Systems of equations are widely used in the economic sector for mathematical modeling of 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 whose right-hand side 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 in the 7th grade general education curriculum is 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. Solving this example is easy and allows you to get 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, equations are added term by term and multiplied by various 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 of the arithmetic operation, one of the coefficients of the variable should 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 a standard 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 is greater than 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 consists in constructing graphs of each equation included in the system on the coordinate axis. The coordinates of the intersection points of the curves will be the general solution of the system.
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 to concisely write a system of linear equations. A matrix is a special type of table 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 is the 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 Gaussian method is difficult for middle school students to understand, but it is one of the most interesting ways to develop the ingenuity of children enrolled in advanced learning programs 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.
A linear equation is an algebraic equation whose total degree of polynomials is equal to one. Solving linear equations is part of the school curriculum, and not the most difficult one. However, some still have difficulty completing this topic. We hope that after reading this material, all difficulties for you will remain in the past. So, let's figure it out. how to solve linear equations.
General form
The linear equation is represented as:
- ax + b = 0, where a and b are any numbers.
Although a and b can be any number, their values affect the number of solutions to the equation. There are several special cases of solution:
- If a=b=0, the equation has an infinite number of solutions;
- If a=0, b≠0, the equation has no solution;
- If a≠0, b=0, the equation has a solution: x = 0.
In the event that both numbers have non-zero values, the equation must be solved to derive 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:
- 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.
- Apply the rule: to find one of the factors (x - in our case), you need to divide the product (-b in our case) by another factor (a - in our case). Thus, you should get an expression of the form: x = -b/a.
That's it - a solution has been found!
Now let's look at a specific example:
- 2x + 4 = 0 - move b, equal to 4 in this case, to the right side
- 2x = -4 - divide b by a (don’t forget about the minus sign)
- x = -4/2 = -2
That's all! Our solution: x = -2.
As you can see, the solution to a linear equation with one variable is quite simple to find, but everything is so simple if we are lucky enough to come across the equation in its general form. In most cases, before solving an equation in the two steps described above, you still need to bring the existing expression to a general form. 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? We will not rush to conclusions, we will act - we will move all the components of our equation to the left side. We get:
- 12x - 2x - 14 - 6 = 0
Now subtract like from like, we get:
- 10x - 20 = 0
Learned? The most linear equation ever! The solution to which is: x = 20/10 = 2.
What if we have this example:
- 12((x + 2)/3) + x) = 12 (1 - 3x/4)
Yes, this is also a linear equation, only more transformations need to be carried out. First, let's open the brackets:
- (12(x+2)/3) + 12x = 12 - 36x/4
- 4(x+2) + 12x = 12 - 36x/4
- 4x + 8 + 12x = 12 - 9x - now we carry out the transfer:
- 25x - 4 = 0 - it remains to find a solution using the already known scheme:
- 25x = 4,
- x = 4/25 = 0.16
As you can see, everything can be solved, the main thing is not to worry, but to act. Remember, if your equation contains only variables of the first degree and numbers, you have a linear equation, which, no matter how it looks initially, can be reduced to a general form and solved. We hope everything works out for you! Good luck!
In this video we will analyze a whole set of linear equations that are solved using the same algorithm - that’s why they are called the simplest.
First, let's define: what is a linear equation and which one is called the simplest?
A linear equation is one in which there is only one variable, and only to the first degree.
The simplest equation means the construction:
All other linear equations are reduced to the simplest using the algorithm:
- Expand parentheses, if any;
- Move terms containing a variable to one side of the equal sign, and terms without a variable to the other;
- Give similar terms to the left and right of the equal sign;
- Divide the resulting equation by the coefficient of the variable $x$.
Of course, this algorithm does not always help. The fact is that sometimes after all these machinations the coefficient of the variable $x$ turns out to be equal to zero. In this case, two options are possible:
- The equation has no solutions at all. For example, when something like $0\cdot x=8$ turns out, i.e. on the left is zero, and on the right is a number other than zero. In the video below we will look at several reasons why this situation is possible.
- The solution is all numbers. The only case when this is possible is when the equation has been reduced to the construction $0\cdot x=0$. It is quite logical that no matter what $x$ we substitute, it will still turn out “zero is equal to zero”, i.e. correct numerical equality.
Now let's see how all this works using real-life examples.
Examples of solving equations
Today we are dealing with linear equations, and only the simplest ones. In general, a linear equation means any equality that contains exactly one variable, and it goes only to the first degree.
Such constructions are solved in approximately the same way:
- First of all, you need to expand the parentheses, if there are any (as in our last example);
- Then combine similar
- Finally, isolate the variable, i.e. move everything connected with the variable—the terms in which it is contained—to one side, and move everything that remains without it to the other side.
Then, as a rule, you need to bring similar ones on each side of the resulting equality, and after that all that remains is to divide by the coefficient of “x”, and we will get the final answer.
In theory, this looks nice and simple, but in practice, even experienced high school students can make offensive mistakes in fairly simple linear equations. Typically, errors are made either when opening brackets or when calculating the “pluses” and “minuses”.
In addition, it happens that a linear equation has no solutions at all, or that the solution is the entire number line, i.e. any number. We will look at these subtleties in today's lesson. But we will start, as you already understood, with the simplest tasks.
Scheme for solving simple linear equations
First, let me once again write the entire scheme for solving the simplest linear equations:
- Expand the brackets, if any.
- We isolate the variables, i.e. We move everything that contains “X’s” to one side, and everything without “X’s” to the other.
- We present similar terms.
- We divide everything by the coefficient of “x”.
Of course, this scheme does not always work; there are certain subtleties and tricks in it, and now we will get to know them.
Solving real examples of simple linear equations
Task No. 1
The first step requires us to open the brackets. But they are not in this example, so we skip this step. In the second step we need to isolate the variables. Please note: we are talking only about individual terms. Let's write it down:
We present similar terms on the left and right, but this has already been done here. Therefore, we move on to the fourth step: divide by the coefficient:
\[\frac(6x)(6)=-\frac(72)(6)\]
So we got the answer.
Task No. 2
We can see the parentheses in this problem, so let's expand them:
Both on the left and on the right we see approximately the same design, but let's act according to the algorithm, i.e. separating the variables:
Here are some similar ones:
At what roots does this work? Answer: for any. Therefore, we can write that $x$ is any number.
Task No. 3
The third linear equation is more interesting:
\[\left(6-x \right)+\left(12+x \right)-\left(3-2x \right)=15\]
There are several brackets here, but they are not multiplied by anything, they are simply preceded by different signs. Let's break them down:
We perform the second step already known to us:
\[-x+x+2x=15-6-12+3\]
Let's do the math:
We carry out the last step - divide everything by the coefficient of “x”:
\[\frac(2x)(x)=\frac(0)(2)\]
Things to Remember When Solving Linear Equations
If we ignore too simple tasks, I would like to say the following:
- As I said above, not every linear equation has a solution - sometimes there are simply no roots;
- Even if there are roots, there may be zero among them - there is nothing wrong with that.
Zero is the same number as the others; you shouldn’t discriminate against it in any way or assume that if you get zero, then you did something wrong.
Another feature is related to the opening of brackets. Please note: when there is a “minus” in front of them, we remove it, but in parentheses we change the signs to opposite. And then we can open it using standard algorithms: we will get what we saw in the calculations above.
Understanding this simple fact will help you avoid making stupid and hurtful mistakes in high school, when doing such things is taken for granted.
Solving complex linear equations
Let's move on to more complex equations. Now the constructions will become more complex and when performing various transformations a quadratic function will appear. However, we should not be afraid of this, because if, according to the author’s plan, we are solving a linear equation, then during the transformation process all monomials containing a quadratic function will certainly cancel.
Example No. 1
Obviously, the first step is to open the brackets. Let's do this very carefully:
Now let's take a look at privacy:
\[-x+6((x)^(2))-6((x)^(2))+x=-12\]
Here are some similar ones:
Obviously, this equation has no solutions, so we’ll write this in the answer:
\[\varnothing\]
or there are no roots.
Example No. 2
We perform the same actions. First step:
Let's move everything with a variable to the left, and without it - to the right:
Here are some similar ones:
Obviously, this linear equation has no solution, so we’ll write it this way:
\[\varnothing\],
or there are no roots.
Nuances of the solution
Both equations are completely solved. Using these two expressions as an example, we were once again convinced that even in the simplest linear equations, everything may not be so simple: there can be either one, or none, or infinitely many roots. In our case, we considered two equations, both simply have no roots.
But I would like to draw your attention to another fact: how to work with parentheses and how to open them if there is a minus sign in front of them. Consider this expression:
Before opening, you need to multiply everything by “X”. Please note: multiplies each individual term. Inside there are two terms - respectively, two terms and multiplied.
And only after these seemingly elementary, but very important and dangerous transformations have been completed, can you open the bracket from the point of view of the fact that there is a minus sign after it. Yes, yes: only now, when the transformations are completed, we remember that there is a minus sign in front of the brackets, which means that everything below simply changes signs. At the same time, the brackets themselves disappear and, most importantly, the front “minus” also disappears.
We do the same with the second equation:
It is not by chance that I pay attention to these small, seemingly insignificant facts. Because solving equations is always a sequence of elementary transformations, where the inability to clearly and competently perform simple actions leads to the fact that high school students come to me and again learn to solve such simple equations.
Of course, the day will come when you will hone these skills to the point of automaticity. You will no longer have to perform so many transformations each time; you will write everything on one line. But while you are just learning, you need to write each action separately.
Solving even more complex linear equations
What we are going to solve now can hardly be called the simplest task, but the meaning remains the same.
Task No. 1
\[\left(7x+1 \right)\left(3x-1 \right)-21((x)^(2))=3\]
Let's multiply all the elements in the first part:
Let's do some privacy:
Here are some similar ones:
Let's complete the last step:
\[\frac(-4x)(4)=\frac(4)(-4)\]
Here is our final answer. And, despite the fact that in the process of solving we had coefficients with a quadratic function, they canceled each other out, which makes the equation linear and not quadratic.
Task No. 2
\[\left(1-4x \right)\left(1-3x \right)=6x\left(2x-1 \right)\]
Let's carefully perform the first step: multiply each element from the first bracket by each element from the second. There should be a total of four new terms after the transformations:
Now let’s carefully perform the multiplication in each term:
Let’s move the terms with “X” to the left, and those without - to the right:
\[-3x-4x+12((x)^(2))-12((x)^(2))+6x=-1\]
Here are similar terms:
Once again we have received the final answer.
Nuances of the solution
The most important note about these two equations is the following: as soon as we begin to multiply brackets that contain more than one term, this is done according to the following rule: we take the first term from the first and multiply with each element from the second; then we take the second element from the first and similarly multiply with each element from the second. As a result, we will have four terms.
About the algebraic sum
With this last example, I would like to remind students what an algebraic sum is. In classical mathematics, by $1-7$ we mean a simple construction: subtract seven from one. In algebra, we mean the following by this: to the number “one” we add another number, namely “minus seven”. This is how an algebraic sum differs from an ordinary arithmetic sum.
As soon as, when performing all the transformations, each addition and multiplication, you begin to see constructions similar to those described above, you simply will not have any problems in algebra when working with polynomials and equations.
Finally, let's look at a couple more examples that will be even more complex than the ones we just looked at, and to solve them we will have to slightly expand our standard algorithm.
Solving equations with fractions
To solve such tasks, we will have to add one more step to our algorithm. But first, let me remind you of our algorithm:
- Open the brackets.
- Separate variables.
- Bring similar ones.
- Divide by the ratio.
Alas, this wonderful algorithm, for all its effectiveness, turns out to be not entirely appropriate when we have fractions in front of us. And in what we will see below, we have a fraction on both the left and the right in both equations.
How to work in this case? Yes, it's very simple! To do this, you need to add one more step to the algorithm, which can be done both before and after the first action, namely, getting rid of fractions. So the algorithm will be as follows:
- Get rid of fractions.
- Open the brackets.
- Separate variables.
- Bring similar ones.
- Divide by the ratio.
What does it mean to “get rid of fractions”? And why can this be done both after and before the first standard step? In fact, in our case, all fractions are numerical in their denominator, i.e. Everywhere the denominator is just a number. Therefore, if we multiply both sides of the equation by this number, we will get rid of fractions.
Example No. 1
\[\frac(\left(2x+1 \right)\left(2x-3 \right))(4)=((x)^(2))-1\]
Let's get rid of the fractions in this equation:
\[\frac(\left(2x+1 \right)\left(2x-3 \right)\cdot 4)(4)=\left(((x)^(2))-1 \right)\cdot 4\]
Please note: everything is multiplied by “four” once, i.e. just because you have two parentheses doesn't mean you have to multiply each one by "four." Let's write down:
\[\left(2x+1 \right)\left(2x-3 \right)=\left(((x)^(2))-1 \right)\cdot 4\]
Now let's expand:
We seclude the variable:
We perform the reduction of similar terms:
\[-4x=-1\left| :\left(-4 \right) \right.\]
\[\frac(-4x)(-4)=\frac(-1)(-4)\]
We have received the final solution, let's move on to the second equation.
Example No. 2
\[\frac(\left(1-x \right)\left(1+5x \right))(5)+((x)^(2))=1\]
Here we perform all the same actions:
\[\frac(\left(1-x \right)\left(1+5x \right)\cdot 5)(5)+((x)^(2))\cdot 5=5\]
\[\frac(4x)(4)=\frac(4)(4)\]
The problem is solved.
That, in fact, is all I wanted to tell you today.
Key points
Key findings are:
- Know the algorithm for solving linear equations.
- Ability to open brackets.
- Don't worry if you have quadratic functions somewhere; most likely, they will be reduced in the process of further transformations.
- There are three types of roots in linear equations, even the simplest ones: one single root, the entire number line is a root, and no roots at all.
I hope this lesson will help you master a simple, but very important topic for further understanding of all mathematics. If something is not clear, go to the site and solve the examples presented there. Stay tuned, many more interesting things await you!
First you need to understand what it is.
There is a simple definition linear equation, which is given in a regular school: “an equation in which the variable occurs only in the first power.” But it is not entirely correct: the equation is not linear, it does not even reduce to that, it reduces to quadratic.
A more precise definition is: linear equation is an equation that, using equivalent transformations can be reduced to the form , where title="a,b in bbR, ~a0">. На деле мы будем приводить это уравнение к виду путём переноса в правую часть и деления обеих частей уравнения на . Осталось разъяснить, какие уравнения и как мы можем привести к такому виду, и, самое главное, что дальше делать с ними, чтобы решить его.!}
In fact, in order to understand whether an equation is linear or not, it must first be simplified, that is, brought to a form where its classification will be unambiguous. Remember, you can do whatever you want with an equation as long as it doesn’t change its roots - that’s what it is. equivalent conversion. The simplest equivalent transformations include:
- opening parentheses
- bringing similar
- multiplying and/or dividing both sides of an equation by a nonzero number
- adding and/or subtracting from both sides of the same number or expression*
*A particular interpretation of the last transformation is the “transfer” of terms from one part to another with a change of sign.
Example 1:
(let's open the brackets)
(add to both parts and subtract/transfer with changing the sign of the number to the left, and the variables to the right)
(let's give similar ones)
(divide both sides of the equation by 3)
So we get an equation that has the same roots as the original one. Let us remind the reader that "solve the equation"- means finding all its roots and proving that there are no others, and "root of the equation"- this is a number that, when substituted for the unknown, will turn the equation into a true equality. Well, in the last equation, finding a number that turns the equation into a true equality is very simple - this is the number. No other number will make an identity from this equation. Answer:
Example 2:
(multiply both sides of the equation by 
, after making sure that we are not multiplying by : title="x3/2"> и title="x3">. То есть если такие корни получатся, то мы их обязаны будем выкинуть.)!}
(let's open the brackets)
(let's move the terms)
(let's give similar ones)
(we divide both parts by )
This is roughly how all linear equations are solved. For younger readers, most likely, this explanation seemed complicated, so we offer a version "linear equations for grade 5"
Learning to solve equations is one of the main tasks that algebra poses for students. Starting with the simplest, when it consists of one unknown, and moving on to more and more complex ones. If you have not mastered the actions that need to be performed with equations from the first group, it will be difficult to understand the others.
To continue the conversation, you need to agree on notation.
General form of a linear equation with one unknown and the principle of its solution
Any equation that can be written like this:
a * x = b,
called linear. This is the general formula. But often in assignments linear equations are written in implicit form. Then it is necessary to perform identical transformations to obtain a generally accepted notation. These actions include:
- opening parentheses;
- moving all terms with a variable value to the left side of the equality, and the rest to the right;
- reduction of similar terms.
In the case where an unknown quantity is in the denominator of a fraction, you need to determine its values at which the expression will not make sense. In other words, you need to know the domain of definition of the equation.
The principle by which all linear equations are solved comes down to dividing the value on the right side of the equation by the coefficient in front of the variable. That is, “x” will be equal to b/a.
Special cases of linear equations and their solutions
During reasoning, moments may arise when linear equations take on one of the special forms. Each of them has a specific solution.
In the first situation:
a * x = 0, and a ≠ 0.
The solution to such an equation will always be x = 0.
In the second case, “a” takes the value equal to zero:
0 * x = 0.
The answer to such an equation will be any number. That is, it has an infinite number of roots.
The third situation looks like this:
0 * x = in, where in ≠ 0.
This equation doesn't make sense. Because there are no roots that satisfy it.
General view of a linear equation with two variables
From its name it becomes clear that there are already two unknown quantities in it. Linear equations in two variables look like this:
a * x + b * y = c.
Since there are two unknowns in the record, the answer will look like a pair of numbers. That is, it is not enough to specify only one value. This will be an incomplete answer. A pair of quantities for which the equation becomes an identity is a solution to the equation. Moreover, in the answer, the variable that comes first in the alphabet is always written down first. Sometimes they say that these numbers satisfy him. Moreover, there can be an infinite number of such pairs.
How to solve a linear equation with two unknowns?
To do this, you just need to select any pair of numbers that turns out to be correct. For simplicity, you can take one of the unknowns equal to some prime number, and then find the second.
When solving, you often have to perform steps to simplify the equation. They are called identity transformations. Moreover, the following properties are always true for equations:
- each term can be moved to the opposite part of the equality by replacing its sign with the opposite one;
- The left and right sides of any equation are allowed to be divided by the same number, as long as it is not equal to zero.
Examples of tasks with linear equations
First task. Solve linear equations: 4x = 20, 8(x - 1) + 2x = 2(4 - 2x); (5x + 15) / (x + 4) = 4; (5x + 15) / (x + 3) = 4.
In the equation that comes first on this list, simply divide 20 by 4. The result will be 5. This is the answer: x = 5.
The third equation requires that an identity transformation be performed. It will consist of opening the brackets and bringing similar terms. After the first step, the equation will take the form: 8x - 8 + 2x = 8 - 4x. Then you need to move all the unknowns to the left side of the equation, and the rest to the right. The equation will look like this: 8x + 2x + 4x = 8 + 8. After adding similar terms: 14x = 16. Now it looks the same as the first one, and its solution is easy to find. The answer will be x=8/7. But in mathematics you are supposed to isolate the whole part from an improper fraction. Then the result will be transformed, and “x” will be equal to one whole and one seventh.
In the remaining examples, the variables are in the denominator. This means that you first need to find out at what values the equations are defined. To do this, you need to exclude numbers at which the denominators go to zero. In the first example it is “-4”, in the second it is “-3”. That is, these values need to be excluded from the answer. After this, you need to multiply both sides of the equality by the expressions in the denominator.
Opening the brackets and bringing similar terms, in the first of these equations we get: 5x + 15 = 4x + 16, and in the second 5x + 15 = 4x + 12. After transformations, the solution to the first equation will be x = -1. The second turns out to be equal to “-3”, which means that the latter has no solutions.
Second task. Solve the equation: -7x + 2y = 5.
Suppose that the first unknown x = 1, then the equation will take the form -7 * 1 + 2y = 5. Moving the factor “-7” to the right side of the equality and changing its sign to plus, it turns out that 2y = 12. This means y =6. Answer: one of the solutions to the equation x = 1, y = 6.
General form of inequality with one variable
All possible situations for inequalities are presented here:
- a * x > b;
- a * x< в;
- a * x ≥b;
- a * x ≤в.
In general, it looks like a simple linear equation, only the equal sign is replaced by an inequality.
Rules for identity transformations of inequalities
Just like linear equations, inequalities can be modified according to certain laws. They boil down to the following:
- any alphabetic or numerical expression can be added to the left and right sides of the inequality, and the sign of the inequality remains the same;
- you can also multiply or divide by the same positive number, this again does not change the sign;
- When multiplying or dividing by the same negative number, the equality will remain true provided that 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 between the brackets you need to write the symbol “and”. 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].