Solving systems of equations by algebraic addition. Examples of systems of linear equations: solution method
With this video, I begin a series of lessons on systems of equations. Today we will talk about solving systems of linear equations addition method This is one of the simplest ways, but at the same time one of the most effective.
The addition method consists of three simple steps:
- Look at the system and choose a variable that has the same (or opposite) coefficients in each equation;
- Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring like terms;
- Solve the new equation obtained after the second step.
If everything is done correctly, then at the output we will get a single equation with one variable- It won't be hard to solve. Then it remains only to substitute the found root in the original system and get the final answer.
However, in practice it is not so simple. There are several reasons for this:
- Solving equations by addition implies that all rows must contain variables with the same/opposite coefficients. What if this requirement is not met?
- Not always, after adding / subtracting equations in this way, we will get a beautiful construction that is easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?
To get an answer to these questions, and at the same time to deal with a few additional subtleties that many students “fall over”, watch my video tutorial:
With this lesson, we begin a series of lectures on systems of equations. And we will start with the simplest of them, namely those that contain two equations and two variables. Each of them will be linear.
Systems is a 7th grade material, but this lesson will also be useful for high school students who want to brush up on their knowledge on this topic.
In general, there are two methods for solving such systems:
- Addition method;
- A method of expressing one variable in terms of another.
Today we will deal with the first method - we will use the method of subtraction and addition. But for this you need to understand the following fact: once you have two or more equations, you can take any two of them and add them together. They are added term by term, i.e. "Xs" are added to "Xs" and similar ones are given, "games" to "games" - similar ones are again given, and what is to the right of the equal sign is also added to each other, and similar ones are also given there.
The results of such machinations will be a new equation, which, if it has roots, they will certainly be among the roots of the original equation. So our task is to do the subtraction or addition in such a way that either $x$ or $y$ disappears.
How to achieve this and what tool to use for this - we will talk about this now.
Solving easy problems using the addition method
So, we are learning to apply the addition method using the example of two simple expressions.
Task #1
\[\left\( \begin(align)& 5x-4y=22 \\& 7x+4y=2 \\\end(align) \right.\]
Note that $y$ has a coefficient of $-4$ in the first equation, and $+4$ in the second. They are mutually opposite, so it is logical to assume that if we add them up, then in the resulting amount, the “games” will mutually annihilate. We add and get:
We solve the simplest construction:
Great, we found the X. What to do with him now? We can substitute it into any of the equations. Let's put it in the first one:
\[-4y=12\left| :\left(-4 \right) \right.\]
Answer: $\left(2;-3\right)$.
Task #2
\[\left\( \begin(align)& -6x+y=21 \\& 6x-11y=-51 \\\end(align) \right.\]
Here, the situation is completely similar, only with the Xs. Let's put them together:
We got the simplest linear equation, let's solve it:
Now let's find $x$:
Answer: $\left(-3;3\right)$.
Important Points
So, we have just solved two simple systems of linear equations using the addition method. Once again the key points:
- If there are opposite coefficients for one of the variables, then it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
- We substitute the found variable into any of the equations of the system to find the second one.
- The final record of the answer can be presented in different ways. For example, like this - $x=...,y=...$, or in the form of coordinates of points - $\left(...;... \right)$. The second option is preferable. The main thing to remember is that the first coordinate is $x$, and the second is $y$.
- The rule to write the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the role of variables is not $x$ and $y$, but, for example, $a$ and $b$.
In the following problems, we will consider the subtraction technique when the coefficients are not opposite.
Solving easy problems using the subtraction method
Task #1
\[\left\( \begin(align)& 10x-3y=5 \\& -6x-3y=-27 \\\end(align) \right.\]
Note that there are no opposite coefficients here, but there are identical ones. Therefore, we subtract the second equation from the first equation:
Now we substitute the value of $x$ into any of the equations of the system. Let's go first:
Answer: $\left(2;5\right)$.
Task #2
\[\left\( \begin(align)& 5x+4y=-22 \\& 5x-2y=-4 \\\end(align) \right.\]
We again see the same coefficient $5$ for $x$ in the first and second equations. Therefore, it is logical to assume that you need to subtract the second from the first equation:
We have calculated one variable. Now let's find the second one, for example, by substituting the value of $y$ into the second construct:
Answer: $\left(-3;-2 \right)$.
Nuances of the solution
So what do we see? In essence, the scheme is no different from the solution of previous systems. The only difference is that we do not add equations, but subtract them. We are doing algebraic subtraction.
In other words, as soon as you see a system consisting of two equations with two unknowns, the first thing you need to look at is the coefficients. If they are the same anywhere, the equations are subtracted, and if they are opposite, the addition method is applied. This is always done so that one of them disappears, and in the final equation that remains after subtraction, only one variable would remain.
Of course, that's not all. Now we will consider systems in which the equations are generally inconsistent. Those. there are no such variables in them that would be either the same or opposite. In this case, to solve such systems, an additional technique is used, namely, the multiplication of each of the equations by a special coefficient. How to find it and how to solve such systems in general, now we will talk about this.
Solving problems by multiplying by a coefficient
Example #1
\[\left\( \begin(align)& 5x-9y=38 \\& 3x+2y=8 \\\end(align) \right.\]
We see that neither for $x$ nor for $y$ the coefficients are not only mutually opposite, but in general they do not correlate in any way with another equation. These coefficients will not disappear in any way, even if we add or subtract the equations from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $y$ variable. To do this, we multiply the first equation by the coefficient of $y$ from the second equation, and the second equation by the coefficient of $y$ from the first equation, without changing the sign. We multiply and get a new system:
\[\left\( \begin(align)& 10x-18y=76 \\& 27x+18y=72 \\\end(align) \right.\]
Let's look at it: for $y$, opposite coefficients. In such a situation, it is necessary to apply the addition method. Let's add:
Now we need to find $y$. To do this, substitute $x$ in the first expression:
\[-9y=18\left| :\left(-9 \right) \right.\]
Answer: $\left(4;-2\right)$.
Example #2
\[\left\( \begin(align)& 11x+4y=-18 \\& 13x-6y=-32 \\\end(align) \right.\]
Again, the coefficients for none of the variables are consistent. Let's multiply by the coefficients at $y$:
\[\left\( \begin(align)& 11x+4y=-18\left| 6 \right. \\& 13x-6y=-32\left| 4 \right. \\\end(align) \right .\]
\[\left\( \begin(align)& 66x+24y=-108 \\& 52x-24y=-128 \\\end(align) \right.\]
Our new system is equivalent to the previous one, but the coefficients of $y$ are mutually opposite, and therefore it is easy to apply the addition method here:
Now find $y$ by substituting $x$ into the first equation:
Answer: $\left(-2;1\right)$.
Nuances of the solution
The key rule here is the following: always multiply only by positive numbers - this will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:
- We look at the system and analyze each equation.
- If we see that neither for $y$ nor for $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: select the variable to get rid of, and then look at the coefficients in these equations. If we multiply the first equation by the coefficient from the second, and multiply the second, corresponding, by the coefficient from the first, then in the end we will get a system that is completely equivalent to the previous one, and the coefficients at $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
- We find one variable.
- We substitute the found variable into one of the two equations of the system and find the second one.
- We write the answer in the form of coordinates of points, if we have variables $x$ and $y$.
But even such a simple algorithm has its own subtleties, for example, the coefficients of $x$ or $y$ can be fractions and other "ugly" numbers. We will now consider these cases separately, because in them you can act in a slightly different way than according to the standard algorithm.
Solving problems with fractional numbers
Example #1
\[\left\( \begin(align)& 4m-3n=32 \\& 0.8m+2.5n=-6 \\\end(align) \right.\]
First, note that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We get $5$. Let's multiply the second equation by $5$:
\[\left\( \begin(align)& 4m-3n=32 \\& 4m+12,5m=-30 \\\end(align) \right.\]
We subtract the equations from each other:
$n$ we found, now we calculate $m$:
Answer: $n=-4;m=5$
Example #2
\[\left\( \begin(align)& 2.5p+1.5k=-13\left| 4 \right. \\& 2p-5k=2\left| 5 \right. \\\end(align )\right.\]
Here, as in the previous system, there are fractional coefficients, however, for none of the variables, the coefficients do not fit into each other by an integer number of times. Therefore, we use the standard algorithm. Get rid of $p$:
\[\left\( \begin(align)& 5p+3k=-26 \\& 5p-12,5k=5 \\\end(align) \right.\]
Let's use the subtraction method:
Let's find $p$ by substituting $k$ into the second construct:
Answer: $p=-4;k=-2$.
Nuances of the solution
That's all optimization. In the first equation, we did not multiply by anything at all, and the second equation was multiplied by $5$. As a result, we have obtained a consistent and even the same equation for the first variable. In the second system, we acted according to the standard algorithm.
But how to find the numbers by which you need to multiply the equations? After all, if we multiply by fractional numbers, we get new fractions. Therefore, the fractions must be multiplied by a number that would give a new integer, and after that, the variables should be multiplied by coefficients, following the standard algorithm.
In conclusion, I would like to draw your attention to the format of the response record. As I already said, since here we don’t have $x$ and $y$ here, but other values, we use a non-standard notation of the form:
Solving complex systems of equations
As a final touch to today's video tutorial, let's look at a couple of really complex systems. Their complexity will consist in the fact that they will contain variables both on the left and on the right. Therefore, to solve them, we will have to apply preprocessing.
System #1
\[\left\( \begin(align)& 3\left(2x-y \right)+5=-2\left(x+3y \right)+4 \\& 6\left(y+1 \right )-1=5\left(2x-1 \right)+8 \\\end(align) \right.\]
Each equation carries a certain complexity. Therefore, with each expression, let's do as with a normal linear construction.
In total, we get the final system, which is equivalent to the original one:
\[\left\( \begin(align)& 8x+3y=-1 \\& -10x+6y=-2 \\\end(align) \right.\]
Let's look at the coefficients of $y$: $3$ fits into $6$ twice, so we multiply the first equation by $2$:
\[\left\( \begin(align)& 16x+6y=-2 \\& -10+6y=-2 \\\end(align) \right.\]
The coefficients of $y$ are now equal, so we subtract the second from the first equation: $$
Now let's find $y$:
Answer: $\left(0;-\frac(1)(3) \right)$
System #2
\[\left\( \begin(align)& 4\left(a-3b \right)-2a=3\left(b+4 \right)-11 \\& -3\left(b-2a \right )-12=2\left(a-5 \right)+b \\\end(align) \right.\]
Let's transform the first expression:
Let's deal with the second:
\[-3\left(b-2a \right)-12=2\left(a-5 \right)+b\]
\[-3b+6a-12=2a-10+b\]
\[-3b+6a-2a-b=-10+12\]
In total, our initial system will take the following form:
\[\left\( \begin(align)& 2a-15b=1 \\& 4a-4b=2 \\\end(align) \right.\]
Looking at the coefficients of $a$, we see that the first equation needs to be multiplied by $2$:
\[\left\( \begin(align)& 4a-30b=2 \\& 4a-4b=2 \\\end(align) \right.\]
We subtract the second from the first construction:
Now find $a$:
Answer: $\left(a=\frac(1)(2);b=0 \right)$.
That's all. I hope this video tutorial will help you understand this difficult topic, namely, solving systems of simple linear equations. There will be many more lessons on this topic further: we will analyze more complex examples, where there will be more variables, and the equations themselves will already be nonlinear. See you soon!
Using the addition method, the equations of the system are added term by term, while 1 or both (several) equations can be multiplied by any number. As a result, they come to an equivalent SLE , where one of the equations has only one variable.
To solve the system term by term addition (subtraction) follow the next steps:
1. We select a variable for which the same coefficients will be made.
2. Now you need to add or subtract the equations and get an equation with one variable.
System solution are the intersection points of the graphs of the function.
Let's look at examples.
Example 1
Given system:
After analyzing this system, you can see that the coefficients of the variable are equal in absolute value and different in sign (-1 and 1). In this case, the equations can be easily added term by term:
The actions that are circled in red are performed in the mind.
The result of termwise addition was the disappearance of the variable y. It is in this and This, in fact, is the meaning of the method - to get rid of the first of the variables.
-4 - y + 5 = 0 → y = 1,
As a system, the solution looks like this:
Answer: x = -4 , y = 1.
Example 2
Given system:
In this example, you can use the "school" method, but it has a rather big minus - when you express any variable from any equation, you will get a solution in ordinary fractions. And solving fractions takes enough time and the probability of making mistakes increases.
Therefore, it is better to use term-by-term addition (subtraction) of equations. Let's analyze the coefficients of the corresponding variables:
Find a number that can be divided by 3 and on 4 , while it is necessary that this number be as small as possible. This least common multiple. If it’s hard for you to find the right number, then you can multiply the coefficients:.
Next step:
Multiply the 1st equation by ,
Multiply the 3rd equation by ,
Algebraic addition method
You can solve a system of equations with two unknowns in various ways - a graphical method or a variable change method.
In this lesson, we will get acquainted with another way of solving systems that you will surely like - this is the algebraic addition method.
And where did the idea come from - to put something in the systems? When solving systems, the main problem is the presence of two variables, because we cannot solve equations with two variables. So, it is necessary to exclude one of them in some legal way. And such legitimate ways are mathematical rules and properties.
One of these properties sounds like this: the sum of opposite numbers is zero. This means that if there are opposite coefficients for one of the variables, then their sum will be equal to zero and we will be able to exclude this variable from the equation. It is clear that we do not have the right to add only the terms with the variable we need. It is necessary to add the equations as a whole, i.e. separately add like terms on the left side, then on the right. As a result, we will get a new equation containing only one variable. Let's take a look at specific examples.
We see that in the first equation there is a variable y, and in the second the opposite number is y. So this equation can be solved by the addition method.
One of the equations is left as it is. Any one you like best.
But the second equation will be obtained by adding these two equations term by term. Those. Add 3x to 2x, add y to -y, add 8 to 7.
We get a system of equations
The second equation of this system is a simple equation with one variable. From it we find x \u003d 3. Substituting the found value into the first equation, we find y \u003d -1.
Answer: (3; - 1).
Design sample:
Solve the system of equations by algebraic addition
There are no variables with opposite coefficients in this system. But we know that both sides of the equation can be multiplied by the same number. Let's multiply the first equation of the system by 2.
Then the first equation will take the form:
Now we see that with the variable x there are opposite coefficients. So, we will do the same as in the first example: we will leave one of the equations unchanged. For example, 2y + 2x \u003d 10. And we get the second by adding.
Now we have a system of equations:
We easily find from the second equation y = 1, and then from the first equation x = 4.
Design sample:
Let's summarize:
We have learned how to solve systems of two linear equations with two unknowns using the algebraic addition method. Thus, we now know three main methods for solving such systems: the graphical method, the change of variable method, and the addition method. Almost any system can be solved using these methods. In more complex cases, a combination of these techniques is used.
List of used literature:
- Mordkovich A.G., Algebra grade 7 in 2 parts, Part 1, Textbook for educational institutions / A.G. Mordkovich. - 10th ed., revised - Moscow, "Mnemosyne", 2007.
- Mordkovich A.G., Algebra grade 7 in 2 parts, Part 2, Task book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, Mnemosyne, 2007.
- HER. Tulchinskaya, Algebra Grade 7. Blitz survey: a guide for students of educational institutions, 4th edition, revised and supplemented, Moscow, Mnemozina, 2008.
- Alexandrova L.A., Algebra Grade 7. Thematic test papers in a new form for students of educational institutions, edited by A.G. Mordkovich, Moscow, "Mnemosyne", 2011.
- Aleksandrova L.A. Algebra 7th grade. Independent work for students of educational institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, "Mnemosyne", 2010.
OGBOU "Education Center for Children with Special Educational Needs in Smolensk"
Distance Education Center
Algebra lesson in 7th grade
Lesson topic: The method of algebraic addition.
- Type of lesson: Lesson of the primary presentation of new knowledge.
The purpose of the lesson: control the level of assimilation of knowledge and skills in solving systems of equations by substitution; formation of skills and abilities for solving systems of equations by the method of addition.
Lesson objectives:
Subject: learn to solve systems of equations with two variables using the addition method.
Metasubject: Cognitive UUD: analyze (highlight the main thing), define concepts, generalize, draw conclusions. Regulatory UUD: determine the goal, problem in educational activities. Communicative UUD: express your opinion, arguing it. Personal UUD: f to form a positive motivation for learning, to create a positive emotional attitude of the student to the lesson and the subject.
Form of work: individual
Lesson steps:
1) Organizational stage.
to organize the work of the student on the topic through the creation of an attitude towards the integrity of thinking and understanding of this topic.
2. Questioning the student on the material given at home, updating knowledge.
Purpose: to check the student's knowledge obtained during homework, to identify mistakes, to work on the mistakes. Review the material from the previous lesson.
3. Learning new material.
one). to form the ability to solve systems of linear equations by adding;
2). develop and improve existing knowledge in new situations;
3). educate the skills of control and self-control, develop independence.
http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html
Purpose: preservation of vision, removal of fatigue from the eyes while working in the lesson.
5. Consolidation of the studied material
Purpose: to test the knowledge, skills and abilities acquired in the lesson
6. The result of the lesson, information about homework, reflection.
Lesson progress (working in a Google electronic document):
1. Today I wanted to start the lesson with the philosophical riddle of Walter.
What is the fastest, but also the slowest, the largest, but also the smallest, the longest and shortest, the most expensive, but also cheaply valued by us?
Time
Let's recall the basic concepts on the topic:
We have a system of two equations.
Let's remember how we solved the systems of equations in the last lesson.
Substitution method
Once again pay attention to the solved system and tell me why we cannot solve each equation of the system without resorting to the substitution method?
Because these are the equations of a system with two variables. We can solve an equation with only one variable.
Only by obtaining an equation with one variable did we manage to solve the system of equations.
3. We proceed to solve the following system:
We choose an equation in which it is convenient to express one variable in terms of another.
There is no such equation.
Those. in this situation, the previously studied method does not suit us. What is the way out of this situation?
Find a new method.
Let's try to formulate the purpose of the lesson.
Learn to solve systems in a new way.
What do we need to do to learn how to solve systems with a new method?
know the rules (algorithm) for solving a system of equations, perform practical tasks
Let's start deriving a new method.
Pay attention to the conclusion we made after solving the first system. We managed to solve the system only after we got a linear equation with one variable.
Look at the system of equations and think about how to get one equation with one variable from the two given equations.
Add equations.
What does it mean to add equations?
Separately, make the sum of the left parts, the sum of the right parts of the equations and equate the resulting sums.
Let's try. We work with me.
13x+14x+17y-17y=43+11
We got a linear equation with one variable.
Have you solved the system of equations?
The solution of the system is a pair of numbers.
How to find u?
Substitute the found value of x into the equation of the system.
Does it matter what equation we put the value of x in?
So the found value of x can be substituted into ...
any equation of the system.
We got acquainted with a new method - the method of algebraic addition.
When solving the system, we discussed the algorithm for solving the system by this method.
We have reviewed the algorithm. Now let's apply it to problem solving.
The ability to solve systems of equations can be useful in practice.
Consider the problem:
The farm has chickens and sheep. How many of those and others if they have 19 heads and 46 legs together?
Knowing that there are 19 chickens and sheep in total, we compose the first equation: x + y \u003d 19
4x is the number of sheep's legs
2y - the number of legs in chickens
Knowing that there are only 46 legs, we compose the second equation: 4x + 2y \u003d 46
Let's make a system of equations:
Let's solve the system of equations using the algorithm for solving by the addition method.
Problem! The coefficients in front of x and y are neither equal nor opposite! What to do?
Let's look at another example!
Let's add one more step to our algorithm and put it in first place: If the coefficients in front of the variables are not the same and not opposite, then we need to equalize the modules for some variable! And then we will act according to the algorithm.
4. Electronic physical education for the eyes: http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html
5. We solve the problem by the method of algebraic addition, fixing the new material and find out how many chickens and sheep were on the farm.
Additional tasks:
6. 
Reflection.
I give grades for my work in class...
6. Used resources-Internet:
Google services for education
Mathematics teacher Sokolova N. N.
With this mathematical program, you can solve a system of two linear equations with two variables using the substitution method and the addition method.
The program not only gives the answer to the problem, but also provides a detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.
This program can be useful for high school students in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it's too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
Rules for Entering Equations
Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q \) etc.
When entering equations you can use brackets. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of the elements.
For example: 6x+1 = 5(x+y)+2
In equations, you can use not only integers, but also fractional numbers in the form of decimal and ordinary fractions.
Rules for entering decimal fractions.
The integer and fractional parts in decimal fractions can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55
Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The integer part is separated from the fraction by an ampersand: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve a system of equations
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A bit of theory.
Solving systems of linear equations. Substitution method
The sequence of actions when solving a system of linear equations by the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression in another equation of the system instead of this variable;
$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$
Let's express from the first equation y through x: y = 7-3x. Substituting the expression 7-3x instead of y into the second equation, we get the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$
It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$
Substituting the number 1 instead of x into the equation y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$
Pair (1;4) - solution of the system
Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.
Solving systems of linear equations by adding
Consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by the substitution method, we pass from a given system to another system equivalent to it, in which one of the equations contains only one variable.
The sequence of actions when solving a system of linear equations by the addition method:
1) multiply the equations of the system term by term, choosing factors so that the coefficients for one of the variables become opposite numbers;
2) add term by term the left and right parts of the equations of the system;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.
Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$
In the equations of this system, the coefficients of y are opposite numbers. Adding term by term the left and right parts of the equations, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$
From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38 \) we get an equation with the variable y: \(11-3y=38 \). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)
Thus, we found the solution to the system of equations by adding: \(x=11; y=-9 \) or \((11; -9) \)
Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both parts of each of the equations of the original symmeme), in which one of the equations contains only one variable.
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