Using this math program you can solve the system of two linear equations with two variable method substitution and addition method.
The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.
Rules for entering equations
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.
When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2
You can use not only integers in equations, but also fractional numbers in the form of decimals and ordinary fractions.
Rules for entering decimal fractions.
Integer and fractional parts in decimals can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering numerical fraction The numerator is separated from the denominator by a division sign: /
Whole part separated from the fraction by an ampersand: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve system of equations
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A little theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$
It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$
Substituting the number 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by addition
Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients for one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. By adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$
From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38\) we get an equation with the variable y: \(11-3y=38\). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)
Thus, we found the solution to the system of equations by addition: \(x=11; y=-9\) or \((11;-9)\)
Taking advantage of the fact that in the equations of the system the coefficients for y are opposite numbers, we reduced its solution to the solution equivalent system(by summing both sides of each of the equations of the original symbol), in which one of the equations contains only one variable.
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Linear equations.
Linear equations are not the most complex topic 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 in 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. In small steps 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 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 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
Take note: to bring the original confusing equation to pleasant view, we used two (only two!) identity transformations– translation left-right with a change of sign and multiplication-division of an equation by the same number. This 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.
Let's say you got it the most elementary 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 indeed equal to zero. But X is missing! And we must write down in the answer, why equal to x. 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, which, when substituted into the original equation, will give us true 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.
Linear equations are a fairly harmless and clear topic school mathematics. But, oddly enough, the number of errors out of the blue when solving linear equations is only slightly less than in other topics - quadratic equations, logarithms, trigonometry and others. The causes of most errors are banal identical transformations of equations. First of all, this is confusion in signs when transferring terms from one part of the equation to another, as well as errors when working with fractions and fractional odds. Yes Yes! Fractions also appear in linear equations! All around. Below we will definitely analyze such evil equations.)
Well, let’s not pull the cat by the tail and let’s start figuring it out, shall we? Then we read and delve into it.)
What is a linear equation? Examples.
Typically the linear equation looks like this:
ax + b = 0,
Where a and b are any numbers. Any kind: integers, fractions, negative, irrational - there can be any!
For example:
7x + 1 = 0 (here a = 7, b = 1)
x – 3 = 0 (here a = 1, b = -3)
x/2 – 1.1 = 0 (here a = 1/2, b = -1.1)
In general, you understand, I hope.) Everything is simple, like in a fairy tale. For the time being... And if you look closely at general record ax+b=0 take a closer look and think a little? After all, a and b are any numbers! And if we have, say, a = 0 and b = 0 (any numbers can be taken!), then what will we get?
0 = 0
But that's not all the fun! What if, say, a = 0, b = -10? Then it turns out to be some kind of nonsense:
0 = 10.
Which is very, very annoying and undermines the trust in mathematics that we have gained through sweat and blood... Especially during tests and exams. But out of these incomprehensible and strange equalities, you also need to find X! Which doesn’t exist at all! And here, even well-prepared students can sometimes fall into what is called a stupor... But don’t worry! In this lesson we will also look at all such surprises. And we will also definitely find an X from such equalities.) Moreover, this same X can be found very, very simply. Yes Yes! Surprising but true.)
Okay, that's understandable. But how can you tell by the appearance of the task that it is a linear equation and not some other equation? Unfortunately, it is not always possible to recognize the type of equation just by appearance. The point is that not only equations of the form ax + b = 0 are called linear, but also any other equations that, in one way or another, can be reduced to this form by identical transformations. How do you know if it adds up or not? Until you can hardly solve the example - almost not at all. This is upsetting. But for some types of equations, you can immediately tell with confidence whether it is linear or not with one quick glance.
To do this, let us turn once again to general structure any linear equation:
ax + b = 0
Please note: in the linear equation Always only variable x is present in the first degree and some numbers! That's all! Nothing else. At the same time, there are no X’s in the square, in the cube, under the root, under the logarithm and other exotic things. And (most importantly!) there are no fractions with X in the denominators! But fractions with numbers in the denominators or division per number- easily!
For example:
This is a linear equation. The equation contains only X's to the first power and numbers. And there are no more X's high degrees- squared, cubed, and so on. Yes, there are fractions here, but at the same time the denominators of the fractions contain only numbers. Namely, two and three. In other words, there is no division by x.
And here is the equation
It can no longer be called linear, although here, too, there are only numbers and X’s to the first power. Because, among other things, there are also fractions with X's in the denominators. And after simplifications and transformations, such an equation can become anything: linear, quadratic - anything.
How to solve linear equations? Examples.
So how do you solve linear equations? Read on and be surprised.) The entire solution of linear equations is based on just two main things. Let's list them.
1) A set of elementary actions and rules of mathematics.
These are using parentheses, opening parentheses, working with fractions, working with negative numbers, multiplication tables, and so on. This knowledge and skills are necessary not only for solving linear equations, but for all mathematics in general. And if you have problems with this, remember junior classes. Otherwise you will have a hard time...
2)
There are only two of them. Yes Yes! Moreover, these very basic identity transformations underlie the solution of not only linear, but generally any mathematical equations! In a word, the solution to any other equation - quadratic, logarithmic, trigonometric, irrational, etc. – as a rule, it begins with these very basic transformations. But the solution of linear equations, in fact, ends with them (transformations). Ready answer.) So don’t be lazy and take a look at the link.) Moreover, linear equations are also analyzed in detail there.
Well, I think it's time to start looking at examples.
To begin with, as a warm-up, let's look at some basic stuff. Without any fractions or other bells and whistles. For example, this equation:
x – 2 = 4 – 5x
This is a classic linear equation. All X's are at most in the first power and there is no division by X anywhere. The solution scheme in such equations is always the same and terribly simple: all terms with X's must be collected on the left, and all terms without X's (i.e. numbers) must be collected on the right. So let's start collecting.
To do this, we launch the first identity transformation. We need to move -5x to the left, and move -2 to the right. With a change of sign, of course.) So we transfer:
x + 5x = 4 + 2
Here you go. Half the battle is done: the X's have been collected into a pile, and so have the numbers. Now we present similar ones on the left, and we count them on the right. We get:
6x = 6
What do we now lack for complete happiness? Yes, so that the pure X remains on the left! And the six gets in the way. How to get rid of it? Now we run the second identity transformation - divide both sides of the equation by 6. And - voila! The answer is ready.)
x = 1
Of course, the example is completely primitive. To get the general idea. Well, let's decide something more significant. For example, let's look at this equation:
Let's look at it in detail.) This is also a linear equation, although it would seem that there are fractions here. But in fractions there is division by two and there is division by three, but there is no division by an expression with an X! So let's decide. Using the same identical transformations, yes.)
What should we do first? With X's - to the left, without X's - to the right? In principle, this is possible. Fly to Sochi via Vladivostok.) Or you can take the shortest route, immediately using a universal and powerful method. If you know the identity transformations, of course.)
First, I ask a key question: what stands out to you most and dislikes most about this equation? 99 out of 100 people will say: fractions! And they will be right.) So let’s get rid of them first. Safe for the equation itself.) Therefore, let's start right away with second identity transformation- from multiplication. What should we multiply the left side by so that the denominator is successfully reduced? That's right, a two. What about the right side? For three! But... Mathematics is a capricious lady. She, you see, requires multiplying both sides only for the same number! Multiplying each part by its own number doesn’t work... What are we going to do? Something... Look for a compromise. In order to satisfy our desires (to get rid of fractions) and not to offend mathematics.) Let’s multiply both parts by six!) That is, by the common denominator of all fractions included in the equation. Then in one fell swoop both the two and the three will be reduced!)
So let's multiply. The entire left side and the entire right side! Therefore, we use parentheses. This is what the procedure itself looks like:
Now we open these same brackets:
Now, representing 6 as 6/1, let's multiply six by each of the fractions on the left and right. This ordinary multiplication fractions, but so be it, I’ll write it down in detail:
And here - attention! I put the numerator (x-3) in brackets! This is all because when multiplying fractions, the numerator is multiplied entirely, entirely! And the x-3 expression must be worked as one integral structure. But if you write the numerator like this:
6x – 3,
But we have everything right and we need to finalize it. What to do next? Open the parentheses in the numerator on the left? In no case! You and I multiplied both sides by 6 to get rid of fractions, and not to worry about opening parentheses. On at this stage we need reduce our fractions. With a feeling of deep satisfaction, we reduce all the denominators and get an equation without any fractions, in a ruler:
3(x-3) + 6x = 30 – 4x
And now the remaining brackets can be opened:
3x – 9 + 6x = 30 – 4x
The equation keeps getting better and better! Now let’s remember again about the first identical transformation. WITH stone-faced We repeat the spell from elementary school: with X - to the left, without X - to the right. And apply this transformation:
3x + 6x + 4x = 30 + 9
We present similar ones on the left and count on the right:
13x = 39
It remains to divide both parts by 13. That is, apply the second transformation again. We divide and get the answer:
x = 3
The job is done. As you can see, in given equation we had to apply the first transformation once (transfer of terms) and the second twice: at the beginning of the solution we used multiplication (by 6) in order to get rid of fractions, and at the end of the solution we used division (by 13) to get rid of the coefficient in front of the X. And the solution to any (yes, any!) linear equation consists of a combination of these same transformations in one sequence or another. Where exactly to start depends on the specific equation. In some places it is more profitable to start with transfer, and in others (as in this example) with multiplication (or division).
We work from simple to complex. Let's now consider outright cruelty. With a bunch of fractions and parentheses. And I’ll tell you how not to overstrain yourself.)
For example, here's the equation:
We look at the equation for a minute, are horrified, but still pull ourselves together! The main problem is where to start? You can add fractions on the right side. You can subtract fractions in parentheses. You can multiply both parts by something. Or divide... So what is still possible? Answer: everything is possible! Mathematics does not prohibit any of the listed actions. And no matter what sequence of actions and transformations you choose, the answer will always be the same - the correct one. Unless, of course, at some step you violate the identity of your transformations and, thereby, make mistakes...
And, in order not to make mistakes, in such sophisticated examples as this one, it is always most useful to evaluate its appearance and figure out in your mind: what can be done in the example so that maximum simplify it in one step?
So let's figure it out. On the left are sixes in the denominators. Personally, I don't like them, and they are very easy to remove. Let me multiply both sides of the equation by 6! Then the sixes on the left will be successfully reduced, the fractions in brackets will not go anywhere yet. Well, that's okay. We'll deal with them a little later.) But on the right, we have the denominators 2 and 3 cancelling. It is with this action (multiplying by 6) that we achieve maximum simplifications in one step!
After multiplication, our whole evil equation becomes like this:
If you don’t understand exactly how this equation came about, then you haven’t understood the analysis of the previous example well. And I tried, by the way...
So, let's reveal:
Now the most logical step would be to isolate the fractions on the left, and send 5x to the right side. At the same time, we will present similar ones on the right side. We get:
Much better already. Now the left side has prepared itself for multiplication. What should we multiply the left side by so that both the five and the four are reduced at once? On 20! But we also have disadvantages on both sides of the equation. Therefore, it will be most convenient to multiply both sides of the equation not by 20, but by -20. Then in one fell swoop both the minuses and the fractions will disappear.
So we multiply:
Anyone who still doesn’t understand this step means that the problem is not in the equations. The problems are in the basics! Let's remember again Golden Rule opening brackets:
If a number is multiplied by some expression in brackets, then this number must be sequentially multiplied by each term of this very expression. Moreover, if the number is positive, then the signs of the expressions are preserved after expansion. If negative, change to the opposite:
a(b+c) = ab+ac
-a(b+c) = -ab-ac
Our cons disappeared after multiplying both sides by -20. And now we multiply the brackets with fractions on the left by quite positive number 20. Therefore, when these brackets are opened, all the signs that were inside them are preserved. But where the brackets in the numerators of fractions come from, I already explained in detail in the previous example.
Now you can reduce fractions:
4(3-5x)-5(3x-2) = 20
Open the remaining brackets. Again, we reveal it correctly. The first brackets are multiplied by the positive number 4 and, therefore, all signs are preserved when they are opened. But the second brackets are multiplied by negative the number is -5 and, therefore, all signs are reversed:
12 - 20x - 15x + 10 = 20
There are mere trifles left. With X's to the left, without X's to the right:
-20x – 15x = 20 – 10 – 12
-35x = -2
That's almost all. On the left you need a pure X, but the number -35 is in the way. So we divide both sides by (-35). Let me remind you that the second identity transformation allows us to multiply and divide both sides by whatever number. Including negative ones.) As long as it’s not zero! Feel free to divide and get the answer:
X = 2/35
This time the X turned out to be fractional. It's OK. Such an example.)
As we can see, the principle of solving linear equations (even the most complicated ones) is quite simple: we take the original equation and, using identical transformations, successively simplify it until we get the answer. With the basics, of course! The main problems here are precisely the failure to follow the basics (for example, there is a minus in front of the brackets, and they forgot to change the signs when expanding), as well as in banal arithmetic. So don't neglect the basics! They are the foundation of all other mathematics!
Some fun things to do when solving linear equations. Or special occasions.
Everything would be fine. However... Among the linear equations there are also such funny pearls that in the process of solving them can drive you into a strong stupor. Even an excellent student.)
For example, here’s an innocuous-looking equation:
7x + 3 = 4x + 5 + 3x - 2
Yawning widely and slightly bored, we collect all the X’s on the left and all the numbers on the right:
7x-4x-3x = 5-2-3
We present similar ones, count and get:
0 = 0
That's it! I gave a sample trick! This equality in itself does not raise objections: zero is really equal to zero. But X is missing! Without a trace! And we must write down in the answer, what is x equal to. Otherwise, the decision does not count, yes.) What to do?
Don't panic! In such non-standard cases, the most general concepts and principles of mathematics. What is an equation? How to solve equations? What does it mean to solve an equation?
Solving an equation means finding All values of the variable x, which, when substituted into original equation will give us the correct equality (identity)!
But we have true equality it's already happened! 0=0, or rather, nowhere!) We can only guess at which X's we get this equality. What kind of X's can be substituted in original equation, if upon substitution all of them will they still be reduced to zero? Haven't you figured it out yet?
Surely! X's can be substituted any!!! Absolutely any. Submit whatever you want. At least 1, at least -23, at least 2.7 - whatever! They will still be reduced and as a result, the pure truth will remain. Try it, substitute it and see for yourself.)
Here's your answer:
x – any number.
In scientific notation this equality is written as follows:
This entry reads like this: “X is any real number.”
Or in another form, at intervals:
Design it the way you like best. This is a correct and completely complete answer!
Now I'm going to change just one number in our original equation. Now let’s solve this equation:
7x + 2 = 4x + 5 + 3x – 2
Again we transfer the terms, count and get:
7x – 4x – 3x = 5 – 2 – 2
0 = 1
And what do you think of this joke? There was an ordinary linear equation, but it became an incomprehensible equality
0 = 1…
Speaking scientific language, we got false equality. But in Russian this is not true. Bullshit. Nonsense.) Because zero is in no way equal to one!
And now let’s figure out again what kind of X’s, when substituted into the original equation, will give us true equality? Which? But none! No matter what X you substitute, everything will still be shortened and everything will remain crap.)
Here is the answer: no solutions.
In mathematical notation, this answer is written like this:
It reads: “X belongs to the empty set.”
Such answers in mathematics also occur quite often: not always do any equations have roots in principle. Some equations may not have roots at all. At all.
Here are two surprises. I hope that now the sudden disappearance of X's from the equation will not leave you perplexed forever. This is quite familiar.)
And then I hear a logical question: will they be in the OGE or the Unified State Exam? On the Unified State Examination in itself as a task - no. Too simple. But in the OGE or in word problems - easily! So now let’s train and decide:
Answers (in disarray): -2; -1; any number; 2; no solutions; 7/13.
Everything worked out? Great! You have a good chance in the exam.
Does something not add up? Hm... Sadness, of course. This means there are still gaps somewhere. Either in the basics or identity transformations. Or it’s just a matter of simple inattention. Read the lesson again. Because this is not a topic that can be so easily dispensed with in mathematics...
Good luck! She will definitely smile at you, believe me!)
Linear equation is algebraic equation, full degree whose polynomials are equal to one. Solving linear equations - part school curriculum, and not the most difficult. However, some still have difficulty completing this topic. We hope after reading this material, all difficulties for you will be a thing of 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 infinite set decisions;
- 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 the final expression for the variable.
How to decide?
Solving a linear equation means finding what the variable is equal to. How to do this? Yes, very simple - using simple algebraic operations and following the transfer rules. 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 in this case 4, to the right
- 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 based on 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!
Equations. To put it another way, the solution of all equations begins with these transformations. When solving linear equations, it (the solution) is based on identity transformations and ends with the final answer.
The case of a non-zero coefficient for an unknown variable.
ax+b=0, a ≠ 0
We move terms with X to one side, and numbers to the other side. Be sure to remember that when transferring the terms to the opposite side equations, you need to change the sign:
ax:(a)=-b:(a)
Let's shorten A at X and we get:
x=-b:(a)
This is the answer. If you need to check if a number is -b:(a) root of our equation, then we need to substitute in initial equation instead of X this is the number:
a(-b:(a))+b=0 ( those. 0=0)
Because this equality is correct, then -b:(a) and truth is the root of the equation.
Answer: x=-b:(a), a ≠ 0.
First example:
5x+2=7x-6
We move members with to one side X, and on the other side the numbers:
5x-7x=-6-2
-2x:(-2)=-8:(-2)
For an unknown factor, we reduced the coefficient and got the answer:
This is the answer. If you need to check whether the number 4 is really the root of our equation, we substitute this number instead of X in the original equation:
5*4+2=7*4-6 ( those. 22=22)
Because this equality is true, then 4 is the root of the equation.
Second example:
Solve the equation:
5x+14=x-49
By transferring the unknowns and numbers into different sides, got:
Divide the parts of the equation by the coefficient at x(by 4) and we get:
Third example:
Solve the equation:
First, we get rid of the irrationality in the coefficient for the unknown by multiplying all terms by:
This form is considered to be simplified, because the number has the root of the number in the denominator. We need to simplify the answer by multiplying the numerator and denominator by same number, we have this:
The case of no solutions.
Solve the equation:
2x+3=2x+7
In front of everyone x our equation will not become a true equality. That is, our equation has no roots.
Answer: there are no solutions.
A special case is an infinite number of solutions.
Solve the equation:
2x+3=2x+3
Moving the X's and numbers in different directions and bringing similar terms, we get the equation:
Here, too, it is not possible to divide both parts by 0, because it is forbidden. However, putting in place X any number, we get the correct equality. That is, every number is a solution to such an equation. Thus, here infinite number decisions.
Answer: an infinite number of solutions.
The case of equality of two complete forms.
ax+b=cx+d
ax-cx=d-b
(a-c)x=d-b
x=(d-b):(a-c)
Answer: x=(d-b):(a-c), If d≠b and a≠c, otherwise there are infinitely many solutions, but if a=c, A d≠b, then there are no solutions.