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. Note: we're talking about only about individual terms. Let's write it down:
We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:
\[\frac(6x)(6)=-\frac(72)(6)\]
So we got the answer.
Task No. 2
We can see the parentheses in this problem, so let's expand them:
Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:
Here are some similar ones:
At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.
Task No. 3
The third linear equation is more interesting:
\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]
There are several 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:
It is obvious that given equation There are 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 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:
- 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 see quadratic functions, most likely, in the process of further transformations they will decrease.
- There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.
I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!
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- 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 want.
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. on 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 a spell from elementary school: 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 with identity transformations before receiving a response. 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 right decision equations.)
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.
Makarova T.P., GBOU secondary school No. 618 Training “Equations” 5th grade
Training for 5th grade on the topic “Equations” in 2 versions
Makarova Tatyana Pavlovna,
Teacher, Secondary School No. 618, Moscow
Contingent: 5th grade
The training is aimed at testing students’ knowledge and skills on the topic “Equations”. The training is intended for 5th grade students for the textbook by N.Ya. Vilenkin, V.I. Zhokhova and others. Textbook for 5th grade. – M.: Mnemosyne, 2013. – 288 p. The test contains two parallel options of equal difficulty, nine tasks each (4 multiple-choice tasks, 3 short-answer tasks, 2 extended-solution tasks).
This training fully complies with the federal state educational standard(second generation), can be used when conducting classroom control, and can also be used by 5th grade students for independent work on this topic.
15 to 25 minutes of class time are allocated to complete the test. Keys included.
Training for 5th grade on the topic “Equations”. Option 1.
№p/p
Exercise
Answer
Solve the equation
574
1124
1114
1024
Find the root of the equation
(156-x )+43=170.
1) The root of an equation is the value of a letter.
2) The root of equation (23 – X) – 21 = 2 is not a natural number.
3) To find the unknown subtrahend, you need to subtract the difference from the minuend.
4) Equation x – x= 0 has exactly one root.
Petya thought of a number. If you add 43 to this number, and add 77 to the resulting amount, you get 258. What number did Petya have in mind?
1) (X + 43) – 77 = 258
2) (X + 43) + 77 = 258
3) (X – 43) + 77 = 258
4) (X – 43) – 77 = 258
Solve the equation: (5· With – 8) : 2 = 121: 11.
Solve the equation: 821 – ( m + 268) = 349.
Find the value of the number A, if 8 A + 9X= 60 and X=4.
Solve the problem using the equation. The library had 125 books on mathematics. After the students took several books and then returned 3 books, there were 116 books. How many books did the students take in total?
Solve the equation:
456 + (X – 367) – 225 =898
Training for 5th grade on the topic “Equations”. Option 2.
№p/p
Exercise
Answer
Part 1. Multiple choice task
Solve the equation 
525
1081
535
1071
Find the root of the equation
942 – (y + 142) = 419.
391
481
1219
381
Please provide numbers true statements:
1) An equation is an equality containing a letter whose value must be found.
2) Any natural number is the root of the equation
3) The root of an equation is the value of the letter at which the correct numerical expression is obtained from the equation.
4) To find the unknown dividend, you need to add a divisor to the quotient.
Dasha thought of a number. If you add 43 to this number and subtract 77 from the resulting amount, you get 258. What number did Dasha have in mind?
1) (X + 43) – 77 = 258
2) (X + 43) + 77 = 258
3) (X – 43) + 77 = 258
4) (X – 43) – 77 = 258
Part 2. Short answer task
Solve the equation: 63: (2· X – 1) = 21: 3.
Solve the equation: 748 – ( b +248) = 300.
Find the value of the number A, if 7 A – 3X= 41 and X=5.
Part 3. Tasks with detailed solutions
Solve the problem using the equation. There were 197 machines in the warehouse. After some were sold and 86 more were brought in, there were still 115 machines left in the warehouse. How many machines were sold in total?
One of the most important skills when admission to 5th grade is the ability to solve simple equations. Since 5th grade is not yet so far from primary school, then there are not so many types of equations that a student can solve. We will introduce you to all the basic types of equations that you need to be able to solve if you want enter a physics and mathematics school.
Type 1: "bulbous"
These are equations that you are almost likely to encounter when admission to any school or a 5th grade club as a separate task. They are easy to distinguish from others: in them the variable is present only once. For example, or.
They are solved very simply: you just need to “get” to the unknown, gradually “removing” everything unnecessary that surrounds it - as if peeling an onion - hence the name. To solve it, just remember a few rules from the second class. Let's list them all:
Addition
- term1 + term2 = sum
- term1 = sum - term2
- term2 = sum - term1
Subtraction
- minuend - subtrahend = difference
- minuend = subtrahend + difference
- subtrahend = minuend - difference
Multiplication
- factor1 * factor2 = product
- factor1 = product: factor2
- factor2 = product: factor1
Division
- dividend: divisor = quotient
- dividend = divisor * quotient
- divisor = dividend: quotient
Let's look at an example of how to apply these rules.
Note that we are dividing 
on and we receive . In this situation, we know the divisor and the quotient. To find the dividend, you need to multiply the divisor by the quotient:
We have become a little closer to ourselves. Now we see that 
is added and it turns out . This means that to find one of the terms, you need to subtract the known term from the sum:
And another “layer” has been removed from the unknown! Now we see a situation with a known value of the product () and one known multiplier ().
Now the situation is “minuend - subtrahend = difference”
And the last step - famous work() and one of the multipliers () 
Type 2: equations with brackets
Equations of this type are most often found in problems - 90% of all problems for admission to 5th grade. Unlike "onion equations" the variable here can appear several times, so it is impossible to solve it using the methods from the previous paragraph. Typical equations: or
The main difficulty is opening the brackets correctly. After you have managed to do this correctly, you should reduce similar terms (numbers to numbers, variables to variables), and after that we get the simplest "onion equation" which we can solve. But first things first.
Expanding parentheses. We will give several rules that should be used in this case. But, as practice shows, the student begins to open the brackets correctly only after 70-80 completed problems. The basic rule is this: any factor outside the brackets must be multiplied by each term inside the brackets. And the minus sign in front of the bracket changes the sign of all the expressions inside. So, the basic rules of disclosure: 
Bringing similar. Here everything is much easier: you need, by transferring the terms through the equal sign, to ensure that on one side there are only terms with the unknown, and on the other - only numbers. The basic rule is this: each term transferred through changes its sign - if it was with, it will become with, and vice versa. After a successful transfer, it is necessary to count the total number of unknowns, the total number on the other side of the equality than the variables, and solve a simple "onion equation".
