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What are "linear equations"
or orally - three friends were given apples each on the basis that Vasya had all the apples he had.
And now you have already decided linear equation
Now let's give this term a mathematical definition.
Linear equation - is an algebraic equation whose total degree of its constituent polynomials is equal to. It looks like this:
Where and are any numbers and
For our case with Vasya and apples, we will write:
- “if Vasya gives the same number of apples to all three friends, he will have no apples left”
"Hidden" linear equations, or the importance of identity transformations
Despite the fact that at first glance everything is extremely simple, when solving equations you need to be careful, because linear equations are called not only equations of this type, but also any equations that can be reduced to this type by transformations and simplifications. For example:
We see what is on the right, which, in theory, already indicates that the equation is not linear. Moreover, if we open the brackets, we will get two more terms in which it will be, but don't rush to conclusions! Before judging whether an equation is linear, it is necessary to make all the transformations and thus simplify the original example. In this case, transformations can change appearance, but not the very essence of the equation.
In other words, the transformation data must be identical or equivalent. There are only two such transformations, but they play very, VERY important role when solving problems. Let's look at both transformations using specific examples.
Transfer left - right.
Let's say we need to solve the following equation:
Also in primary school we were told: “with X’s - to the left, without X’s - to the right.” What expression with an X is on the right? That's right, but not how not. And this is important, because if this is misunderstood, it would seem simple question, the wrong answer comes out. What expression with an X is on the left? Right, .
Now that we have figured this out, we move all the terms with unknowns to the left side, and everything that is known to the right, remembering that if there is no sign in front of the number, for example, then the number is positive, that is, there is a sign in front of it “ "
Transferred? What did you get?
All that remains to be done is to bring similar terms. We present:
So, we have successfully analyzed the first identical transformation, although I am sure that you knew it and actively used it without me. The main thing is not to forget about the signs of numbers and change them to the opposite ones when transferring through the equal sign!
Multiplication-division.
Let's start right away with an example
Let’s look and think: what don’t we like about this example? The unknown is all in one part, the known is in another, but something is stopping us... And this something is a four, because if it didn’t exist, everything would be perfect - x is equal to a number - exactly as we need !
How can you get rid of it? We can’t move it to the right, because then we need to move the entire multiplier (we can’t take it and tear it away), and moving the entire multiplier also doesn’t make sense...
It's time to remember about division, so let's divide everything by! Everything - this means both the left and the right side. This way and only this way! What are we doing?
Here is the answer.
Let's now look at another example:
Can you guess what needs to be done in this case? That's right, multiply the left and right sides by! What answer did you receive? Right. .
Surely you already knew everything about identity transformations. Consider that we have simply refreshed this knowledge in your memory and it is time for something more - For example, to solve our big example:
As we said earlier, looking at it, you cannot say that this equation is linear, but we need to open the brackets and carry out identical transformations. So let's get started!
To begin with, we recall the formulas for abbreviated multiplication, in particular, the square of the sum and the square of the difference. If you don’t remember what it is and how the parentheses are opened, I strongly recommend reading the topic, as these skills will be useful to you when solving almost all the examples encountered in the exam.
Revealed? Let's compare:
Now it's time to bring similar terms. Do you remember how in those same elementary grades they told us “don’t put flies and cutlets together”? Here I remind you of this. We add everything separately - the factors that have, the factors that have, and the remaining factors that do not have unknowns. When you bring similar terms, move all unknowns to the left, and all that is known to the right. What did you get?
As you can see, the X's in the square have disappeared and we see something completely normal. linear equation. All that remains is to find it!
And finally, I’ll say one more very important thing about identity transformations - identity transformations are applicable not only for linear equations, but also for quadratic, fractional rational and others. You just need to remember that when we transfer factors through the equal sign, we change the sign to the opposite one, and when dividing or multiplying by some number, we multiply/divide both sides of the equation by the SAME number.
What else did you take away from this example? That by looking at an equation it is not always possible to directly and accurately determine whether it is linear or not. It is necessary to first completely simplify the expression, and only then judge what it is.
Linear equations. Examples.
Here are a couple more examples for you to practice on your own - determine whether the equation is linear and if so, find its roots:
Answers:
1. Is.
2. Is not.
Let's open the brackets and present similar terms:
Let's perform an identical transformation - divide the left and right sides into:
We see that the equation is not linear, so there is no need to look for its roots.
3. Is.
Let's perform an identical transformation - multiply the left and right sides by to get rid of the denominator.
Think about why it is so important that? If you know the answer to this question, move on to further solving the equation; if not, be sure to look into the topic so as not to make mistakes in more complex examples. By the way, as you can see, the situation is impossible. Why?
So, let's go ahead and rearrange the equation:
If you managed everything without difficulty, let's talk about linear equations with two variables.
Linear equations in two variables
Now let's move on to a little more complex - linear equations with two variables.
Linear equations with two variables have the form:
Where, and - any numbers and.
As you can see, the only difference is that another variable is added to the equation. And so everything is the same - there are no x squared, no division by a variable, etc. and so on.
What kind of life example can I give you... Let’s take the same Vasya. Let's say he decided that he would give each of 3 friends the same number of apples, and keep the apples for himself. How many apples does Vasya need to buy if he gives each friend an apple? What about? What if by?
The relationship between the number of apples that each person will receive and the total number of apples that needs to be purchased will be expressed by the equation:
- - the number of apples that a person will receive (, or, or);
- - the number of apples that Vasya will take for himself;
- - how many apples does Vasya need to buy, taking into account the number of apples per person?
Solving this problem, we get that if Vasya gives one friend an apple, then he needs to buy pieces, if he gives apples, etc.
And generally speaking. We have two variables. Why not plot this relationship on a graph? We build and mark the value of ours, that is, points, with coordinates, and!
As you can see, they depend on each other linear, hence the name of the equations - “ linear».
Let's abstract from apples and look at various equations graphically. Look carefully at the two graphs constructed - a straight line and a parabola, specified by arbitrary functions:
Find and mark the corresponding points in both pictures.
What did you get?
You see that on the graph of the first function alone corresponds one, that is, they also linearly depend on each other, which cannot be said about the second function. Of course, you can argue that in the second graph the x - also corresponds, but this is only one point, that is special case, since you can still find one that matches more than just one. And the constructed graph does not resemble a line in any way, but is a parabola.
I repeat, one more time: the graph of a linear equation must be a STRAIGHT line.
With the fact that the equation will not be linear if we go to any degree - this is clear using the example of a parabola, although you can build a few more simple graphs for yourself, for example or. But I assure you - none of them will be a STRAIGHT LINE.
Do not believe? Build it and then compare it with what I got:
What happens if we divide something by, for example, some number? Will there be a linear relationship and? Let's not argue, but let's build! For example, let's build a graph of a function.
Somehow it doesn’t look like it’s constructed as a straight line... accordingly, the equation is not linear.
Let's summarize:
- Linear equation - is an algebraic equation in which the total degree of its constituent polynomials is equal.
- Linear equation with one variable has the form:
, where and are any numbers;
Linear equation with two variables:
, where, and are any numbers. - It is not always possible to immediately determine whether an equation is linear or not. Sometimes, in order to understand this, it is necessary to perform identical transformations: move left/right similar members, not forgetting to change the sign, or multiply/divide both sides of the equation by the same number.
LINEAR EQUATIONS. BRIEFLY ABOUT THE MAIN THINGS
1. Linear equation
This is an algebraic equation in which the total degree of its constituent polynomials is equal.
2. Linear equation with one variable has the form:
Where and are any numbers;
3. Linear equation with two variables has the form:
Where, and - any numbers.
4. Identity transformations
To determine whether an equation is linear or not, it is necessary to perform identical transformations:
- move similar terms left/right, not forgetting to change the sign;
- multiply/divide both sides of the equation by the same number.
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successful passing the Unified State Exam, for admission to college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
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Find problems and solve them!
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 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, it is necessary 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 thing 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].
When solving linear equations, we strive to find the root, that is, the value for the variable that will turn the equation into a correct equality.
To find the root of the equation you need equivalent transformations bring the equation given to us to the form
\(x=[number]\)
This number will be the root.
That is, we transform the equation, making it simpler with each step, until we reduce it to a completely primitive equation “x = number”, where the root is obvious. The most frequently used transformations when solving linear equations are the following:
For example: add \(5\) to both sides of the equation \(6x-5=1\)
\(6x-5=1\) \(|+5\)
\(6x-5+5=1+5\)
\(6x=6\)
Please note that we could get the same result faster by simply writing the five on the other side of the equation and changing its sign. Actually, this is exactly how the school “transfer through equals with a change of sign to the opposite” is done.
2. Multiplying or dividing both sides of an equation by the same number or expression.
For example: divide the equation \(-2x=8\) by minus two
\(-2x=8\) \(|:(-2)\)
\(x=-4\)
Typically this step is performed at the very end, when the equation has already been reduced to the form \(ax=b\), and we divide by \(a\) to remove it from the left.
3. Using the properties and laws of mathematics: opening parentheses, bringing similar terms, reducing fractions, etc.
Add \(2x\) left and right
Subtract \(24\) from both sides of the equation
We present similar terms again
Now we divide the equation by \(-3\), thereby removing the front X on the left side.
Answer : \(7\)
The answer has been found. However, let's check it out. If seven is really a root, then substituting it instead of X into the original equation should result in the correct equality - same numbers left and right. Let's try.
Examination:
\(6(4-7)+7=3-2\cdot7\)
\(6\cdot(-3)+7=3-14\)
\(-18+7=-11\)
\(-11=-11\)
It worked out. This means that seven is indeed the root of the original linear equation.
Don’t be lazy to check the answers you found by substitution, especially if you are solving an equation on a test or exam.
The question remains - how to determine what to do with the equation at the next step? How exactly to convert it? Divide by something? Or subtract? And what exactly should I subtract? Divide by what?
The answer is simple:
Your goal is to bring the equation to the form \(x=[number]\), that is, on the left is x without coefficients and numbers, and on the right is only a number without variables. Therefore, look at what is stopping you and do the opposite of what the interfering component does.
To better understand this, let's look at the solution of the linear equation \(x+3=13-4x\) step by step.
Let's think: how does this equation differ from \(x=[number]\)? What's stopping us? What's wrong?
Well, firstly, the three interferes, since on the left there should be only a lone X, without numbers. What does the troika “do”? Added to X. So, to remove it - subtract the same three. But if we subtract three from the left, we must subtract it from the right so that the equality is not violated.
\(x+3=13-4x\) \(|-3\)
\(x+3-3=13-4x-3\)
\(x=10-4x\)
Fine. Now what's stopping you? \(4x\) on the right, because there should only be numbers there. \(4x\) deducted- we remove by adding.
\(x=10-4x\) \(|+4x\)
\(x+4x=10-4x+4x\)
Now we present similar terms on the left and right.
It's almost ready. All that remains is to remove the five on the left. What is she doing"? Multiplies on x. So let's remove it division.
\(5x=10\) \(|:5\)
\(\frac(5x)(5)\) \(=\)\(\frac(10)(5)\)
\(x=2\)
The solution is complete, the root of the equation is two. You can check by substitution.
notice, that most often there is only one root in linear equations. However, two special cases may occur.
Special case 1 – there are no roots in a linear equation.
Example . Solve the equation \(3x-1=2(x+3)+x\)
Solution :
Answer : no roots.
In fact, the fact that we will come to such a result was visible earlier, even when we received \(3x-1=3x+6\). Think about it: how can \(3x\) from which we subtracted \(1\), and \(3x\) to which we added \(6\) be equal? Obviously, no way, because they did the same thing different actions! It is clear that the results will vary.
Special case 2 – a linear equation has an infinite number of roots.
Example . Solve linear equation \(8(x+2)-4=12x-4(x-3)\)Solution :
Answer : any number.
This, by the way, was noticeable even earlier, at the stage: \(8x+12=8x+12\). Indeed, left and right are the same expressions. Whatever X you substitute, it will be the same number both there and there.
More complex linear equations.
The original equation does not always immediately look like a linear one; sometimes it is “masked” as other, more complex equations. However, in the process of transformation, the disguise disappears.
Example . Find the root of the equation \(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\)
Solution :
|
\(2x^(2)-(x-4)^(2)=(3+x)^(2)-15\) |
It would seem that there is an x squared here - this is not a linear equation! But don't rush. Let's apply |
|
|
\(2x^(2)-(x^(2)-8x+16)=9+6x+x^(2)-15\) |
Why is the expansion result \((x-4)^(2)\) in parentheses, but the result \((3+x)^(2)\) is not? Because there is a minus in front of the first square, which will change all the signs. And in order not to forget about this, we take the result in brackets, which we now open. |
|
|
\(2x^(2)-x^(2)+8x-16=9+6x+x^(2)-15\) |
We present similar terms |
|
|
\(x^(2)+8x-16=x^(2)+6x-6\) |
||
|
\(x^(2)-x^(2)+8x-6x=-6+16\) |
We present similar ones again. |
|
|
Like this. It turns out that the original equation is quite linear, and the X squared is nothing more than a screen to confuse us. :) We complete the solution by dividing the equation by \(2\), and we get the answer. |
Answer : \(x=5\)
Example . Solve linear equation \(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6 )\)
Solution :
|
\(\frac(x+2)(2)\) \(-\) \(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\) |
The equation doesn’t look linear, it’s some kind of fractions... However, let’s get rid of the denominators by multiplying both sides of the equation by the common denominator of all – six |
|
|
\(6\cdot\)\((\frac(x+2)(2)\) \(-\) \(\frac(1)(3))\) \(=\) \(\frac( 9+7x)(6)\) \(\cdot 6\) |
Expand the bracket on the left |
|
|
\(6\cdot\)\(\frac(x+2)(2)\) \(-\) \(6\cdot\)\(\frac(1)(3)\) \(=\) \(\frac(9+7x)(6)\) \(\cdot 6\) |
Now let's reduce the denominators |
|
|
\(3(x+2)-2=9+7x\) |
Now it looks like a regular linear one! Let's finish it. |
|
|
By translating through equals we collect X's on the right and numbers on the left |
||
|
Well, dividing the right and left sides by \(-4\), we get the answer |
Answer : \(x=-1.25\)
- An equality with a variable is called an equation.
- Solving an equation means finding its many roots. An equation may have one, two, several, many roots, or none at all.
- Each value of a variable at which a given equation turns into a true equality is called a root of the equation.
- Equations that have the same roots are called equivalent equations.
- Any term of the equation can be transferred from one part of the equality to another, while changing the sign of the term to the opposite.
- If both sides of an equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given equation.
Examples. Solve the equation.
1. 1.5x+4 = 0.3x-2.
1.5x-0.3x = -2-4. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the following property was used:
1.2x = -6. Similar terms were given according to the rule:
x = -6 : 1.2. Both sides of the equality were divided by the coefficient of the variable, since
x = -5. Divided according to the rule for dividing a decimal fraction by decimal:
To divide a number by a decimal fraction, you need to move the commas in the dividend and divisor as many digits to the right as there are after the decimal point in the divisor, and then divide by a natural number:
6 : 1,2 = 60 : 12 = 5.
Answer: 5.
2. 3∙ (2x-9) = 4 ∙ (x-4).
6x-27 = 4x-16. We opened the brackets using the distributive law of multiplication relative to subtraction: (a-b) ∙ c = a ∙ c-b ∙ c.
6x-4x = -16+27. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the following property was used: any term of the equation can be transferred from one part of the equality to another, thereby changing the sign of the term to the opposite.
2x = 11. Similar terms were given according to the rule: to bring similar terms, you need to add their coefficients and multiply the resulting result by their common letter part (i.e., add their common letter part to the result obtained).
x = 11 : 2. Both sides of the equality were divided by the coefficient of the variable, since If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given equation.
Answer: 5,5.
3. 7x- (3+2x)=x-9.
7x-3-2x = x-9. We opened the brackets according to the rule for opening brackets preceded by a “-” sign: if there is a “-” sign in front of the brackets, then remove the brackets, the “-” sign and write the terms in the brackets with opposite signs.
7x-2x-x = -9+3. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the following property was used: any term of the equation can be transferred from one part of the equality to another, thereby changing the sign of the term to the opposite.
4x = -6. Similar terms were given according to the rule: to bring similar terms, you need to add their coefficients and multiply the resulting result by their common letter part (i.e., add their common letter part to the result obtained).
x = -6 : 4. Both sides of the equality were divided by the coefficient of the variable, since If both sides of the equation are multiplied or divided by the same non-zero number, you get an equation equivalent to the given equation.
Answer: -1,5.
3 ∙ (x-5) = 7 ∙ 12 — 4 ∙ (2x-11). We multiplied both sides of the equation by 12 - the lowest common denominator for the denominators of these fractions.
3x-15 = 84-8x+44. We opened the brackets using the distributive law of multiplication relative to subtraction: In order to multiply the difference of two numbers by a third number, you can separately multiply the minuend and separately subtract by the third number, and then subtract the second result from the first result, i.e.(a-b) ∙ c = a ∙ c-b ∙ c.
3x+8x = 84+44+15. We collected the terms containing the variable on the left side of the equality, and the free terms on the right side of the equality. In this case, the following property was used: any term of the equation can be transferred from one part of the equality to another, thereby changing the sign of the term to the opposite.
Systems of equations have been widely used in the economic industry with mathematical modeling various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.
Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.
A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.
Linear equation
Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.
Types of systems of linear equations
The simplest examples are considered to be systems of linear equations with two variables X and Y.
F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.
Solve system of equations - this means finding values (x, y) at which the system turns into a true equality or establishing that suitable values x and y do not exist.
A pair of values (x, y), written as the coordinates of a point, is called a solution to a system of linear equations.
If systems have one common solution or no solution exists, they are called equivalent.
Homogeneous systems of linear equations are systems right part which is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.
The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.
When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.
Simple and complex methods for solving systems of equations
There is no general analytical method for solving such systems; all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.
The main task when teaching solution methods is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method
Solving examples of systems of linear equations of the 7th grade program secondary school quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.
Solving systems using the substitution method
The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system
Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:
As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solution this example does not cause difficulties and allows you to obtain the Y value. Last step This is a check of the received values.
It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also inappropriate.
Solution of an example of a system of linear inhomogeneous equations:
Solution using algebraic addition
When searching for solutions to systems using the addition method, they perform term-by-term addition and multiplication of equations by different numbers. The ultimate goal of mathematical operations is an equation in one variable.
Application of this method requires practice and observation. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.
Solution algorithm:
- Multiply both sides of the equation by a certain number. As a result arithmetic action one of the coefficients of the variable must become equal to 1.
- Add the resulting expression term by term and find one of the unknowns.
- Substitute the resulting value into the 2nd equation of the system to find the remaining variable.
Method of solution by introducing a new variable
A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.
The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.
The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to the standard one quadratic trinomial. You can solve a polynomial by finding the discriminant.
It is necessary to find the value of the discriminant using the well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore D=100. If the discriminant Above zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant less than zero, then there is only one solution: x= -b / 2*a.
The solution for the resulting systems is found by addition.
Visual method for solving systems
Suitable for 3 equation systems. The method consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the points of intersection of the curves and will be general decision systems.
The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.
As can be seen from the example, for each line two points were constructed, the values of the variable x were chosen arbitrarily: 0 and 3. Based on the values of x, the values for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.
The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.
The following example requires finding a graphical solution to a system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.
As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.
The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.
The matrix and its varieties
Matrices are used for short note systems of linear equations. A matrix is a table special type filled with numbers. n*m has n - rows and m - columns.
A matrix is square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with an infinitely possible number of rows. A matrix with ones along one of the diagonals and other zero elements is called identity.
An inverse matrix is a matrix when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.
Rules for converting a system of equations into a matrix
In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.
A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.
The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.
When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.
Options for finding the inverse matrix
The formula for finding the inverse matrix is quite simple: K -1 = 1 / |K|, where K -1 - inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.
The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option, there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.
Solving examples of systems of linear equations using the matrix method
The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number of variables and equations.
In the example, a nm are the coefficients of the equations, the matrix is a vector x n are variables, and b n are free terms.
Solving systems using the Gaussian method
IN higher mathematics The Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variables of systems with a large number of linear equations.
The Gauss method is very similar to solutions by substitution and algebraic addition, but is more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By means of algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.
After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.
In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:
As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.
Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.
The Gauss method is difficult for students to understand high school, but is one of the most interesting ways to develop the ingenuity of children studying under the program in-depth study in math and physics classes.
For ease of recording, calculations are usually done as follows:
The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.
First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.
The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.
This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.
The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of human activity, while others exist for educational purposes.