How to solve division equations. Solving simple linear equations. Solving linear equations. Examples

An equation with one unknown, which, after opening the brackets and bringing similar terms, takes the form

ax + b = 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we’ll figure out how to solve these linear equations.

For example, all equations:

2x + 3= 7 – 0.5x; 0.3x = 0; x/2 + 3 = 1/2 (x – 2) - linear.

The value of the unknown that turns the equation into a true equality is called decision or root of the equation .

For example, if in the equation 3x + 7 = 13 instead of the unknown x we ​​substitute the number 2, we obtain the correct equality 3 2 +7 = 13. This means that the value x = 2 is the solution or root of the equation.

And the value x = 3 does not turn the equation 3x + 7 = 13 into a true equality, since 3 2 +7 ≠ 13. This means that the value x = 3 is not a solution or a root of the equation.

Solving any linear equations reduces to solving equations of the form

ax + b = 0.

Let's move the free term from the left side of the equation to the right, changing the sign in front of b to the opposite, we get

If a ≠ 0, then x = ‒ b/a .

Example 1. Solve the equation 3x + 2 =11.

Let's move 2 from the left side of the equation to the right, changing the sign in front of 2 to the opposite, we get
3x = 11 – 2.

Let's do the subtraction, then
3x = 9.

To find x, you need to divide the product by a known factor, that is
x = 9:3.

This means that the value x = 3 is the solution or root of the equation.

Answer: x = 3.

If a = 0 and b = 0, then we get the equation 0x = 0. This equation has infinitely many solutions, since when we multiply any number by 0 we get 0, but b is also equal to 0. The solution to this equation is any number.

Example 2. Solve the equation 5(x – 3) + 2 = 3 (x – 4) + 2x ‒ 1.

Let's expand the brackets:
5x – 15 + 2 = 3x – 12 + 2x ‒ 1.

5x – 3x ‒ 2x = – 12 ‒ 1 + 15 ‒ 2.

Here are some similar terms:
0x = 0.

Answer: x - any number.

If a = 0 and b ≠ 0, then we get the equation 0x = - b. This equation has no solutions, since when we multiply any number by 0 we get 0, but b ≠ 0.

Example 3. Solve the equation x + 8 = x + 5.

Let’s group terms containing unknowns on the left side, and free terms on the right side:
x – x = 5 – 8.

Here are some similar terms:
0х = ‒ 3.

Answer: no solutions.

On Figure 1 shows a diagram for solving a linear equation

Let's draw up a general scheme for solving equations with one variable. Let's consider the solution to Example 4.

Example 4. Suppose we need to solve the equation

1) Multiply all terms of the equation by the least common multiple of the denominators, equal to 12.

2) After reduction we get
4 (x – 4) + 3 2 (x + 1) ‒ 12 = 6 5 (x – 3) + 24x – 2 (11x + 43)

3) To separate terms containing unknown and free terms, open the brackets:
4x – 16 + 6x + 6 – 12 = 30x – 90 + 24x – 22x – 86.

4) Let us group in one part the terms containing unknowns, and in the other - free terms:
4x + 6x – 30x – 24x + 22x = ‒ 90 – 86 + 16 – 6 + 12.

5) Let us present similar terms:
- 22х = - 154.

6) Divide by – 22, We get
x = 7.

As you can see, the root of the equation is seven.

Generally such equations can be solved using the following scheme:

a) bring the equation to its integer form;

b) open the brackets;

c) group the terms containing the unknown in one part of the equation, and the free terms in the other;

d) bring similar members;

e) solve an equation of the form aх = b, which was obtained after bringing similar terms.

However, this scheme is not necessary for every equation. When solving many simpler equations, you have to start not from the first, but from the second ( Example. 2), third ( Example. 13) and even from the fifth stage, as in example 5.

Example 5. Solve the equation 2x = 1/4.

Find the unknown x = 1/4: 2,
x = 1/8 .

Let's look at solving some linear equations found in the main state exam.

Example 6. Solve the equation 2 (x + 3) = 5 – 6x.

2x + 6 = 5 – 6x

2x + 6x = 5 – 6

Answer: - 0.125

Example 7. Solve the equation – 6 (5 – 3x) = 8x – 7.

– 30 + 18x = 8x – 7

18x – 8x = – 7 +30

Answer: 2.3

Example 8. Solve the equation

3(3x – 4) = 4 7x + 24

9x – 12 = 28x + 24

9x – 28x = 24 + 12

Example 9. Find f(6) if f (x + 2) = 3 7's

Solution

Since we need to find f(6), and we know f (x + 2),
then x + 2 = 6.

We solve the linear equation x + 2 = 6,
we get x = 6 – 2, x = 4.

If x = 4 then
f(6) = 3 7-4 = 3 3 = 27

Answer: 27.

If you still have questions or want to understand solving equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!

TutorOnline also recommends watching a new video lesson from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.

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A decision in absentia, in addition to the exceptional methods of decision provided for by law, can be canceled by the same court, with the resumption of consideration of the case on the merits at the request of the defendant, if he can prove that his failure to appear at the court hearing was caused by valid reasons.

It is possible to review a decision that has entered into legal force in cassation if the court reinstated the cassation deadline missed for a good reason.

Exclusivity property:

The property of exclusivity is the impossibility of re-applying to the court with a claim, complaint, statement, in a case between the same parties or their legal successors, on the same subject and based on the same circumstances (grounds of action), if there is a decision that has entered into legal force.

If, after the entry into force of the decision by which periodic payments are collected from the defendant, circumstances affecting the determination of the amount of payments or their duration change, then each party has the right, by filing a new claim, to demand a change in the amount and timing of payments.

In this case, new demands become the subject of consideration by the court, a new decision is made, which enters into legal force according to the general rules.

Submitting an identical application for consideration is also unacceptable when, during the initial consideration, the dispute between the parties was finally resolved by a ruling on the approval of a settlement agreement or on the applicant’s refusal of his claims. A second appeal to the court is not allowed if the proceedings are terminated.

Mandatory property:

Mandatory means that government bodies, officials, organizations and citizens are obliged to subordinate their activities to the content of the decision.

The Code of Civil Procedure emphasizes that the decision is binding throughout the territory of the Russian Federation, and in cases provided for by law, the courts of the Russian Federation can turn to foreign courts with a request to enforce decisions.

State bodies and officials are obliged to take the necessary actions to formalize and register the rights established by a court decision that has entered into legal force.

A court decision, after entering into legal force, must be executed voluntarily by the obligated persons, and, in necessary cases, forcibly by the executive bodies.

The need to implement the actions provided for in the decision is called the enforceability of decisions.

It is an integral part of obligation. The concept of obligation is broader than enforceability; it also covers the obligation of all persons and organizations that do not have a direct legal interest in a given case to take into account the authority of the court decision and contribute to its implementation.

Decisions in all cases are binding, but not all of them require execution, since they cannot be enforced. For example, decisions on claims for recognition do not require specific actions to protect the right challenged by the defendant. For them to be binding, it is sufficient for the court to recognize certain circumstances or legal relations (example: establishing paternity, recognizing the right of authorship, etc.).

Decisions on claims for recognition may have a prejudicial effect in a case regarding a claim for award. For example, the decision to establish paternity has prejudicial significance for the case of a claim for the recovery of alimony. Also, the decision to recognize the right of authorship is mandatory for the court in the case of collecting royalties from the publishing house.

The Family Code of the Russian Federation, in addition to family law issues, introduces several procedural rules regarding the actions (responsibilities) of the court after a decision is made. For example, the IC indicates that the court is obliged, within 3 days from the date of entry into legal force of the court decision on divorce, to send an extract from this decision to the civil registry office at the place of state registration of the marriage.

Family law requires the court to take certain actions to enforce the decision. After entering into legal force, court decisions acquire properties derived from the essence of legal force, the quality of prejudiciality (predecision).

Prejudiciality means that the relations and facts established by the court and recorded by the decision cannot be refuted during their secondary study by judicial and administrative bodies.

Prejudice comes down to the rules:

1. The court, administrative bodies, acting as jurisdictional bodies, re-analyzing the facts and relations, in whole or in part, the content of which was established by the court in a decision that has entered into legal force, are obliged to base their decisions on these facts and relations in the same form in which they were established , that is, the facts already established in the court decision are not proven again.

2. A party that bases its claims on legal relations that were fully or partially the subject of a court decision that has entered into legal force does not have to repeatedly prove the existence of these legal relations, the content of the elements of its components, as well as the legal facts underlying the parties’ claims.

Relations and facts are considered valid and are not subject to proof as long as the legal force of the decision is in effect, that is, until the decision is canceled. The other party, objecting to the applicant’s request, cannot present evidence to refute the facts and circumstances previously established by the court, as well as demand that the court examine them and attach them to the case.

3. If the subject of the study is a relationship whose content is established by a decision that has entered into legal force, then predetermination, that is, prejudiciality, applies to legal relations in full in any part of it in the form in which it was the subject of judicial research.

A decision that has entered into legal force has prejudicial significance in the consideration of a criminal case. A verdict in a criminal case that has entered into legal force is obligatory for the court considering the case on the civil legal consequences of the actions of a person in relation to whom the court verdict was made on the issues of whether this action took place and whether it was committed by this person.

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 give similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.

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

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

Scheme for solving simple linear equations

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

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

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

Solving real examples of simple linear equations

Task No. 1

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

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

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

So we got the answer.

Task No. 2

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

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

Here are some similar ones:

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

Task No. 3

The third linear equation is more interesting:

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

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

We perform the second step already known to us:

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

Let's do the math:

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

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

Things to Remember When Solving Linear Equations

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

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

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

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

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

Solving complex linear equations

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

Example No. 1

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

Now let's take a look at privacy:

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

Here are some similar ones:

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

\[\varnothing\]

or there are no roots.

Example No. 2

We perform the same actions. First step:

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

Here are some similar ones:

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

\[\varnothing\],

or there are no roots.

Nuances of the solution

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

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

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

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

We do the same with the second equation:

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

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

Solving even more complex linear equations

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

Task No. 1

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

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

Let's do some privacy:

Here are some similar ones:

Let's complete the last step:

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

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

Task No. 2

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

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

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

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

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

Here are similar terms:

Once again we have received the final answer.

Nuances of the solution

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

About the algebraic sum

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

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

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

Solving equations with fractions

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

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

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

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

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

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

Example No. 1

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

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

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

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

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

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

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

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

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

Example No. 2

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

Here we perform all the same actions:

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

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

The problem is solved.

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

Key points

Key findings are:

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

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