Linear equations. Solution, examples.

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Linear equations.

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

A linear equation is usually defined as an equation of the form:

ax + b = 0 where a and b- any numbers.

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

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

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

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

But that's not all! If, say, a=0, a b=5, it turns out something quite absurd:

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

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

A linear equation can be clearly recognized in some cases. Say, if we have an equation in which there are only unknowns in the first degree, yes numbers. And the equation doesn't fractions divided by unknown , it is important! And division by number, or a numeric fraction - that's it! For example:

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

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

It turns out that it is impossible to find out a linear equation in some intricate example until you almost solve it. It's upsetting. But in assignments, as a rule, they don’t ask about the form of the equation, right? In tasks, equations are ordered decide. This makes me happy.)

Solution of linear equations. Examples.

The entire solution of linear equations consists of identical transformations of equations. By the way, these transformations (as many as two!) underlie the solutions all equations of mathematics. In other words, the decision any The equation begins with these same transformations. In the case of linear equations, it (the solution) on these transformations ends with a full-fledged answer. It makes sense to follow the link, right?) Moreover, there are also examples of solving linear equations.

Let's start with the simplest example. Without any pitfalls. Let's say we need to solve the following equation.

x - 3 = 2 - 4x

This is a linear equation. Xs are all to the first power, there is no division by X. But, actually, we don't care what the equation is. We need to solve it. The scheme here is simple. Collect everything with x's on the left side of the equation, everything without x's (numbers) on the right.

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

x + 4x = 2 + 3

We give similar, we consider:

What do we need to be completely happy? Yes, so that there is a clean X on the left! Five gets in the way. Get rid of the five with second identical transformation of equations. Namely, we divide both parts of the equation by 5. We get a ready-made answer:

An elementary example, of course. This is for a warm-up.) It is not very clear why I recalled identical transformations here? OK. We take the bull by the horns.) Let's decide something more impressive.

For example, here is this equation:

Where do we start? With X - to the left, without X - to the right? Could be so. Little steps along the long road. And you can immediately, in a universal and powerful way. Unless, of course, in your arsenal there are identical transformations of equations.

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

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

Expanding the brackets:

Note! Numerator (x+2) I took in brackets! This is because when multiplying fractions, the numerator is multiplied by the whole, entirely! And now you can reduce fractions and reduce:

Opening the remaining parentheses:

Not an example, but pure pleasure!) Now we recall the spell from the lower grades: with x - to the left, without x - to the right! And apply this transformation:

Here are some like:

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

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

Take note: to bring the original confusing equation to a pleasant form, we used two (only two!) identical transformations- translation left-right with a change of sign and multiplication-division of the equation by the same number. This is the universal way! We will work in this way any equations! Absolutely any. That is why I keep repeating these identical transformations all the time.)

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

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

Special cases in solving linear equations.

Surprise first.

Suppose you come across an elementary equation, something like:

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

Slightly bored, we transfer with X to the left, without X - to the right ... With a change of sign, everything is chin-chinar ... We get:

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

We believe, and ... oh my! We get:

In itself, this equality is not objectionable. Zero really zero. But X is gone! And we must write in the answer, what x is equal to. Otherwise, the solution doesn't count, yes...) A dead end?

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

But we have the correct equality already happened! 0=0, where really?! It remains to figure out at what x's this is obtained. What values ​​of x can be substituted into initial equation if these x's still shrink to zero? Come on?)

Yes!!! Xs can be substituted any! What do you want. At least 5, at least 0.05, at least -220. They will still shrink. If you don't believe me, you can check it.) Substitute any x values ​​in initial equation and calculate. All the time will be pure truth: 0=0, 2=2, -7.1=-7.1 and so on.

Here is your answer: x is any number.

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

Surprise second.

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

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

After the same identical transformations, we get something intriguing:

Like this. Solved a linear equation, got a strange equality. Mathematically speaking, we have wrong equality. And speaking plain language, this is not true. Rave. But nevertheless, this nonsense is quite a good reason for right decision equations.)

Again, we think from general rules. What x, when substituted into the original equation, will give us correct equality? Yes, none! There are no such xes. Whatever you substitute, everything will be reduced, nonsense will remain.)

Here is your answer: there are no solutions.

This is also a perfectly valid answer. In mathematics, such answers often occur.

Like this. Now, I hope, the loss of Xs in the process of solving any (not only linear) equation will not bother you at all. The matter is familiar.)

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

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Service for solving equations online will help you solve any equation. Using our site, you will not only get the answer to the equation, but also see detailed solution, that is, a step-by-step display of the process of obtaining the result. Our service will be useful for high school students and their parents. Students will be able to prepare for tests, exams, test their knowledge, and parents will be able to control the solution of mathematical equations by their children. The ability to solve equations is a mandatory requirement for students. The service will help you self-learn and improve your knowledge in the field of mathematical equations. With it, you can solve any equation: quadratic, cubic, irrational, trigonometric, etc. online service but priceless, because in addition to the correct answer, you get a detailed solution to each equation. Benefits of solving equations online. You can solve any equation online on our website absolutely free of charge. The service is fully automatic, you do not have to install anything on your computer, you just need to enter the data and the program will issue a solution. Any calculation errors or typographical errors are excluded. It is very easy to solve any equation online with us, so be sure to use our site to solve any kind of equations. You only need to enter the data and the calculation will be completed in seconds. The program works independently, without human intervention, and you get an accurate and detailed answer. Solution of the equation in general form. In such an equation, the variable coefficients and the desired roots are interconnected. The highest power of a variable determines the order of such an equation. Based on this, for the equations use various methods and theorems for finding solutions. Solving equations of this type means finding the desired roots in a general form. Our service allows you to solve even the most complex algebraic equation online. You can get both the general solution of the equation and the private solution for the ones you specified. numerical values coefficients. To solve an algebraic equation on the site, it is enough to correctly fill in only two fields: the left and right parts of the given equation. Algebraic equations with variable coefficients have an infinite number of solutions, and by setting certain conditions, particular ones are selected from the set of solutions. Quadratic equation. The quadratic equation has the form ax^2+bx+c=0 for a>0. The solution of equations of a square form implies finding the values ​​of x, at which the equality ax ^ 2 + bx + c \u003d 0 is satisfied. To do this, the value of the discriminant is found by the formula D=b^2-4ac. If the discriminant is less than zero, then the equation has no real roots (the roots are from the field of complex numbers), if it is zero, then the equation has one real root, and if the discriminant is greater than zero, then the equation has two real roots, which are found by the formula: D \u003d -b + -sqrt / 2a. To solve a quadratic equation online, you just need to enter the coefficients of such an equation (whole numbers, fractions or decimal values). If there are subtraction signs in the equation, you must put a minus in front of the corresponding terms of the equation. You can also solve a quadratic equation online depending on the parameter, that is, the variables in the coefficients of the equation. Our online service for finding common solutions perfectly copes with this task. Linear equations. To solve linear equations (or systems of equations), four main methods are used in practice. Let's describe each method in detail. Substitution method. Solving equations using the substitution method requires expressing one variable in terms of the others. After that, the expression is substituted into other equations of the system. Hence the name of the solution method, that is, instead of a variable, its expression through the rest of the variables is substituted. In practice, the method requires complex calculations, although it is easy to understand, so solving such an equation online will save time and make calculations easier. You just need to specify the number of unknowns in the equation and fill in the data from linear equations, then the service will make the calculation. Gauss method. The method is based on the simplest transformations of the system in order to arrive at an equivalent triangular system. The unknowns are determined one by one from it. In practice, it is required to solve such an equation online with detailed description, thanks to which you will master well the Gauss method for solving systems of linear equations. Write down the system of linear equations in the correct format and take into account the number of unknowns in order to correctly solve the system. Cramer's method. This method solves systems of equations in cases where the system has only decision. The main mathematical operation here is the calculation of matrix determinants. The solution of equations by the Cramer method is carried out online, you get the result instantly with a complete and detailed description. It is enough just to fill the system with coefficients and choose the number of unknown variables. matrix method. This method consists in collecting the coefficients of the unknowns in matrix A, the unknowns in column X, and the free terms in column B. Thus, the system of linear equations is reduced to matrix equation of the form AxX=B. This equation has a unique solution only if the determinant of the matrix A is non-zero, otherwise the system has no solutions, or an infinite number of solutions. The solution of equations by the matrix method is to find the inverse matrix A.

In the 7th grade mathematics course, they first meet with equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why it is out of sight whole line problems in which certain conditions are introduced on the coefficients of the equation that restrict them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although in USE materials and at the entrance examinations problems of this kind are encountered more and more often.

Which equation will be called an equation with two variables?

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

Consider the equation 2x - y = 1. It turns into a true equality at x = 2 and y = 3, so this pair of variable values ​​is a solution to the equation under consideration.

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

An equation with two unknowns can:

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

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

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

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

The main methods for solving equations with two variables are methods based on the decomposition of expressions into factors, the selection of a full square, the use of the properties of a quadratic equation, the boundedness of expressions, and evaluation methods. The equation, as a rule, is transformed into a form from which a system for finding unknowns can be obtained.

Factorization

Example 1

Solve the equation: xy - 2 = 2x - y.

Solution.

We group the terms for the purpose of factoring:

(xy + y) - (2x + 2) = 0. Take out the common factor from each bracket:

y(x + 1) – 2(x + 1) = 0;

(x + 1)(y - 2) = 0. We have:

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

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

Zero is not negative numbers

Example 2

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

Solution.

Grouping:

(9x 2 - 12x + 4) + (4y 2 - 12y + 9) = 0. Now each parenthesis can be collapsed using the square difference formula.

(3x - 2) 2 + (2y - 3) 2 = 0.

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

So x = 2/3 and y = 3/2.

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

Evaluation method

Example 3

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

Solution.

In each bracket, select the full square:

((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Estimate the meaning of the expressions in brackets.

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

(x + 1) 2 + 1 = 1 and (y - 2) 2 + 2 = 2, so x = -1, y = 2.

Answer: (-1; 2).

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

Example 4

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

Solution.

Let's solve the equation as a quadratic one with respect to x. Let's find the discriminant:

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

Answer: (3; 4).

Often in equations with two unknowns indicate restrictions on variables.

Example 5

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

Solution.

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

Answer: no roots.

Example 6

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

Solution.

Let's select the full squares in each bracket:

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

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

Example 7

For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Answer the smallest amount.

Solution.

Select full squares:

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

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

(x - y) 2 = 36 and (y + 2) 2 = 1

(x - y) 2 = 1 and (y + 2) 2 = 36.

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

Answer: -17.

Do not despair if you have difficulties when solving equations with two unknowns. With a little practice, you will be able to master any equation.

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Instruction

Substitution method Express one variable and substitute it into another equation. You can express any variable you like. For example, express "y" from the second equation:
x-y=2 => y=x-2 Then plug everything into the first equation:
2x+(x-2)=10 Move everything without x to the right side and count:
2x+x=10+2
3x=12 Next, for "x, divide both sides of the equation by 3:
x=4. So, you have found "x. Find "at. To do this, substitute "x" into the equation from which you expressed "y:
y=x-2=4-2=2
y=2.

Make a check. To do this, substitute the resulting values ​​into the equations:
2*4+2=10
4-2=2
Unknown found correctly!

How to add or subtract equations Get rid of any variable at once. In our case, this is easier to do with "y.
Since in “y” is “+” and in the second “-”, then you can perform the addition operation, i.e. We add the left side to the left, and the right side to the right:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4 Substitute "x" into any equation and find "y:
2*4+y=10
8+y=10
y=10-8
y=2 According to the 1st method, you can find what you found correctly.

If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have "2x", and in the second just "x. In order for the addition or “x to decrease, multiply the second equation by 2:
x-y=2
2x-2y=4 Then subtract the second equation from the first equation:
2x+y-(2x-2y)=10-4
2x+y-2x+2y=6
3y=6
find y \u003d 2 "x by expressing from any equation, i.e.
x=4

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Tip 2: How to solve a linear equation with two variables

The equation, written in general form ax + by + c \u003d 0, is called a linear equation with two variables. Such an equation itself contains an infinite number of solutions, so in problems it is always supplemented with something - one more equation or limiting conditions. Depending on the conditions provided by the problem, solve a linear equation with two variables should different ways.

You will need

• - linear equation with two variables;
• - second equation or additional terms.

Instruction

Given a system of two linear equations, solve it as follows. Choose one of the equations in which the coefficients before variables smaller and express one of the variables, for example, x. Then plug that value containing y into the second equation. In the resulting equation there will be only one variable y, move all parts with y to the left side, and the free ones to the right. Find y and substitute in any of the original equations, find x.

There is another way to solve a system of two equations. Multiply one of the equations by a number so that the coefficient in front of one of the variables, for example, in front of x, is the same in both equations. Then subtract one of the equations from the other (if the right hand side is not 0, remember to subtract the right hand side in the same way). You will see that the x variable has disappeared and only one y remains. Solve the resulting equation, and substitute the found value of y into any of the original equalities. Find x.

The third way to solve a system of two linear equations is graphical. Draw a coordinate system and draw graphs of two straight lines, the equations of which are indicated in your system. To do this, substitute any two x values ​​\u200b\u200binto the equation and find the corresponding y - these will be the coordinates of the points belonging to the line. It is most convenient to find the intersection with the coordinate axes - just substitute the values ​​x=0 and y=0. The coordinates of the point of intersection of these two lines will be the tasks.

If there is only one linear equation in the conditions of the problem, then you are given additional conditions due to which you can find a solution. Read the problem carefully to find these conditions. If a variables x and y are distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y is hiding the number of , apples, etc. – then the values ​​can only be . If x is the age of the son, it is clear that he cannot be older than father, so specify it in the task conditions.

Sources:

• how to solve an equation with one variable

By itself the equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.

You will need

• - a system of three equations with three unknowns.

Instruction

If two of the three systems have only two of the three unknowns, try expressing some variables in terms of the others and plugging them into the equation with three unknown. Your goal with this is to turn it into a normal the equation with the unknown. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.

Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of by or a variable so that two unknowns are reduced at once. If there is such an opportunity, use it, most likely, the subsequent decision will not be difficult. Do not forget that when multiplying by a number, you must multiply both the left side and the right side. Similarly, when subtracting equations, remember that the right hand side must also be subtracted.

If the previous methods did not help, use in a general way solutions of any equations with three unknown. To do this, rewrite the equations in the form a11x1 + a12x2 + a13x3 \u003d b1, a21x1 + a22x2 + a23x3 \u003d b2, a31x1 + a32x2 + a33x3 \u003d b3. Now make a matrix of coefficients at x (A), a matrix of unknowns (X) and a matrix of free ones (B). Pay attention, multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix, a matrix of free members, that is, A * X \u003d B.

Find the matrix A to the power (-1) after finding , note that it should not be equal to zero. After that, multiply the resulting matrix by matrix B, as a result you will get the desired matrix X, indicating all the values.

You can also find a solution to a system of three equations using the Cramer method. To do this, find the third-order determinant ∆ corresponding to the matrix of the system. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values ​​of the free terms instead of the values ​​of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.

Sources:

• solutions of equations with three unknowns

Solving a system of equations is complex and exciting. The more complex the system, the more interesting it is to solve. Most often in mathematics high school there are systems of equations with two unknowns, but in higher mathematics there can be more variables. Systems can be solved in several ways.

Instruction

The most common method for solving a system of equations is substitution. To do this, you need to express one variable through another and substitute it into the second the equation systems, thus bringing the equation to one variable. For example, given the equations: 2x-3y-1=0; x+y-3=0.

It is convenient to express one of the variables from the second expression, transferring everything else to the right side of the expression, not forgetting to change the sign of the coefficient: x = 3-y.

We open the brackets: 6-2y-3y-1 \u003d 0; -5y + 5 \u003d 0; y \u003d 1. The resulting value of y is substituted into the expression: x \u003d 3-y; x \u003d 3-1; x \u003d 2.

In the first expression, all members are 2, you can take 2 out of the bracket to the distributive property of multiplication: 2 * (2x-y-3) = 0. Now both parts of the expression can be reduced by this number, and then express y, since the modulo coefficient for it is equal to one: -y \u003d 3-2x or y \u003d 2x-3.

Just as in the first case, we substitute this expression into the second the equation and we get: 3x+2*(2x-3)-8=0;3x+4x-6-8=0;7x-14=0;7x=14;x=2. Substitute the resulting value into the expression: y=2x -3;y=4-3=1.

We see that the coefficient at y is the same in value, but different in sign, therefore, if we add these equations, we will completely get rid of y: 4x + 3x-2y + 2y-6-8 \u003d 0; 7x-14 \u003d 0; x=2. We substitute the value of x into any of the two equations of the system and get y=1.

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Bisquare the equation represents the equation fourth degree general form which is represented by the expression ax^4 + bx^2 + c = 0. Its solution is based on the use of the method of substitution of unknowns. In this case, x^2 is replaced by another variable. Thus, the result is an ordinary square the equation, which is to be solved.

Instruction

Solve a square the equation resulting from the substitution. To do this, first calculate the value in accordance with the formula: D = b^2 ? 4ac. In this case, the variables a, b, c are the coefficients of our equation.

Find the roots of the biquadratic equation. To do this, take the square root of the solutions obtained. If there was one decision, then there will be two - positive and negative meaning square root. If there were two solutions, the biquadratic equation would have four roots.

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One of classical ways solving systems of linear equations is the Gauss method. It consists in the successive exclusion of variables, when the system of equations is converted into a step system with the help of simple transformations, from which all variables are sequentially found, starting with the last ones.

Instruction

First, bring the system of equations into such a form when all the unknowns will be in a strictly defined order. For example, all unknown Xs will come first in each line, all Ys will come after X, all Zs will come after Y, and so on. There should be no unknowns on the right side of each equation. Mentally determine the coefficients in front of each unknown, as well as the coefficients on the right side of each equation.

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.

To begin with, let's define: what is a linear equation and which of them should be called the simplest?

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

The simplest equation means the construction:

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

1. Open brackets, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Bring like 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 you get something like $0\cdot x=8$, i.e. on the left is zero, and on the right is a non-zero number. 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.

And now let's see how it all works on the example of real problems.

Examples of solving equations

Today we deal 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 open the parentheses, if any (as in our last example);
2. Then bring similar
3. Finally, isolate the variable, i.e. everything that is connected with the variable - the terms in which it is contained - is transferred to one side, and everything that remains without it is transferred to the other side.

Then, as a rule, you need to bring similar on each side of the resulting equality, and after that it remains only to divide by the coefficient at "x", and we will get the final answer.

In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Usually, mistakes are made either when opening brackets, or when counting "pluses" and "minuses".

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

Scheme for solving simple linear equations

To begin with, let me once again write the entire scheme for solving the simplest linear equations:

1. Expand the parentheses, if any.
2. Seclude variables, i.e. everything that contains "x" is transferred to one side, and without "x" - to the other.
3. We present similar terms.
4. We divide everything by the coefficient at "x".

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

Solving real examples of simple linear equations

Task #1

In the first step, we are required 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. Note: we are talking only about individual terms. Let's write:

We give similar terms on the left and on the right, but this has already been done here. Therefore, we proceed to the fourth step: divide by a factor:

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

Here we got the answer.

Task #2

In this task, we can observe the brackets, so let's expand them:

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

Here are some like:

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

Task #3

The third linear equation is already more interesting:

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

There are a few brackets here, but they are not multiplied by anything, they just stand in front of them 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 calculate:

We perform the last step - we divide everything by the coefficient at "x":

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

Things to Remember When Solving Linear Equations

If we ignore too simple tasks, then 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, zero can get in among them - there is nothing wrong with that.

Zero is the same number as the rest, you should not somehow discriminate it or assume that if you get zero, then you did something wrong.

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

Understanding this simple fact will keep you from making stupid and hurtful mistakes in high school when doing such things is taken for granted.

Solving complex linear equations

Let's move on to more complex equations. Now the constructions will become more complicated and a quadratic function will appear when performing various transformations. However, you should not be afraid of this, because if, according to the author's intention, we solve a linear equation, then in the process of transformation all monomials containing a quadratic function will necessarily be reduced.

Example #1

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

Now let's take privacy:

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

Here are some like:

Obviously, this equation has no solutions, so in the answer we write as follows:

\[\variety \]

or no roots.

Example #2

We perform the same steps. First step:

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

Here are some like:

Obviously, this linear equation has no solution, so we write it like this:

\[\varnothing\],

or no roots.

Nuances of the solution

Both equations are completely solved. On the example of these two expressions, we once again made sure that even in the simplest linear equations, everything can be not so simple: there can be either one, or none, or infinitely many. In our case, we considered two equations, in both there are simply no roots.

But I would like to draw your attention to another fact: how to work with brackets and how to expand 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: multiply each individual term. Inside there are two terms - respectively, two terms and is multiplied.

And only after these seemingly elementary, but very important and dangerous transformations have been completed, can the bracket be opened from the point of view that there is a minus sign after it. Yes, yes: only now, when the transformations are done, we remember that there is a minus sign in front of the brackets, which means that everything below just 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 no coincidence 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 learn to solve such simple equations again.

Of course, the day will come when you will hone these skills to automatism. You no longer have to perform so many transformations each time, you will write everything in 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 #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 a retreat:

Here are some like:

Let's do 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, however, they mutually annihilated, which makes the equation exactly linear, not square.

Task #2

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

Let's do the first step carefully: multiply every element in the first bracket by every element in the second. In total, four new terms should be obtained after transformations:

And now carefully perform the multiplication in each term:

Let's move the terms with "x" to the left, and without - to the right:

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

Here are similar terms:

We have received a definitive answer.

Nuances of the solution

The most important remark about these two equations is as follows: as soon as we start multiplying brackets in which there is a term greater than it, then this is done according to next 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 get four terms.

On the algebraic sum

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

In conclusion, 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 a fraction

To solve such tasks, one more step will have to be added to our algorithm. But first, I will remind our algorithm:

1. Open brackets.
2. Separate variables.
3. Bring similar.
4. Divide by a factor.

Alas, this wonderful algorithm, for all its efficiency, is not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on the left and on 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 performed both before the first action and after it, namely, to get rid of fractions. Thus, the algorithm will be as follows:

1. Get rid of fractions.
2. Open brackets.
3. Separate variables.
4. Bring similar.
5. Divide by a factor.

What does it mean to "get rid of fractions"? And why is it possible to do this both after and before the first standard step? In fact, in our case, all fractions are numeric in terms of the denominator, i.e. everywhere the denominator is just a number. Therefore, if we multiply both parts of the equation by this number, then we will get rid of fractions.

Example #1

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

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

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

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

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

Now let's open it:

We perform seclusion of a variable:

We carry out the reduction of similar terms:

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

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

We have received the final solution, we pass to the second equation.

Example #2

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

Here we perform all the same actions:

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

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

Problem solved.

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

Key points

The key findings are as follows:

• Know the algorithm for solving linear equations.
• Ability to open brackets.
• Do not worry if you have quadratic functions somewhere, most likely, in the process of further transformations, they will be reduced.
• The roots in linear equations, even the simplest ones, are of three types: one single root, the entire number line is a root, there are 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, solve the examples presented there. Stay tuned, there are many more interesting things waiting for you!