers 8 and 2. The value of this expression is 10

8 + 2 = 10

We got equality. Now you can express any number from this equality through other numbers included in the same equality. For example, let's express the number 2.

To express the number 2, you need to ask the question: “what must be done with the numbers 10 and 8 to get the number 2.” It is clear that to obtain the number 2, you need to subtract the number 8 from the number 10.

That's what we do. We write down the number 2 and through the equal sign we say that to obtain this number 2 we subtracted the number 8 from the number 10:

2 = 10 − 8

We expressed the number 2 from the equality 8 + 2 = 10. As you can see from the example, there is nothing complicated about this.

When solving equations, in particular when expressing one number in terms of others, it is convenient to replace the equal sign with the word “ There is" . This must be done mentally, and not in the expression itself.

So, expressing the number 2 from the equality 8 + 2 = 10, we got the equality 2 = 10 − 8. This equality can be read as follows:

2 There is 10 − 8

That is, a sign = replaced by the word "is". Moreover, the equality 2 = 10 − 8 can be translated from mathematical language into full human language. Then it can be read as follows:

Number 2 There is difference between number 10 and number 8

Number 2 There is difference between number 10 and number 8.

But we will limit ourselves to only replacing the equal sign with the word “is,” and we will not always do this. Elementary expressions can be understood without translating mathematical language into human language.

Let us return the resulting equality 2 = 10 − 8 to its original state:

8 + 2 = 10

Let's express the number 8 this time. What needs to be done with the remaining numbers to get the number 8? That's right, you need to subtract 2 from the number 10

8 = 10 − 2

Let us return the resulting equality 8 = 10 − 2 to its original state:

8 + 2 = 10

This time we will express the number 10. But it turns out that there is no need to express the ten, since it has already been expressed. It is enough to swap the left and right parts, then we get what we need:

10 = 8 + 2

Example 2. Consider the equality 8 − 2 = 6

Let us express the number 8 from this equality. To express the number 8, the remaining two numbers must be added:

8 = 6 + 2

Let us return the resulting equality 8 = 6 + 2 to its original state:

8 − 2 = 6

Let's express the number 2 from this equality. To express the number 2, you need to subtract 6 from 8

2 = 8 − 6

Example 3. Consider the equality 3 × 2 = 6

Let's express the number 3. To express the number 3, you need 6 divided by 2

Let's return the resulting equality to its original state:

3 × 2 = 6

Let us express the number 2 from this equality. To express the number 2, you need 6 divided by 3

Example 4. Consider the equality

Let us express the number 15 from this equality. To express the number 15, you need to multiply the numbers 3 and 5

15 = 3 × 5

Let us return the resulting equality 15 = 3 × 5 to its original state:

Let us express the number 5 from this equality. To express the number 5, you need 15 divided by 3

Rules for finding unknowns

Let's consider several rules for finding unknowns. They may be familiar to you, but it doesn’t hurt to repeat them again. In the future, they can be forgotten, as we learn to solve equations without applying these rules.

Let's return to the first example, which we looked at in the previous topic, where in the equality 8 + 2 = 10 we needed to express the number 2.

In the equality 8 + 2 = 10, the numbers 8 and 2 are the terms, and the number 10 is the sum.

To express the number 2, we did the following:

2 = 10 − 8

That is, from the sum of 10 we subtracted the term 8.

Now imagine that in the equality 8 + 2 = 10, instead of the number 2 there is a variable x

8 + x = 10

In this case, the equality 8 + 2 = 10 becomes the equation 8 + x= 10 and the variable x unknown term

Our task is to find this unknown term, that is, to solve the equation 8 + x= 10 . To find an unknown term, the following rule is provided:

To find the unknown term, you need to subtract the known term from the sum.

Which is basically what we did when we expressed two in the equality 8 + 2 = 10. To express term 2, we subtracted another term 8 from the sum 10

2 = 10 − 8

Now, to find the unknown term x, we must subtract the known term 8 from the sum 10:

x = 10 − 8

If you calculate the right side of the resulting equality, you can find out what the variable is equal to x

x = 2

We have solved the equation. Variable value x equals 2. To check the value of a variable x sent to the original equation 8 + x= 10 and substitute x. It is advisable to do this with any solved equation, since you cannot be absolutely sure that the equation has been solved correctly:

As a result

The same rule would apply if the unknown term was the first number 8.

x + 2 = 10

In this equation x is the unknown term, 2 is the known term, 10 is the sum. To find an unknown term x, you need to subtract the known term 2 from the sum 10

x = 10 − 2

x = 8

Let's return to the second example from the previous topic, where in the equality 8 − 2 = 6 it was necessary to express the number 8.

In the equality 8 − 2 = 6, the number 8 is the minuend, the number 2 is the subtrahend, and the number 6 is the difference

To express the number 8, we did the following:

8 = 6 + 2

That is, we added the difference of 6 and the subtracted 2.

Now imagine that in the equality 8 − 2 = 6, instead of the number 8, there is a variable x

x − 2 = 6

In this case the variable x takes on the role of the so-called unknown minuend

To find an unknown minuend, the following rule is provided:

To find the unknown minuend, you need to add the subtrahend to the difference.

This is what we did when we expressed the number 8 in the equality 8 − 2 = 6. To express the minuend of 8, we added the subtrahend of 2 to the difference of 6.

Now, to find the unknown minuend x, we must add the subtrahend 2 to the difference 6

x = 6 + 2

If you calculate the right side, you can find out what the variable is equal to x

x = 8

Now imagine that in the equality 8 − 2 = 6, instead of the number 2, there is a variable x

8 − x = 6

In this case the variable x takes on the role unknown subtrahend

To find an unknown subtrahend, the following rule is provided:

To find the unknown subtrahend, you need to subtract the difference from the minuend.

This is what we did when we expressed the number 2 in the equality 8 − 2 = 6. To express the number 2, we subtracted the difference 6 from the minuend 8.

Now, to find the unknown subtrahend x, you again need to subtract the difference 6 from the minuend 8

x = 8 − 6

We calculate the right side and find the value x

x = 2

Let's return to the third example from the previous topic, where in the equality 3 × 2 = 6 we tried to express the number 3.

In the equality 3 × 2 = 6, the number 3 is the multiplicand, the number 2 is the multiplier, the number 6 is the product

To express the number 3 we did the following:

That is, we divided the product of 6 by the factor of 2.

Now imagine that in the equality 3 × 2 = 6, instead of the number 3 there is a variable x

x× 2 = 6

In this case the variable x takes on the role unknown multiplicand.

To find an unknown multiplicand, the following rule is provided:

To find an unknown multiplicand, you need to divide the product by the factor.

This is what we did when we expressed the number 3 from the equality 3 × 2 = 6. We divided the product 6 by the factor 2.

Now to find the unknown multiplicand x, you need to divide the product 6 by the factor 2.

Calculating the right side allows us to find the value of a variable x

x = 3

The same rule applies if the variable x is located instead of the multiplier, not the multiplicand. Let's imagine that in the equality 3 × 2 = 6, instead of the number 2 there is a variable x.

In this case the variable x takes on the role unknown multiplier. To find an unknown factor, the same procedure is provided as for finding an unknown multiplicand, namely, dividing the product by a known factor:

To find an unknown factor, you need to divide the product by the multiplicand.

This is what we did when we expressed the number 2 from the equality 3 × 2 = 6. Then to get the number 2 we divided the product of 6 by its multiplicand 3.

Now to find the unknown factor x We divided the product of 6 by the multiplicand of 3.

Calculating the right side of the equality allows you to find out what x is equal to

x = 2

The multiplicand and the multiplier together are called factors. Since the rules for finding a multiplicand and a multiplier are the same, we can formulate general rule finding the unknown factor:

To find an unknown factor, you need to divide the product by the known factor.

For example, let's solve the equation 9 × x= 18. Variable x is an unknown factor. To find this unknown factor, you need to divide the product 18 by the known factor 9

Let's solve the equation x× 3 = 27. Variable x is an unknown factor. To find this unknown factor, you need to divide the product 27 by the known factor 3

Let's return to the fourth example from the previous topic, where in an equality we needed to express the number 15. In this equality, the number 15 is the dividend, the number 5 is the divisor, and the number 3 is the quotient.

To express the number 15 we did the following:

15 = 3 × 5

That is, we multiplied the quotient of 3 by the divisor of 5.

Now imagine that in the equality, instead of the number 15, there is a variable x

In this case the variable x takes on the role unknown dividend.

To find an unknown dividend, the following rule is provided:

To find the unknown dividend, you need to multiply the quotient by the divisor.

This is what we did when we expressed the number 15 from the equality. To express the number 15, we multiply the quotient of 3 by the divisor of 5.

Now, to find the unknown dividend x, you need to multiply the quotient 3 by the divisor 5

x= 3 × 5

x .

x = 15

Now imagine that in the equality, instead of the number 5, there is a variable x .

In this case the variable x takes on the role unknown divisor.

To find an unknown divisor, the following rule is provided:

This is what we did when we expressed the number 5 from the equality. To express the number 5, we divide the dividend 15 by the quotient 3.

And now to find unknown divisor x, you need to divide the dividend 15 by the quotient 3

Let's calculate the right side of the resulting equality. This way we find out what the variable is equal to x .

x = 5

So, to find unknowns, we studied the following rules:

• To find the unknown term, you need to subtract the known term from the sum;
• To find the unknown minuend, you need to add the subtrahend to the difference;
• To find the unknown subtrahend, you need to subtract the difference from the minuend;
• To find an unknown multiplicand, you need to divide the product by the factor;
• To find an unknown factor, you need to divide the product by the multiplicand;
• To find an unknown dividend, you need to multiply the quotient by the divisor;
• To find an unknown divisor, you need to divide the dividend by the quotient.

Components

We will call components the numbers and variables included in the equality

So, the components of addition are terms And sum

The subtraction components are minuend, subtrahend And difference

The components of multiplication are multiplicand, factor And work

The components of division are the dividend, divisor and quotient.

Depending on which components we are dealing with, the corresponding rules for finding unknowns will apply. We studied these rules in the previous topic. When solving equations, it is advisable to know these rules by heart.

Example 1. Find the root of the equation 45 + x = 60

45 - term, x- unknown term, 60 - sum. We are dealing with the components of addition. We recall that to find the unknown term, you need to subtract the known term from the sum:

x = 60 − 45

Let's calculate the right side and get the value x equal to 15

x = 15

So the root of the equation is 45 + x= 60 is equal to 15.

Most often, an unknown term must be reduced to a form in which it can be expressed.

Example 2. Solve the equation

Here, unlike the previous example, the unknown term cannot be expressed immediately, since it contains a coefficient of 2. Our task is to bring this equation to a form in which it could be expressed x

IN in this example We are dealing with the components of addition—the terms and the sum. 2 x is the first term, 4 is the second term, 8 is the sum.

In this case, term 2 x contains a variable x. After finding the value of the variable x term 2 x will take a different look. Therefore, term 2 x can be completely taken as an unknown term:

Now we apply the rule for finding the unknown term. Subtract the known term from the sum:

Let's calculate the right side of the resulting equation:

We have a new equation. Now we are dealing with the components of multiplication: the multiplicand, the multiplier, and the product. 2 - multiplicand, x- multiplier, 4 - product

In this case, the variable x is not just a multiplier, but an unknown multiplier

To find this unknown factor, you need to divide the product by the multiplicand:

Let's calculate the right side and get the value of the variable x

To check, send the found root to the original equation and substitute x

Example 3. Solve the equation 3x+ 9x+ 16x= 56

Immediately express the unknown x it is forbidden. First you need to bring this equation to a form in which it could be expressed.

We present on the left side of this equation:

We are dealing with the components of multiplication. 28 - multiplicand, x- multiplier, 56 - product. At the same time x is an unknown factor. To find an unknown factor, you need to divide the product by the multiplicand:

From here x equals 2

Equivalent equations

In the previous example, when solving the equation 3x + 9x + 16x = 56 , we have given similar terms on the left side of the equation. As a result, we obtained a new equation 28 x= 56 . Old equation 3x + 9x + 16x = 56 and the resulting new equation 28 x= 56 is called equivalent equations, since their roots coincide.

Equations are called equivalent if their roots coincide.

Let's check it out. For the equation 3x+ 9x+ 16x= 56 we found the root equal to 2. Let's first substitute this root into the equation 3x+ 9x+ 16x= 56 , and then into equation 28 x= 56, which was obtained by bringing similar terms on the left side of the previous equation. We must get the correct numerical equalities

According to the order of operations, multiplication is performed first:

Let's substitute root 2 into the second equation 28 x= 56

We see that both equations have the same roots. So the equations 3x+ 9x+ 16x= 6 and 28 x= 56 are indeed equivalent.

To solve the equation 3x+ 9x+ 16x= 56 We used one of them - reduction of similar terms. The correct identity transformation of the equation allowed us to obtain the equivalent equation 28 x= 56, which is easier to solve.

From identical transformations to at the moment We only know how to reduce fractions, add similar terms, take the common factor out of brackets, and also open brackets. There are other conversions you should be aware of. But for general idea about identical transformations of equations, the topics we have studied are quite sufficient.

Let's consider some transformations that allow us to obtain the equivalent equation

If you add the same number to both sides of the equation, you get an equation equivalent to the given one.

and similarly:

If you subtract the same number from both sides of an equation, you get an equation equivalent to the given one.

In other words, the root of the equation will not change if the same number is added to (or subtracted from both sides) the same number.

Example 1. Solve the equation

Subtract 10 from both sides of the equation

We got equation 5 x= 10 . We are dealing with the components of multiplication. To find an unknown factor x, you need to divide the product 10 by the known factor 5.

and substitute x found value 2

We got the correct numerical equality. This means the equation is solved correctly.

Solving the equation we subtracted the number 10 from both sides of the equation. As a result, we obtained an equivalent equation. The root of this equation, like the equation is also equal to 2

Example 2. Solve equation 4( x+ 3) = 16

Subtract the number 12 from both sides of the equation

There will be 4 left on the left side x, and on the right side the number 4

We got equation 4 x= 4 . We are dealing with the components of multiplication. To find an unknown factor x, you need to divide the product 4 by the known factor 4

Let's return to the original equation 4( x+ 3) = 16 and substitute x found value 1

We got the correct numerical equality. This means the equation is solved correctly.

Solving equation 4( x+ 3) = 16 we subtracted the number 12 from both sides of the equation. As a result, we obtained the equivalent equation 4 x= 4 . The root of this equation, like equation 4( x+ 3) = 16 is also equal to 1

Example 3. Solve the equation

Let's expand the brackets on the left side of the equality:

Add the number 8 to both sides of the equation

Let us present similar terms on both sides of the equation:

There will be 2 left on the left side x, and on the right side the number 9

In the resulting equation 2 x= 9 we express the unknown term x

Let's return to the original equation and substitute x found value 4.5

We got the correct numerical equality. This means the equation is solved correctly.

Solving the equation we added the number 8 to both sides of the equation. As a result, we got an equivalent equation. The root of this equation, like the equation also equal to 4.5

The next rule that allows us to obtain an equivalent equation is as follows

If you move a term in an equation from one part to another, changing its sign, you will get an equation equivalent to the given one.

That is, the root of the equation will not change if we move a term from one part of the equation to another, changing its sign. This property is one of the important and one of the often used when solving equations.

Consider the following equation:

The root of this equation is equal to 2. Let us substitute x this root and check whether the numerical equality is correct

The result is a correct equality. This means that the number 2 is indeed the root of the equation.

Now let's try to experiment with the terms of this equation, moving them from one part to another, changing the signs.

For example, term 3 x is located on the left side of the equation. Let's move it to the right side, changing the sign to the opposite:

The result is an equation 12 = 9x − 3x . on the right side of this equation:

x is an unknown factor. Let's find this well-known factor:

From here x= 2 . As you can see, the root of the equation has not changed. So the equations are 12 + 3 x = 9x And 12 = 9x − 3x are equivalent.

In fact, this transformation is a simplified method of the previous transformation, where the same number was added (or subtracted) to both sides of the equation.

We said that in the equation 12 + 3 x = 9x term 3 x was moved to the right side, changing sign. In reality, the following happened: term 3 was subtracted from both sides of the equation x

Then similar terms were given on the left side and the equation was obtained 12 = 9x − 3x. Then similar terms were again given, but on the right side, and the equation 12 = 6 was obtained x.

But the so-called “transfer” is more convenient for such equations, which is why it has become so widespread. When solving equations, we will often use this particular transformation.

The equations 12 + 3 are also equivalent x= 9x And 3x− 9x= −12 . This time in the equation 12 + 3 x= 9x term 12 was moved to the right side, and term 9 x to the left. We should not forget that the signs of these terms were changed during the transfer

The next rule that allows us to obtain an equivalent equation is as follows:

If both sides of the equation are multiplied or divided by the same number, not equal to zero, you get an equation equivalent to the given one.

In other words, the roots of an equation will not change if both sides are multiplied or divided by the same number. This action is often used when you need to solve an equation containing fractional expressions.

First, let's look at examples in which both sides of the equation will be multiplied by the same number.

Example 1. Solve the equation

When solving equations containing fractional expressions, it is customary to first simplify the equation.

In this case, we are dealing with just such an equation. To simplify this equation, both sides can be multiplied by 8:

We remember that for , we need to multiply the numerator of a given fraction by this number. We have two fractions and each of them is multiplied by the number 8. Our task is to multiply the numerators of the fractions by this number 8

Now the interesting part happens. The numerators and denominators of both fractions contain a factor of 8, which can be reduced by 8. This will allow us to get rid of the fractional expression:

As a result, the simplest equation remains

Well, it’s not hard to guess that the root of this equation is 4

x found value 4

The result is a correct numerical equality. This means the equation is solved correctly.

When solving this equation, we multiplied both sides by 8. As a result, we got the equation. The root of this equation, like the equation, is 4. This means that these equations are equivalent.

The factor by which both sides of the equation are multiplied is usually written before the part of the equation, and not after it. So, solving the equation, we multiplied both sides by a factor of 8 and got the following entry:

This did not change the root of the equation, but if we had done this while at school, we would have been reprimanded, since in algebra it is customary to write a factor before the expression with which it is multiplied. Therefore, it is advisable to rewrite the multiplication of both sides of the equation by a factor of 8 as follows:

Example 2. Solve the equation

On the left side, the factors of 15 can be reduced by 15, and on the right side, the factors of 15 and 5 can be reduced by 5

Let's open the brackets on the right side of the equation:

Let's move the term x from the left side of the equation to the right side, changing the sign. And we move term 15 from the right side of the equation to the left side, again changing the sign:

We present similar terms in both sides, we get

We are dealing with the components of multiplication. Variable x

Let's return to the original equation and substitute x found value 5

The result is a correct numerical equality. This means the equation is solved correctly. When solving this equation, we multiplied both sides by 15. Further performing identical transformations, we obtained the equation 10 = 2 x. The root of this equation, like the equation equals 5. This means that these equations are equivalent.

Example 3. Solve the equation

On the left side you can reduce two threes, and the right side will be equal to 18

The simplest equation remains. We are dealing with the components of multiplication. Variable x is an unknown factor. Let's find this well-known factor:

Let's return to the original equation and substitute x found value 9

The result is a correct numerical equality. This means the equation is solved correctly.

Example 4. Solve the equation

Multiply both sides of the equation by 6

Let's open the brackets on the left side of the equation. On the right side, the factor 6 can be raised to the numerator:

Let’s reduce what can be reduced on both sides of the equations:

Let's rewrite what we have left:

Let's use the transfer of terms. Terms containing the unknown x, we group on the left side of the equation, and the terms free of unknowns - on the right:

Let us present similar terms in both parts:

Now let's find the value of the variable x. To do this, divide the product 28 by the known factor 7

From here x= 4.

Let's return to the original equation and substitute x found value 4

The result is a correct numerical equation. This means the equation is solved correctly.

Example 5. Solve the equation

Let's open the parentheses on both sides of the equation where possible:

Multiply both sides of the equation by 15

Let's open the brackets on both sides of the equation:

Let's reduce what can be reduced on both sides of the equation:

Let's rewrite what we have left:

Let's expand the brackets where possible:

Let's use the transfer of terms. We group the terms containing the unknown on the left side of the equation, and the terms free of unknowns on the right. Do not forget that during the transfer, the terms change their signs to the opposite:

Let us present similar terms on both sides of the equation:

Let's find the value x

The resulting answer contains a whole part:

Let's return to the original equation and substitute x found value

It turns out to be a rather cumbersome expression. Let's use variables. Let's put the left side of the equality into a variable A, and the right side of the equality into a variable B

Our task is to make sure whether the left side is equal to the right. In other words, prove the equality A = B

Let's find the value of the expression in variable A.

Variable value A equals . Now let's find the value of the variable B. That is, the value of the right side of our equality. If it is also equal, then the equation will be solved correctly

We see that the value of the variable B, as well as the value of variable A is . This means that the left side is equal to the right side. From this we conclude that the equation is solved correctly.

Now let's try not to multiply both sides of the equation by the same number, but to divide.

Consider the equation 30x+ 14x+ 14 = 70x− 40x+ 42 . Let's solve it using the usual method: we group terms containing unknowns on the left side of the equation, and terms free of unknowns - on the right. Next, performing the known identity transformations, we find the value x

Let's substitute the found value 2 instead x into the original equation:

Now let's try to separate all the terms of the equation 30x+ 14x+ 14 = 70x− 40x+ 42 by some number. We note that all terms of this equation have a common factor of 2. We divide each term by it:

Let's perform a reduction in each term:

Let's rewrite what we have left:

Let's solve this equation using the well-known identity transformations:

We got root 2. So the equations 15x+ 7x+ 7 = 35x− 20x+ 21 And 30x+ 14x+ 14 = 70x− 40x+ 42 are equivalent.

Dividing both sides of the equation by the same number allows you to remove the unknown from the coefficient. In the previous example when we got equation 7 x= 14, we needed to divide the product 14 by the known factor 7. But if we had freed the unknown from the factor 7 on the left side, the root would have been found immediately. To do this, it was enough to divide both sides by 7

We will also use this method often.

Multiplication by minus one

If both sides of the equation are multiplied by minus one, you get an equation equivalent to this one.

This rule follows from the fact that multiplying (or dividing) both sides of an equation by the same number does not change the root of the given equation. This means that the root will not change if both its parts are multiplied by −1.

This rule allows you to change the signs of all components included in the equation. What is this for? Again, to get an equivalent equation that is easier to solve.

Consider the equation. What is the root of this equation?

Add the number 5 to both sides of the equation

Let's look at similar terms:

Now let's remember about. What is the left side of the equation? This is the product of minus one and a variable x

That is, the minus sign in front of the variable x does not refer to the variable itself x, but to one, which we do not see, since coefficient 1 is not usually written down. This means that the equation actually looks like this:

We are dealing with the components of multiplication. To find X, you need to divide the product −5 by the known factor −1.

or divide both sides of the equation by −1, which is even simpler

So the root of the equation is 5. To check, let's substitute it into the original equation. Do not forget that in the original equation the minus is in front of the variable x refers to an invisible unit

The result is a correct numerical equation. This means the equation is solved correctly.

Now let's try to multiply both sides of the equation by minus one:

After opening the brackets, the expression is formed on the left side, and the right side will be equal to 10

The root of this equation, like the equation, is 5

This means that the equations are equivalent.

Example 2. Solve the equation

In this equation, all components are negative. It is more convenient to work with positive components than with negative ones, so let’s change the signs of all components included in the equation. To do this, multiply both sides of this equation by −1.

It is clear that when multiplied by −1, any number will change its sign to the opposite. Therefore, the procedure of multiplying by −1 and opening the brackets is not described in detail, but the components of the equation with opposite signs are immediately written down.

Thus, multiplying an equation by −1 can be written in detail as follows:

or you can simply change the signs of all components:

The result will be the same, but the difference will be that we will save ourselves time.

So, multiplying both sides of the equation by −1, we get the equation. Let's solve this equation. Subtract 4 from both sides and divide both sides by 3

When the root is found, the variable is usually written on the left side, and its value on the right, which is what we did.

Example 3. Solve the equation

Let's multiply both sides of the equation by −1. Then all components will change their signs to opposite ones:

Subtract 2 from both sides of the resulting equation x and give similar terms:

Let's add one to both sides of the equation and give similar terms:

Equating to zero

We recently learned that if we move a term in an equation from one part to another, changing its sign, we will get an equation equivalent to the given one.

What happens if you move from one part to another not just one term, but all the terms? That's right, in the part where all the terms were taken away there will be zero left. In other words, there will be nothing left.

As an example, consider the equation. Let's solve this equation as usual - we will group the terms containing unknowns in one part, and leave the numerical terms free of unknowns in the other. Next, performing the known identity transformations, we find the value of the variable x

Now let's try to solve the same equation by equating all its components to zero. To do this, we move all the terms from the right side to the left, changing the signs:

Let us present similar terms on the left side:

Add 77 to both sides and divide both sides by 7

An alternative to the rules for finding unknowns

Obviously, knowing about identical transformations of equations, you don’t have to memorize the rules for finding unknowns.

For example, to find the unknown in an equation, we divided the product 10 by the known factor 2

But if you divide both sides of the equation by 2, the root will be found immediately. On the left side of the equation in the numerator the factor 2 and in the denominator the factor 2 will be reduced by 2. And the right side will be equal to 5

We solved equations of the form by expressing the unknown term:

But you can use the identical transformations that we studied today. In the equation, term 4 can be moved to the right side by changing the sign:

On the left side of the equation, two twos will cancel out. The right side will be equal to 2. Hence .

Or you could subtract 4 from both sides of the equation. Then you would get the following:

In the case of equations of the form, it is more convenient to divide the product by a known factor. Let's compare both solutions:

The first solution is much shorter and neater. The second solution can be significantly shortened if you do the division in your head.

However, it is necessary to know both methods and only then use the one you prefer.

When there are several roots

An equation can have multiple roots. For example the equation x(x+ 9) = 0 has two roots: 0 and −9.

In Eq. x(x+ 9) = 0 it was necessary to find such a value x at which the left side would be equal to zero. The left side of this equation contains the expressions x And (x+9), which are factors. From the product laws we know that a product is equal to zero if at least one of the factors is equal to zero (either the first factor or the second).

That is, in Eq. x(x+ 9) = 0 equality will be achieved if x will be equal to zero or (x+9) will be equal to zero.

x= 0 or x + 9 = 0

By setting both of these expressions to zero, we can find the roots of the equation x(x+ 9) = 0 . The first root, as can be seen from the example, was found immediately. To find the second root you need to solve the elementary equation x+ 9 = 0 . It is easy to guess that the root of this equation is −9. Checking shows that the root is correct:

−9 + 9 = 0

Example 2. Solve the equation

This equation has two roots: 1 and 2. The left side of the equation is the product of the expressions ( x− 1) and ( x− 2) . And the product is equal to zero if at least one of the factors is equal to zero (or the factor ( x− 1) or factor ( x − 2) ).

Let's find something like this x under which the expressions ( x− 1) or ( x− 2) become zero:

We substitute the found values ​​one by one into the original equation and make sure that for these values ​​the left-hand side is equal to zero:

When there are infinitely many roots

An equation can have infinitely many roots. That is, by substituting any number into such an equation, we get the correct numerical equality.

Example 1. Solve the equation

The root of this equation is any number. If you open the brackets on the left side of the equation and add similar terms, you get the equality 14 = 14. This equality will be obtained for any x

Example 2. Solve the equation

The root of this equation is any number. If you open the brackets on the left side of the equation, you get the equality 10x + 12 = 10x + 12. This equality will be obtained for any x

When there are no roots

It also happens that the equation has no solutions at all, that is, it has no roots. For example, the equation has no roots, since for any value x, the left side of the equation will not be equal to the right side. For example, let . Then the equation will take the following form

Example 2. Solve the equation

Let's expand the brackets on the left side of the equality:

Let's look at similar terms:

We see that the left side is not equal to the right side. And this will be the case for any value. y. For example, let y = 3 .

Letter equations

An equation can contain not only numbers with variables, but also letters.

For example, the formula for finding speed is a literal equation:

This equation describes the speed of a body during uniformly accelerated motion.

A useful skill is the ability to express any component included in a letter equation. For example, to determine distance from an equation, you need to express the variable s .

Multiply both sides of the equation by t

Variables on the right side t let's cut it by t

In the resulting equation, we swap the left and right sides:

We now have the formula for finding the distance that we studied earlier.

Let's try to determine time from the equation. To do this you need to express the variable t .

Multiply both sides of the equation by t

Variables on the right side t let's cut it by t and rewrite what we have left:

In the resulting equation v×t = s divide both parts into v

Variables on the left v let's cut it by v and rewrite what we have left:

We have the formula for determining time, which we studied earlier.

Let's assume the train speed is 50 km/h

v= 50 km/h

And the distance is 100 km

s= 100 km

Then the letter will take the following form

Time can be found from this equation. To do this you need to be able to express the variable t. You can use the rule for finding an unknown divisor by dividing the dividend by the quotient and thus determining the value of the variable t

or you can use identical transformations. First multiply both sides of the equation by t

Then divide both sides by 50

Example 2 x

Subtract from both sides of the equation a

Let's divide both sides of the equation by b

a + bx = c, then we will have a ready-made solution. It will be enough to substitute the required values ​​into it. Those values ​​that will be substituted for letters a, b, c usually called parameters. And equations of the form a + bx = c called equation with parameters. Depending on the parameters, the root will change.

Let's solve the equation 2 + 4 x= 10 . It looks like a letter equation a + bx = c. Instead of performing identical transformations, we can use a ready-made solution. Let's compare both solutions:

We see that the second solution is much simpler and shorter.

For a ready-made solution, it is necessary to make a small remark. Parameter b must not be zero (b ≠ 0), since division by zero by is allowed.

Example 3. A literal equation is given. Express from this equation x

Let's open the brackets on both sides of the equation

Let's use the transfer of terms. Parameters containing a variable x, we group on the left side of the equation, and parameters free from this variable - on the right.

On the left side we take the factor out of brackets x

Let's divide both sides into the expression a − b

On the left side, the numerator and denominator can be reduced by a − b. This is how the variable is finally expressed x

Now, if we come across an equation of the form a(x − c) = b(x + d), then we will have a ready-made solution. It will be enough to substitute the required values ​​into it.

Let's say we are given the equation 4(x− 3) = 2(x+ 4) . It looks like an equation a(x − c) = b(x + d). Let's solve it in two ways: using identical transformations and using a ready-made solution:

For convenience, let's take it out of the equation 4(x− 3) = 2(x+ 4) parameter values a, b, c, d . This will allow us not to make a mistake when substituting:

As in the previous example, the denominator here should not be equal to zero ( a − b ≠ 0) . If we encounter an equation of the form a(x − c) = b(x + d) in which the parameters a And b will be the same, we can say without solving it that this equation has no roots, since the difference of identical numbers is zero.

For example, the equation 2(x − 3) = 2(x + 4) is an equation of the form a(x − c) = b(x + d). In Eq. 2(x − 3) = 2(x + 4) parameters a And b identical. If we start solving it, we will come to the conclusion that the left side will not be equal to the right side:

Example 4. A literal equation is given. Express from this equation x

Let's bring the left side of the equation to a common denominator:

Multiply both sides by a

On the left side x let's put it out of brackets

Divide both sides by the expression (1 − a)

Linear equations with one unknown

The equations discussed in this lesson are called linear equations of the first degree with one unknown.

If the equation is given to the first degree, does not contain division by the unknown, and also does not contain roots from the unknown, then it can be called linear. We have not yet studied powers and roots, so in order not to complicate our lives, we will understand the word “linear” as “simple”.

Most of the equations solved in this lesson ultimately came down to a simple equation in which you had to divide the product by a known factor. For example, this is equation 2( x+ 3) = 16 . Let's solve it.

Let's open the brackets on the left side of the equation, we get 2 x+ 6 = 16. Let's move term 6 to the right side, changing the sign. Then we get 2 x= 16 − 6. Calculate the right side, we get 2 x= 10. To find x, divide the product 10 by the known factor 2. Hence x = 5.

Equation 2( x+ 3) = 16 is linear. It comes down to equation 2 x= 10, to find the root of which it was necessary to divide the product by a known factor. This simplest equation is called linear equation of the first degree with one unknown in canonical form. The word "canonical" is synonymous with the words "simple" or "normal".

A linear equation of the first degree with one unknown in canonical form is called an equation of the form ax = b.

Our resulting equation 2 x= 10 is a linear equation of the first degree with one unknown in canonical form. This equation has the first degree, one unknown, it does not contain division by the unknown and does not contain roots from the unknown, and it is presented in canonical form, that is, in the simplest form in which the value can be easily determined x. Instead of parameters a And b our equation contains the numbers 2 and 10. But such an equation can also contain other numbers: positive, negative or equal to zero.

If in a linear equation a= 0 and b= 0, then the equation has infinitely many roots. Indeed, if a equal to zero and b equals zero, then the linear equation ax= b will take the form 0 x= 0 . For any value x the left side will be equal to the right side.

If in a linear equation a= 0 and b≠ 0, then the equation has no roots. Indeed, if a equal to zero and b is equal to some number that is not equal to zero, say the number 5, then the equation ax = b will take the form 0 x= 5 . The left side will be zero, and the right side will be five. And zero is not equal to five.

If in a linear equation a≠ 0, and b equals any number, then the equation has one root. It is determined by dividing the parameter b per parameter a

Indeed, if a equal to some number that is not zero, say the number 3, and b equal to some number, say the number 6, then the equation will take the form .
From here.

There is another form of writing a linear equation of the first degree with one unknown. It looks like this: ax−b= 0 . This is the same equation as ax = b

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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. Everyone else.)

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 appropriate sections.

I’ll say right away that sometimes the equations of the first three types they will cheat you so much 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 - It 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 changing appearance 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. etc.

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 identical 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, the result is, of course, two.

That's it.

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 is the expression with an X 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.

This lesson discusses in detail the procedure for performing arithmetic operations in expressions without and with brackets. Students are given the opportunity, while completing assignments, to determine whether the meaning of expressions depends on the order in which arithmetic operations are performed, to find out whether the order of arithmetic operations is different in expressions without parentheses and with parentheses, to practice applying the learned rule, to find and correct errors made when determining the order of actions.

In life, we constantly perform some kind of action: we walk, study, read, write, count, smile, quarrel and make peace. We perform these actions in in different order. Sometimes they can be swapped, sometimes not. For example, when getting ready for school in the morning, you can first do exercises, then make your bed, or vice versa. But you can’t go to school first and then put on clothes.

In mathematics, is it necessary to perform arithmetic operations in a certain order?

Let's check

Let's compare the expressions:
8-3+4 and 8-3+4

We see that both expressions are exactly the same.

Let's perform actions in one expression from left to right, and in the other from right to left. You can use numbers to indicate the order of actions (Fig. 1).

Rice. 1. Procedure

In the first expression, we will first perform the subtraction operation and then add the number 4 to the result.

In the second expression, we first find the value of the sum, and then subtract the resulting result 7 from 8.

We see that the meanings of the expressions are different.

Let's conclude: The order in which arithmetic operations are performed cannot be changed.

Let's learn the rule for performing arithmetic operations in expressions without parentheses.

If an expression without parentheses includes only addition and subtraction or only multiplication and division, then the actions are performed in the order in which they are written.

Let's practice.

Consider the expression

This expression contains only addition and subtraction operations. These actions are called first stage actions.

We perform the actions from left to right in order (Fig. 2).

Rice. 2. Procedure

Consider the second expression

This expression contains only multiplication and division operations - These are the actions of the second stage.

We perform the actions from left to right in order (Fig. 3).

Rice. 3. Procedure

In what order are arithmetic operations performed if the expression contains not only addition and subtraction, but also multiplication and division?

If an expression without parentheses includes not only the operations of addition and subtraction, but also multiplication and division, or both of these operations, then first perform in order (from left to right) multiplication and division, and then addition and subtraction.

Let's look at the expression.

Let's think like this. This expression contains the operations of addition and subtraction, multiplication and division. We act according to the rule. First, we perform in order (from left to right) multiplication and division, and then addition and subtraction. Let's arrange the order of actions.

Let's calculate the value of the expression.

18:2-2*3+12:3=9-6+4=3+4=7

In what order are arithmetic operations performed if there are parentheses in an expression?

If an expression contains parentheses, the value of the expressions in the parentheses is evaluated first.

Let's look at the expression.

30 + 6 * (13 - 9)

We see that in this expression there is an action in parentheses, which means we will perform this action first, then multiplication and addition in order. Let's arrange the order of actions.

30 + 6 * (13 - 9)

Let's calculate the value of the expression.

30+6*(13-9)=30+6*4=30+24=54

How should one reason to correctly establish the order of arithmetic operations in a numerical expression?

Before starting calculations, you need to look at the expression (find out whether it contains parentheses, what actions it contains) and only then perform the actions in the following order:

1. actions written in brackets;

2. multiplication and division;

3. addition and subtraction.

The diagram will help you remember this simple rule (Fig. 4).

Rice. 4. Procedure

Let's practice.

Let's consider the expressions, establish the order of actions and perform calculations.

43 - (20 - 7) +15

32 + 9 * (19 - 16)

We will act according to the rule. The expression 43 - (20 - 7) +15 contains operations in parentheses, as well as addition and subtraction operations. Let's establish a procedure. The first action is to perform the operation in parentheses, and then, in order from left to right, subtraction and addition.

43 - (20 - 7) +15 =43 - 13 +15 = 30 + 15 = 45

The expression 32 + 9 * (19 - 16) contains operations in parentheses, as well as multiplication and addition operations. According to the rule, we first perform the action in parentheses, then multiplication (we multiply the number 9 by the result obtained by subtraction) and addition.

32 + 9 * (19 - 16) =32 + 9 * 3 = 32 + 27 = 59

In the expression 2*9-18:3 there are no parentheses, but there are multiplication, division and subtraction operations. We act according to the rule. First, we perform multiplication and division from left to right, and then subtract the result obtained from division from the result obtained by multiplication. That is, the first action is multiplication, the second is division, and the third is subtraction.

2*9-18:3=18-6=12

Let's find out whether the order of actions in the following expressions is correctly defined.

37 + 9 - 6: 2 * 3 =

18: (11 - 5) + 47=

7 * 3 - (16 + 4)=

Let's think like this.

37 + 9 - 6: 2 * 3 =

There are no parentheses in this expression, which means that we first perform multiplication or division from left to right, then addition or subtraction. In this expression, the first action is division, the second is multiplication. The third action should be addition, the fourth - subtraction. Conclusion: the procedure is determined correctly.

Let's find the value of this expression.

37+9-6:2*3 =37+9-3*3=37+9-9=46-9=37

Let's continue to talk.

The second expression contains parentheses, which means that we first perform the action in parentheses, then from left to right multiplication or division, addition or subtraction. We check: the first action is in parentheses, the second is division, the third is addition. Conclusion: the procedure is defined incorrectly. Let's correct the errors and find the value of the expression.

18:(11-5)+47=18:6+47=3+47=50

This expression also contains parentheses, which means that we first perform the action in parentheses, then from left to right multiplication or division, addition or subtraction. Let's check: the first action is in parentheses, the second is multiplication, the third is subtraction. Conclusion: the procedure is defined incorrectly. Let's correct the errors and find the value of the expression.

7*3-(16+4)=7*3-20=21-20=1

Let's complete the task.

Let's arrange the order of actions in the expression using the learned rule (Fig. 5).

Rice. 5. Procedure

We don't see numerical values, therefore we will not be able to find the meaning of the expressions, but we will practice applying the learned rule.

We act according to the algorithm.

The first expression contains parentheses, which means the first action is in parentheses. Then from left to right multiplication and division, then from left to right subtraction and addition.

The second expression also contains parentheses, which means we perform the first action in parentheses. After that, from left to right, multiplication and division, after that, subtraction.

Let's check ourselves (Fig. 6).

Rice. 6. Procedure

Today in class we learned about the rule for the order of actions in expressions without and with brackets.

References

1. M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 1. - M.: “Enlightenment”, 2012.
2. M.I. Moreau, M.A. Bantova and others. Mathematics: Textbook. 3rd grade: in 2 parts, part 2. - M.: “Enlightenment”, 2012.
3. M.I. Moro. Math lessons: Methodical recommendations for the teacher. 3rd grade. - M.: Education, 2012.
4. Regulatory document. Monitoring and evaluation of learning outcomes. - M.: “Enlightenment”, 2011.
5. “School of Russia”: Programs for primary school. - M.: “Enlightenment”, 2011.
6. S.I. Volkova. Mathematics: Test work. 3rd grade. - M.: Education, 2012.
7. V.N. Rudnitskaya. Tests. - M.: “Exam”, 2012.

1. Festival.1september.ru ().
2. Sosnovoborsk-soobchestva.ru ().
3. Openclass.ru ().

Homework

1. Determine the order of actions in these expressions. Find the meaning of the expressions.

2. Determine in what expression this order of actions is performed:

1. multiplication; 2. division;. 3. addition; 4. subtraction; 5. addition. Find the meaning of this expression.

3. Make up three expressions in which the following order of actions is performed:

1. multiplication; 2. addition; 3. subtraction

1. addition; 2. subtraction; 3. addition

1. multiplication; 2. division; 3. addition

Find the meaning of these expressions.

An equation is an equality containing a letter whose value must be found.

In equations, the unknown is usually represented by a lowercase Latin letter. The most commonly used letters are “x” [ix] and “y” [y].

• Root of the equation- this is the value of the letter at which the correct numerical equality is obtained from the equation.
• Solve the equation- means to find all its roots or make sure that there are no roots.

Having solved the equation, we always write down a check after the answer.

Information for parents

Dear parents, we draw your attention to the fact that elementary school and in 5th grade, children DO NOT know the topic “Negative Numbers”.

Therefore, they must solve equations using only the properties of addition, subtraction, multiplication, and division. Methods for solving equations for grade 5 are given below.

Do not try to explain the solution of equations by transferring numbers and letters from one part of the equation to another with a change in sign.

You can brush up on concepts related to addition, subtraction, multiplication and division in the lesson “Laws of Arithmetic”.

Solving addition and subtraction equations

How to find the unknown
term

How to find the unknown
minuend

How to find the unknown
subtrahend

To find the unknown term, you need to subtract the known term from the sum.

To find the unknown minuend, you need to add the subtrahend to the difference.

To find the unknown subtrahend, you need to subtract the difference from the minuend.

x + 9 = 15
x = 15 − 9
x=6
Examination

x − 14 = 2
x = 14 + 2
x = 16
Examination

16 − 2 = 14
14 = 14

5 − x = 3
x = 5 − 3
x = 2
Examination

Solving multiplication and division equations

How to find an unknown
factor

How to find the unknown
dividend

How to find an unknown
divider

To find an unknown factor, you need to divide the product by the known factor.

To find the unknown dividend, you need to multiply the quotient by the divisor.

To find an unknown divisor, you need to divide the dividend by the quotient.

y 4 = 12
y=12:4
y=3
Examination

y: 7 = 2
y = 2 7
y=14
Examination

8:y=4
y=8:4
y=2
Examination

An equation is an equality containing a letter whose sign must be found. The solution to an equation is the set of letter values ​​that turns the equation into a true equality:

Recall that to solve equation you need to transfer the terms with the unknown to one part of the equality, and the numerical terms to the other, bring similar ones and get the following equality:

From the last equality we determine the unknown according to the rule: “one of the factors is equal to the quotient divided by the second factor.”

Because rational numbers a and b can have the same and different signs, then the sign of the unknown is determined by the rules for dividing rational numbers.

Procedure for solving linear equations

The linear equation must be simplified by opening the brackets and performing the second step operations (multiplication and division).

Move the unknowns to one side of the equal sign, and the numbers to the other side of the equal sign, obtaining an equality identical to the given one,

Bring similar ones to the left and right of the equal sign, obtaining an equality of the form ax = b.

Calculate the root of the equation (find the unknown X from equality x = b : a),

Check by substituting the unknown into the given equation.

If we obtain an identity in a numerical equality, then the equation is solved correctly.

Special cases of solving equations

1. If equation given a product equal to 0, then to solve it we use the property of multiplication: “the product is equal to zero if one of the factors or both factors are equal to zero.”

27 (x - 3) = 0
27 is not equal to 0, which means x - 3 = 0

The second example has two solutions to the equation, since
this is a second degree equation:

If the coefficients of the equation are ordinary fractions, then first of all we need to get rid of the denominators. To do this:

Find the common denominator;

Determine additional factors for each term of the equation;

Multiply the numerators of fractions and integers by additional factors and write all terms of the equation without denominators (the common denominator can be discarded);

Move the terms with unknowns to one side of the equation, and the numerical terms to the other from the equal sign, obtaining an equivalent equality;

Bring similar members;

Basic properties of equations

In any part of the equation, you can add similar terms or open a parenthesis.

Any term of the equation can be transferred from one part of the equation to another by changing its sign to the opposite.

Both sides of the equation can be multiplied (divided) by the same number, except 0.

In the example above, all its properties were used to solve the equation.

Multiplication equations

1) To develop the ability to build an algorithm using the example of constructing an algorithm for solving simple multiplication equations, to develop the ability to use the constructed algorithm when solving an equation.

2) Train your computing skills and solve word problems.

Mental operations necessary at the design stage: analysis, synthesis, comparison, analogy.

Stage 1. Motivation for learning activities

1) motivate students to educational activities,

2) determine the content framework of the lesson.

Organization educational process at stage 1:

— What topic are we studying in mathematics lessons now? (Multiplication and division)

— In what tasks do we use these actions? (In solving examples, problems)

— Do you want to know what other tasks there are in which we can use these actions? (Yes)

Guys, look who came to our lesson today? Did you recognize them? What do you know about these heroes? (...)

(Question marks appear). What's happening? The Koloboks are puzzled and upset. They wanted to complete the task, but for the first time they failed. They don't know how to discover new knowledge. Shall we help? (...)

Is it possible to get to work with the same mood as the koloboks? (It’s impossible, there will be no result)

Let's smile at each other and wish each other good luck! Well, let's act according to the plan for discovering new knowledge. You know him well.

Stage 2. Updating knowledge and fixing difficulties in a trial action

1) updating the studied methods of action sufficient for construction, their verbal and symbolic fixation and generalization;

2) actualization of mental and cognitive processes, sufficient for the construction of new knowledge;

3) motivation for a trial educational action and its independent implementation;

4) students’ recording of individual difficulties in completing the test educational action or its justification.

Organization of the educational process at stage 2:

1) Updating formulas for finding the area and unknown side of a rectangle.

Where do we start? (With repetition). Should we repeat everything we know? (No, only what is useful for us to discover new knowledge)



- What do you need to find in this task? (Area of ​​rectangle)

— How to find the area of ​​a rectangle? (To find the area of ​​a rectangle, multiply the length by the width)

The area formula appears.

Students complete the task.

-What is the area? (18 sq. m)

- Who got a different answer?

- What is your mistake?

- How to find known side rectangle? (To find the unknown side of a rectangle, divide the area by the known side)

— A formula for finding the unknown side of a rectangle appears.

— Create an inverse problem in which you need to find the length of a rectangle (...)

— Let's write down the solution to the inverse problem.

The student who composed the inverse problem solves it on the board: 18:3=6(m) – length

- Now create another inverse problem.

The student who composed the inverse problem solves it on the board: 18:6=3 (m) – width

Who didn’t make mistakes in this task? Put yourself a + sign on the route sheet next to the repetition. Who made the mistake? Why did the error occur? Do you understand the reason? Correct the error. What will you set for yourself? (? and +).

2) Updating the algorithm for solving addition and subtraction equations.

— Write down: the sum of X + 5 is equal to 7. What can you call this entry? (Equation)

— What is an equation? (An equality in which there is an unknown number is called an equation)

- What will help us solve this equation? (Standard for solving addition equations)

One student at the blackboard commenting. (I’ll designate the components of the equation, underline the parts, circle the whole (sum). I see that the part is unknown. To find the unknown part, you need to subtract the known part from the sum.

Who didn’t make mistakes in this task? Put yourself a + sign on the route sheet next to the repetition. Who made the mistake? Why did the error occur? Do you understand the reason? Correct the error. What will you set for yourself? (- and +).

- Why did we repeat this? (This will be useful for us to discover new knowledge)

- Which next step? (Test action) What is it for? (To understand what we don't know)

The teacher gives students cards with a task for a trial action:

— What task needs to be completed? (Solve the equation)



- With what action? (With multiplication)

- What's new in this task? (We did not solve multiplication equations)

Try this task. (30 sec.)

— Who didn’t complete the task?

What were you unable to do? (We couldn't solve the equation)

- Who found the root of the equation? What results did you get?

The teacher records the results on the board next to the trial action.

- Justify your opinion.

What can't you do? (We cannot justify our answer.)

You have a problem. (difficulty). Let's put... (question mark) next to the trial action on the route sheet.

— What is the next step in the lesson? (Figure out what our problem is)

- And since a difficulty has arisen, you need to... (Stop and think)

Stage 3. Identifying the location and cause of the problem

1) restore the operations performed and record the location of the difficulty;

2) correlate your actions with the method of action used and, on this basis, identify and record in external speech the cause of the difficulty.

Organization of the educational process at stage 3:

-What task did you have to complete? (We had to solve a multiplication equation)

- How did you reason while performing the test action? (We tried to use a well-known algorithm for solving equations...)

- What is the difficulty? (The algorithm is not suitable)

Why did the difficulty arise? (We don't have a way to solve multiplication equations)

Do you understand what you don't know? (Yes). Put a + sign on your route sheet next to the third step.

Stage 4. Building a project to get out of a problem

1) agree and record the purpose and topic of the lesson;

2) build a plan and determine the means to achieve the goal.

Organization of the educational process at stage 4:

- We realized what we don’t know, now we can... (Discover the method ourselves)

First you need to set a goal. If you don’t know how to solve multiplication equations, then your goal is... (Discover a way to solve such equations)

- Formulate the topic of our lesson (...)

Write the topic on the board:

- We will act like real detectives. Let's draw up an action plan. Slide

- Let's think about what can help us. Remember, you repeated at the very beginning of the lesson. (Algorithm for solving addition equations, formula for finding area)

- What formula can help us? (Formula for finding the area and unknown side of a rectangle)

— Let's try to apply the formula for the area of ​​a rectangle.

— I suggest using the algorithm you know for solving addition equations.

Algorithm.

2. I distinguish the whole and the parts.
3. What is unknown?
4. I apply the rule.
5. I find unknown x.
6. What in this algorithm clearly does not suit you? (1 point)
7. When you had addition equations, you related their components to parts and wholes using line segments. What did you correlate the components of multiplication with? (With area)
8. What will you use instead of the segment? (Rectangle model)

Let's replace item 1 with Let's denote the components of the equation using the rectangle model.

— Do the remaining points of the algorithm suit you?

— Using this algorithm, can you try to solve the equation?

— What can we do to make it convenient to always use this rule? (Let's write the rule in general view)

Let's write the rule in general form.

- What means will we use?

Let's try to apply the formula for the area of ​​a rectangle...

Tools: rectangle model, algorithm.

Stage 5. Implementation of the completed project

1) implement the constructed project in accordance with the plan;

2) fix ways of writing expressions on the standard;

3) organize recording of overcoming the difficulty;

4) organize clarification of the general nature of the new knowledge.

Organization of the educational process at stage 5:

I suggest you work in groups. State the rules for working in groups.

Rules for working in groups

1. There must be a person in charge in the group.

2. One speaks, others listen.

3. Express your disagreement politely.

4. Everyone must work.

Students form groups.

- Carry out the plan in groups.

The person in charge from each group receives a task.

1. I will use a rectangle model and plot the components of the equation on the model.

2. I will apply the rule for the area of ​​a rectangle. (To find the unknown side of a rectangle, divide the area by the known side)

3. Find the root of the equation

We marked the numbers on the rectangle model. It can be seen that the side of the rectangle is unknown. To find the unknown side of a rectangle, you need to divide the area by the known side. We performed the calculations and found the root of the equation, x=5.

— What remains to be done according to plan? (Write the equation in general form)

— How to write the equation in general form? (Using letters of the Latin alphabet)

— How do you designate in the equation the numbers that are the sides of the rectangle? (We emphasize)

— I propose to take the number, which is the area, into a rectangle, why is this convenient? (Reminds me of the formula we use)

— Will it be necessary to create a different standard for the case where x is in the place of another factor? (No)

- Why? (You can use the commutative property of multiplication)

— How to check your discovery? What keys to knowledge do we have? (Look in textbook)

Open your textbooks on page 1. Read the rule.

Well done! You helped the koloboks. Slide (applause).

Let's now return to the trial action.

Complete what is needed on the board.

Were you able to overcome the difficulty? (Yes). Let's put a + sign on the route sheet.

On a regular board, under the step “I’ll find a way myself,” attach new standards.

What can you do now with the help of your new knowledge? (Solve equations)

Stage 6. Primary consolidation

1) organize children’s assimilation of a new method of action when solving multiplication equations with their pronunciation in external speech.

Organization of the educational process at stage 6:

1) Frontal work. On the board, the left side is the algorithm, the right side is the equation + model.

2) 4 x=8; 3 x=9; x · 4=12.

3) The teacher opens a task for consolidation on the board. Students go up to the board one by one and complete the task with commentary. Comment option:

- First, I will mark the area of ​​the rectangle with a square, and I will underline the sides. In this equation, the side of the rectangle is unknown. This means that the area of ​​the rectangle must be divided by the known side. Eight divided by 4 is 2, x equals 2.

Further execution of the task is commented in the same way.

Physical exercises for the eyes.

We'll take a little rest. and we will find the answer to everything.
Let's stand on our toes and stretch our arms up.
Hands on the waist, bend forward.
Now let's jump and sit down!

Now everyone has rested, and there is a new concern:

You need to do excellent pair work.

The teacher distributes cards with a task for pairs to work on.

Students complete tasks in pairs with comments. The check is organized according to the D-7 model.

— Check your results.

Correct the errors. Who didn’t make mistakes in this task? Put yourself a + sign on the route sheet next to the 5th step. Who made the mistake? Why did the error occur? Do you understand the reason? Correct the error. What will you set for yourself? (? and +)

— What is the next step in the lesson? (Test ourselves to see if we can handle it on our own)

Stage 7. Self-monitoring with self-test against a standard

1) train the ability to self-control and self-esteem;

2) test your ability to solve multiplication equations.

Organization of the educational process at stage 7:

- Complete these equations yourself. Students do independent work on cards

— The check is organized according to the D-8 standard.

- Draw a conclusion. (Need more practice.)

- Draw a conclusion. (We learned everything well.)

- Who didn’t make mistakes in this task? Put yourself a + sign on the route sheet next to the 5th step. Who made the mistake? Why did the error occur? Do you understand the reason? Correct the error. What will you set for yourself? (? and +).

Stage 8. Inclusion in the knowledge system and repetition

1) include new knowledge in the knowledge system;

2) train the ability to solve problems.

Organization of the educational process at stage 8:

— What do you need to know to correctly solve multiplication equations? (Multiplication and division tables, area formula). I suggest you solve problem No. 4 p.2.

Students complete the task. The check is organized according to the D-9 model.

-Which of you made a mistake?

- What is the mistake? (In choosing a rule, in calculations, ...)

Stage 9. Reflection on learning activities in the classroom

Goals:

1) record new content learned in the lesson;

2) evaluate your work and the work of the class in the lesson;

4) outline directions for future educational activities;

3) discuss homework.

Organization of the educational process at stage 9:

— What goal did you set for yourself? (...)

— Have you achieved your goal? (Prove it)

— I suggest you evaluate your work in class. Take another look at your lesson plans, see how many positives you have.

— On a regular board there is a picture of koloboks individually. One smiles. Those of you who think you understand and remember new topic, take exclamation marks and attach them next to the smiling Kolobok. Those who are still not sure about something, who still have questions, who have made mistakes in independent work– attach question mark next to the serious Kolobok. You will practice and you will definitely overcome your difficulty.

- You worked very well today, but does this mean that you don’t need to train anymore? (I need to do my homework)

xn--i1abbnckbmcl9fb.xn--p1ai

Solving equations by multiplication

An unknown quantity can be related to a known quantity not only by a + or - sign, but also by divided by some amount, as in this equation: $\frac = b$.

Here the solution cannot be found, as in previous examples, by transferring a term in the equation. But if both terms of the equation multiply on a, the equation takes the form
$x = ab.$

That is, the denominator of the fraction on the left side cancels. This can be proven by the properties of fractions.

When the unknown quantity divided by a known value, the equation is solved by multiplication each side by this known amount.

The same transfers must be made in this case as in the previous examples. However, we must remember that it is necessary to multiply every term of the equation.

Example 1. Solve the equation $\frac + a = b + d$
Multiply both sides by $c$
The product will be $x + ac = bc + cd$
And $x = bc + cd - ac$.

Example 1: Solve the equation $\frac + d = h$
Multiply by $a + b$ $x + ad + bd = ah + bh$.
And $x = ag + bh - ad - bd.$

When unknown value is in denominator fractions, the equation is solved in a similar way, that is, by multiplying the equation by the denominator.

Example 3: Solve the equation $\frac + 7 = 8$
Multiplying by $10 - x$ $6 + 70 - 7x = 80 - 8x$
Then $x = 4$.

Although this is not necessary, but it is often very convenient to get rid of the denominator of a fraction consisting only of famous quantities This can be done in a similar way, when we get rid of the denominator, which includes the unknown quantity.

Let's take for example $\frac = \frac +\frac $
Multiply by a $x = \frac +\frac $
Multiply by b $bx = ad + \frac $
Multiply by c $bcx = acd + abh$.

Or, we can multiply by the product of all denominators at once.

In the same equation $\frac = \frac +\frac $
Multiply terms by abc $\frac = \frac +\frac $

After reducing each identical value in one fraction, we get $bcx = acd + abh$, as in the previous version. From here,

In the equation we can get rid of fractions, multiplying each side of the equation by all denominators.

When getting rid of fractions in an equation, you must ensure that the signs and coefficients of each fraction are written correctly when opening the brackets

Cheat card “Solving equations. How to find the unknown", multiplication and division, 11x20 cm

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Russia is one of the ten most reading countries in the world! The interest in reading among our compatriots is growing from year to year, which is good news, because it is a wonderful and very useful habit.

By studying various literature, you can get a lot of valuable information, broaden your horizons, vocabulary and become erudite. Moreover, the book is great way relax and have fun. Let the Cheat Sheet “Solving Equations. How to find the unknown", multiplication and division, 11x20 cm will be another useful publication in your collection.

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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 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 beautiful 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 this step. In the second step we need to isolate the variables. Please 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 parentheses, but they are not multiplied by anything, they are simply preceded by various 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 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 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 algebraic sum. 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 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 got final decision, 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!