Solving simple linear equations. How to Solve Linear Equations

In this article we will consider the principle of solving such equations as linear equations. Let's write down the definition of these equations and set the general form. We will analyze all the conditions for finding solutions to linear equations, using, among other things, practical examples.

Please note that the material below contains information on linear equations with one variable. Linear equations with two variables are discussed in a separate article.

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What is a linear equation

Definition 1

Linear equation is an equation written as follows:
a x = b, Where x– variable, a And b- some numbers.

This formulation was used in the algebra textbook (7th grade) by Yu.N. Makarychev.

Example 1

Examples of linear equations would be:

3 x = 11(equation with one variable x at a = 5 And b = 10);

− 3 , 1 y = 0 ( linear equation with variable y, Where a = - 3, 1 And b = 0);

x = − 4 And − x = 5.37(linear equations, where the number a written explicitly and equal to 1 and - 1, respectively. For the first equation b = - 4 ; for the second - b = 5.37) and so on.

Different educational materials may have different definitions. For example, Vilenkin N.Ya. Linear equations also include those equations that can be transformed into the form a x = b by transferring terms from one part to another with a change of sign and bringing similar terms. If we follow this interpretation, the equation 5 x = 2 x + 6 – also linear.

But the algebra textbook (7th grade) by Mordkovich A.G. gives the following description:

Definition 2

A linear equation in one variable x is an equation of the form a x + b = 0, Where a And b– some numbers called coefficients of a linear equation.

Example 2

An example of linear equations of this type could be:

3 x − 7 = 0 (a = 3 , b = − 7) ;

1, 8 y + 7, 9 = 0 (a = 1, 8, b = 7, 9).

But there are also examples of linear equations that we have already used above: of the form a x = b, For example, 6 x = 35.

We will immediately agree that in this article by a linear equation with one variable we will understand the equation written a x + b = 0, Where x– variable; a, b – coefficients. We see this form of a linear equation as the most justified, since linear equations are algebraic equations of the first degree. And the other equations indicated above, and the equations given by equivalent transformations in the form a x + b = 0, we define as equations that reduce to linear equations.

With this approach, the equation 5 x + 8 = 0 is linear, and 5 x = − 8- an equation that reduces to a linear one.

Principle of solving linear equations

Let's look at how to determine whether a given linear equation will have roots and, if so, how many and how to determine them.

Definition 3

The fact of the presence of roots of a linear equation is determined by the values of the coefficients a And b. Let's write down these conditions:

at a ≠ 0 linear equation has a single root x = - b a ;
at a = 0 And b ≠ 0 a linear equation has no roots;
at a = 0 And b = 0 a linear equation has infinitely many roots. Essentially, in this case, any number can become the root of a linear equation.

Let's give an explanation. We know that in the process of solving an equation it is possible to transform a given equation into an equivalent one, which means that it has the same roots as the original equation, or also has no roots. We can make the following equivalent transformations:

transfer a term from one part to another, changing the sign to the opposite;
multiply or divide both sides of an equation by the same number that is not zero.

Thus, we transform the linear equation a x + b = 0, moving the term b from the left side to the right side with a change of sign. We get: a · x = − b .

So, we divide both sides of the equation by a non-zero number A, resulting in an equality of the form x = - b a . That is, when a ≠ 0, original equation a x + b = 0 is equivalent to the equality x = - b a, in which the root - b a is obvious.

By contradiction it is possible to demonstrate that the root found is the only one. Let us designate the found root - b a as x 1 . Let us assume that there is another root of the linear equation with the designation x 2 . And of course: x 2 ≠ x 1, and this, in turn, based on the definition equal numbers through the difference, is equivalent to the condition x 1 − x 2 ≠ 0 . Taking into account the above, we can create the following equalities by substituting the roots:
a x 1 + b = 0 and a x 2 + b = 0.
The property of numerical equalities makes it possible to perform term-by-term subtraction of parts of equalities:

a x 1 + b − (a x 2 + b) = 0 − 0, from here: a · (x 1 − x 2) + (b − b) = 0 and onwards a · (x 1 − x 2) = 0 . Equality a · (x 1 − x 2) = 0 is incorrect because it was previously specified that a ≠ 0 And x 1 − x 2 ≠ 0 . The resulting contradiction serves as proof that when a ≠ 0 linear equation a x + b = 0 has only one root.

Let us justify two more clauses of the conditions containing a = 0 .

When a = 0 linear equation a x + b = 0 will be written as 0 x + b = 0. The property of multiplying a number by zero gives us the right to assert that whatever number is taken as x, substituting it into equality 0 x + b = 0, we get b = 0 . The equality is valid for b = 0; in other cases, when b ≠ 0, equality becomes false.

So when a = 0 and b = 0 , any number can become the root of a linear equation a x + b = 0, since when these conditions are met, substituting instead x any number, we get the correct numerical equality 0 = 0 . When a = 0 And b ≠ 0 linear equation a x + b = 0 will not have roots at all, since when the specified conditions are met, substituting instead x any number, we get an incorrect numerical equality b = 0.

All the above considerations give us the opportunity to write down an algorithm that makes it possible to find a solution to any linear equation:

by the type of record we determine the values of the coefficients a And b and analyze them;
at a = 0 And b = 0 the equation will have infinitely many roots, i.e. any number will become the root of the given equation;
at a = 0 And b ≠ 0
at a, different from zero, we begin searching for the only root of the original linear equation:

let's move the coefficient b to the right side with a change of sign to the opposite, bringing the linear equation to the form a · x = − b ;
divide both sides of the resulting equality by the number a, which will give us the desired root of the given equation: x = - b a.

Actually, the described sequence of actions is the answer to the question of how to find a solution to a linear equation.

Finally, let us clarify that equations of the form a x = b are solved using a similar algorithm with the only difference that the number b in such a notation has already been transferred to the required part of the equation, and with a ≠ 0 you can immediately divide the parts of an equation by a number a.

Thus, to find a solution to the equation a x = b, we use the following algorithm:

at a = 0 And b = 0 the equation will have infinitely many roots, i.e. any number can become its root;
at a = 0 And b ≠ 0 the given equation will have no roots;
at a, not equal to zero, both sides of the equation are divided by the number a, which makes it possible to find the only root that is equal to b a.

Examples of solving linear equations

Example 3

Linear equation needs to be solved 0 x − 0 = 0.

Solution

By writing the given equation we see that a = 0 And b = − 0(or b = 0, which is the same). Thus, a given equation can have an infinite number of roots or any number.

Answer: x– any number.

Example 4

It is necessary to determine whether the equation has roots 0 x + 2, 7 = 0.

Solution

From the record we determine that a = 0, b = 2, 7. Thus, the given equation will have no roots.

Answer: the original linear equation has no roots.

Example 5

Given a linear equation 0.3 x − 0.027 = 0. It needs to be resolved.

Solution

By writing the equation we determine that a = 0, 3; b = - 0.027, which allows us to assert that the given equation has a single root.

Following the algorithm, we move b to the right side of the equation, changing the sign, we get: 0.3 x = 0.027. Next, we divide both sides of the resulting equality by a = 0, 3, then: x = 0, 027 0, 3.

Let's divide decimal fractions:

0.027 0.3 = 27 300 = 3 9 3 100 = 9 100 = 0.09

The result obtained is the root of the given equation.

Let us briefly write the solution as follows:

0.3 x - 0.027 = 0.0.3 x = 0.027, x = 0.027 0.3, x = 0.09.

Answer: x = 0.09.

For clarity, we present the solution to the writing equation a x = b.

Example N

The given equations are: 1) 0 x = 0 ; 2) 0 x = − 9 ; 3) - 3 8 x = - 3 3 4 . They need to be resolved.

Solution

All given equations correspond to the entry a x = b. Let's look at them one by one.

In the equation 0 x = 0, a = 0 and b = 0, which means: any number can be the root of this equation.

In the second equation 0 x = − 9: a = 0 and b = − 9, thus, this equation will have no roots.

Based on the form of the last equation - 3 8 · x = - 3 3 4, we write the coefficients: a = - 3 8, b = - 3 3 4, i.e. the equation has a single root. Let's find him. Let's divide both sides of the equation by a, resulting in: x = - 3 3 4 - 3 8. Let's simplify the fraction by applying the rule of dividing negative numbers and then converting the mixed number to common fraction and dividing ordinary fractions:

3 3 4 - 3 8 = 3 3 4 3 8 = 15 4 3 8 = 15 4 8 3 = 15 8 4 3 = 10

Let us briefly write the solution as follows:

3 8 · x = - 3 3 4 , x = - 3 3 4 - 3 8 , x = 10 .

Answer: 1) x– any number, 2) the equation has no roots, 3) x = 10.

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Linear equations. Solution, examples.

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

Linear equations.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solving linear equations. Examples.

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

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

x - 3 = 2 - 4x

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

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

x + 4x = 2 + 3

Here are similar ones, we consider:

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

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

For example, here's the equation:

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

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

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

Expanding the brackets:

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

Expand the remaining brackets:

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

Here are some similar ones:

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

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

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

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

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

Special cases in solving linear equations.

First surprise.

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

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

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

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

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

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

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

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

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

Here's your answer: x - any number.

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

Second surprise.

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

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

After the same identical transformations, we get something intriguing:

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

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

Here's your answer: there are no solutions.

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

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

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

If you like this site...

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

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

A linear equation is an algebraic equation whose total degree of polynomials is equal to one. Solving linear equations is part of the school curriculum, and not the most difficult one. However, some still have difficulty completing this topic. We hope that after reading this material, all difficulties for you will remain in the past. So, let's figure it out. how to solve linear equations.

General form

The linear equation is represented as:

ax + b = 0, where a and b are any numbers.

Although a and b can be any number, their values affect the number of solutions to the equation. There are several special cases of solution:

If a=b=0, the equation has an infinite number of solutions;
If a=0, b≠0, the equation has no solution;
If a≠0, b=0, the equation has a solution: x = 0.

In the event that both numbers have non-zero values, the equation must be solved to derive final expression for a variable.

How to decide?

Solving a linear equation means finding what the variable is equal to. How to do this? Yes, it’s very simple - using simple algebraic operations and following the rules of transfer. If the equation appears in front of you in general form, you are in luck; all you need to do is:

Move b to right side equation, not forgetting to change the sign (translation rule!), thus, from an expression of the form ax + b = 0, an expression of the form should be obtained: ax = -b.
Apply the rule: to find one of the factors (x - in our case), you need to divide the product (-b in our case) by another factor (a - in our case). Thus, you should get an expression of the form: x = -b/a.

That's it - a solution has been found!

Now let's look at a specific example:

2x + 4 = 0 - move b, equal to 4 in this case, to the right side
2x = -4 - divide b by a (don’t forget about the minus sign)
x = -4/2 = -2

That's all! Our solution: x = -2.

As you can see, the solution to a linear equation with one variable is quite simple to find, but everything is so simple if we are lucky enough to come across the equation in its general form. In most cases, before solving the equation in the two steps described above, you also need to reduce the existing expression to general appearance. However, this is also not an extremely difficult task. Let's look at some special cases using examples.

Solving special cases

First, let's look at the cases that we described at the beginning of the article and explain what it means to have an infinite number of solutions and no solution.

If a=b=0, the equation will look like: 0x + 0 = 0. Performing the first step, we get: 0x = 0. What does this nonsense mean, you exclaim! After all, no matter what number you multiply by zero, you always get zero! Right! That's why they say that the equation has an infinite number of solutions - no matter what number you take, the equality will be true, 0x = 0 or 0 = 0.
If a=0, b≠0, the equation will look like: 0x + 3 = 0. Perform the first step, we get 0x = -3. Nonsense again! It is obvious that this equality will never be true! That's why they say that the equation has no solutions.
If a≠0, b=0, the equation will look like: 3x + 0 = 0. Performing the first step, we get: 3x = 0. What is the solution? It's easy, x = 0.

Lost in translation

The described special cases are not all that linear equations can surprise us with. Sometimes the equation is difficult to identify at first glance. Let's look at an example:

12x - 14 = 2x + 6

Is this a linear equation? What about the zero on the right side? Let's not rush to conclusions, let's act - let's transfer all the components of our equation into left side. We get:

12x - 2x - 14 - 6 = 0

Now subtract like from like, we get:

10x - 20 = 0

Learned? The most linear equation ever! The solution to which is: x = 20/10 = 2.

What if we have this example:

12((x + 2)/3) + x) = 12 (1 - 3x/4)

Yes, this is also a linear equation, only more transformations need to be carried out. First, let's open the brackets:

(12(x+2)/3) + 12x = 12 - 36x/4
4(x+2) + 12x = 12 - 36x/4
4x + 8 + 12x = 12 - 9x - now we carry out the transfer:
25x - 4 = 0 - it remains to find a solution using the already known scheme:
25x = 4,
x = 4/25 = 0.16

As you can see, everything can be solved, the main thing is not to worry, but to act. Remember, if your equation contains only variables of the first degree and numbers, you have a linear equation, which, no matter how it looks initially, can be reduced to a general form and solved. We hope everything works out for you! Good luck!

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:

Expand parentheses, if any;
Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
Give similar terms to the left and right of the equal sign;
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:

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.
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:

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

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

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

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

Scheme for solving simple linear equations

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

Expand the brackets, if any.
We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
We present similar terms.
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 simple design: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

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

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

Solving equations with fractions

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

Open the brackets.
Separate variables.
Bring similar ones.
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:

Get rid of fractions.
Open the brackets.
Separate variables.
Bring similar ones.
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!

Learning to solve equations is one of the main tasks that algebra poses for students. Starting with the simplest, when it consists of one unknown, and moving on to more and more complex ones. If you have not mastered the actions that need to be performed with equations from the first group, it will be difficult to understand the others.

To continue the conversation, you need to agree on notation.

General form of a linear equation with one unknown and the principle of its solution

Any equation that can be written like this:

a * x = b,

called linear. This is the general formula. But often in assignments linear equations are written in implicit form. Then it is necessary to perform identical transformations to obtain a generally accepted notation. These actions include:

opening parentheses;
moving all terms with a variable value to the left side of the equality, and the rest to the right;
reduction of similar terms.

In the case where an unknown quantity is in the denominator of a fraction, you need to determine its values at which the expression will not make sense. In other words, you need to know the domain of definition of the equation.

The principle by which all linear equations are solved comes down to dividing the value on the right side of the equation by the coefficient in front of the variable. That is, “x” will be equal to b/a.

Special cases of linear equations and their solutions

During reasoning, moments may arise when linear equations take on one of the special forms. Each of them has a specific solution.

In the first situation:

a * x = 0, and a ≠ 0.

The solution to such an equation will always be x = 0.

In the second case, “a” takes the value equal to zero:

0 * x = 0.

The answer to such an equation will be any number. That is, it has an infinite number of roots.

The third situation looks like this:

0 * x = in, where in ≠ 0.

This equation doesn't make sense. Because there are no roots that satisfy it.

General view of a linear equation with two variables

From its name it becomes clear that there are already two unknown quantities in it. Linear equations in two variables look like this:

a * x + b * y = c.

Since there are two unknowns in the record, the answer will look like a pair of numbers. That is, it is not enough to specify only one value. This will be an incomplete answer. A pair of quantities for which the equation becomes an identity is a solution to the equation. Moreover, in the answer, the variable that comes first in the alphabet is always written down first. Sometimes they say that these numbers satisfy him. Moreover, there can be an infinite number of such pairs.

How to solve a linear equation with two unknowns?

To do this, you just need to select any pair of numbers that turns out to be correct. For simplicity, you can take one of the unknowns equal to some prime number, and then find the second.

When solving, you often have to perform steps to simplify the equation. They are called identity transformations. Moreover, the following properties are always true for equations:

each term can be moved to the opposite part of the equality by replacing its sign with the opposite one;
The left and right sides of any equation are allowed to be divided by the same number, as long as it is not equal to zero.

Examples of tasks with linear equations

First task. Solve linear equations: 4x = 20, 8(x - 1) + 2x = 2(4 - 2x); (5x + 15) / (x + 4) = 4; (5x + 15) / (x + 3) = 4.

In the equation that comes first on this list, simply divide 20 by 4. The result will be 5. This is the answer: x = 5.

The third equation requires that an identity transformation be performed. It will consist of opening the brackets and bringing similar terms. After the first step, the equation will take the form: 8x - 8 + 2x = 8 - 4x. Then you need to move all the unknowns to the left side of the equation, and the rest to the right. The equation will look like this: 8x + 2x + 4x = 8 + 8. After adding similar terms: 14x = 16. Now it looks the same as the first one, and its solution is easy to find. The answer will be x=8/7. But in mathematics you are supposed to isolate the whole part from an improper fraction. Then the result will be transformed, and “x” will be equal to one whole and one seventh.

In the remaining examples, the variables are in the denominator. This means that you first need to find out at what values the equations are defined. To do this, you need to exclude numbers at which the denominators go to zero. In the first example it is “-4”, in the second it is “-3”. That is, these values need to be excluded from the answer. After this, you need to multiply both sides of the equality by the expressions in the denominator.

Opening the brackets and bringing similar terms, in the first of these equations we get: 5x + 15 = 4x + 16, and in the second 5x + 15 = 4x + 12. After transformations, the solution to the first equation will be x = -1. The second turns out to be equal to “-3”, which means that the latter has no solutions.

Second task. Solve the equation: -7x + 2y = 5.

Suppose that the first unknown x = 1, then the equation will take the form -7 * 1 + 2y = 5. Moving the factor “-7” to the right side of the equality and changing its sign to plus, it turns out that 2y = 12. This means y =6. Answer: one of the solutions to the equation x = 1, y = 6.

General form of inequality with one variable

All possible situations for inequalities are presented here:

a * x > b;
a * x< в;
a * x ≥b;
a * x ≤в.

In general, it looks like a simple linear equation, only the equal sign is replaced by an inequality.

Rules for identity transformations of inequalities

Just like linear equations, inequalities can be modified according to certain laws. They boil down to the following:

any alphabetic or numerical expression can be added to the left and right sides of the inequality, and the sign of the inequality remains the same;
you can also multiply or divide by the same positive number, this again does not change the sign;
when multiplying or dividing by the same thing a negative number equality will remain true provided the inequality sign is reversed.

General view of double inequalities

The following inequalities can be presented in problems:

V< а * х < с;
c ≤ a * x< с;
V< а * х ≤ с;
c ≤ a * x ≤ c.

It is called double because it is limited by inequality signs on both sides. It is solved using the same rules as ordinary inequalities. And finding the answer comes down to a series of identical transformations. Until the simplest is obtained.

Features of solving double inequalities

The first of them is its image on the coordinate axis. There is no need to use this method for simple inequalities. But in difficult cases it may simply be necessary.

To depict an inequality, you need to mark on the axis all the points that were obtained during the reasoning. These are invalid values, which are indicated by punctured dots, and values from inequalities obtained after transformations. Here, too, it is important to draw the dots correctly. If the inequality is strict, that is< или >, then these values are punched out. In non-strict inequalities, the points must be shaded.

Then it is necessary to indicate the meaning of the inequalities. This can be done using shading or arcs. Their intersection will indicate the answer.

The second feature is related to its recording. There are two options offered here. The first is ultimate inequality. The second is in the form of intervals. It happens with him that difficulties arise. The answer in spaces always looks like a variable with a membership sign and parentheses with numbers. Sometimes there are several spaces, then you need to write the “and” symbol between the brackets. These signs look like this: ∈ and ∩. Spacing brackets also play a role. The round one is placed when the point is excluded from the answer, and the rectangular one includes this value. The infinity sign is always in parentheses.

Examples of solving inequalities

1. Solve the inequality 7 - 5x ≥ 37.

After simple transformations, we get: -5x ≥ 30. Dividing by “-5” we can get the following expression: x ≤ -6. This is already the answer, but it can be written in another way: x ∈ (-∞; -6].

2. Solve double inequality -4< 2x + 6 ≤ 8.

First you need to subtract 6 everywhere. You get: -10< 2x ≤ 2. Теперь нужно разделить на 2. Неравенство примет вид: -5 < x ≤ 1. Изобразив ответ на числовой оси, сразу можно понять, что результатом будет промежуток от -5 до 1. Причем первая точка исключена, а вторая включена. То есть ответ у неравенства такой: х ∈ (-5; 1].