What is called solving a linear equation. Linear equations: formulas and examples

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

\(-2x=8\) \(|:(-2)\)
\(x=-4\)

Usually this step is performed at the very end, when the equation is already 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

Here we go again similar terms

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 common denominator 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.

In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.

First, let's define: what is a linear equation and which one is called the simplest?

A linear equation is one in which there is only one variable, and only to the first degree.

The simplest equation means the construction:

All other linear equations are reduced to the simplest using the algorithm:

1. Expand parentheses, if any;
2. Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
3. Give similar terms to the left and right of the equal sign;
4. Divide the resulting equation by the coefficient of the variable $x$.

Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:

1. The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
2. The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.

Now let's see how all this works using real-life examples.

Examples of solving equations

Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.

Such constructions are solved in approximately the same way:

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

Then, as a rule, you need to 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:

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

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

Solving real examples of simple linear equations

Task No. 1

The first step requires us to open the brackets. But they are not in this example, so we skip them this stage. 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 parentheses, but they are not multiplied by anything, they are simply preceded by various 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 allow you to avoid making stupid and offensive mistakes in high school, when doing such actions 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 steps 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 algebraic sum. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.

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

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

Solving equations with fractions

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

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

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

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

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

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

Example No. 1

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

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

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

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

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

Now let's expand:

We seclude the variable:

We perform the reduction of similar terms:

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

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

We got final decision, 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!

Systems of equations are widely used in economic sector at 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 ​​of 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 way to solve similar systems, all methods are based on numerical solutions. IN school course mathematics, such methods as permutation, algebraic addition, substitution, as well as graphical and matrix method, solution by Gaussian method.

The main task when teaching solution methods is to teach how to correctly analyze the system and find optimal algorithm solutions 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. The last step is to check the obtained 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 mathematical operations is an equation with one variable.

For Applications this method practice and observation are required. 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:

1. Multiply both sides of the equation by a certain number. As a result arithmetic operation one of the coefficients of the variable must become equal to 1.
2. Add the resulting expression term by term and find one of the unknowns.
3. 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 discriminant value by known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. IN 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 is less than zero, then there is one solution: x = -b / 2*a.

The solution for the resulting systems is found by the addition method.

Visual method for solving systems

Suitable for 3 equation systems. The method is to build on coordinate axis graphs of each equation included in the system. 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 graphic solution systems 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 infinite possible number lines. 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 big amount 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 variable systems with a large number of linear equations.

Gauss's method is very similar to solutions using substitutions and algebraic addition, but 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 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 Gaussian 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.

Linear equation is algebraic equation, the total degree of which polynomials is 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 do not have 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, it’s very simple - using simple algebraic operations and following the rules of transfer. If the equation appears in front of you in general form, you are in luck; all you need to do is:

1. Move b to the right side of the equation, not forgetting to change the sign (transfer rule!), thus, from an expression of the form ax + b = 0, an expression of the form should be obtained: ax = -b.
2. 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:

1. 2x + 4 = 0 - move b equal to in this case 4, to the right
2. 2x = -4 - divide b by a (don’t forget about the minus sign)
3. 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:

1. (12(x+2)/3) + 12x = 12 - 36x/4
2. 4(x+2) + 12x = 12 - 36x/4
3. 4x + 8 + 12x = 12 - 9x - now we carry out the transfer:
4. 25x - 4 = 0 - it remains to find a solution using the already known scheme:
5. 25x = 4,
6. 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!