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What is the method of algebraic addition. Method of addition in solving systems of equations

Method algebraic addition

You can solve a system of equations with two unknowns in various ways- graphical method or variable replacement method.

In this lesson we will get acquainted with another method of solving systems that you will probably like - this is the method of algebraic addition.

Where did the idea of ​​putting something in systems come from? When solving systems main problem is the presence of two variables, because we do not know how to solve equations with two variables. This means that one of them must be excluded in some legal way. And such legal ways are mathematical rules and properties.

One of these properties is: the sum of opposite numbers is zero. This means that if one of the variables has opposite coefficients, 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 terms with the variable we need. You need to add the entire equations, i.e. separately add similar terms on the left side, then on the right. As a result, we get a new equation containing only one variable. Let's look at what has been said with specific examples.

We see that in the first equation there is a variable y, and in the second there is the opposite number -y. This means that this equation can be solved by addition.

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. We add 3x with 2x, we add y with -y, we add 8 with 7.

We obtain a system of equations

The second equation of this system is a simple equation with one variable. From it we find x = 3. Substituting the found value into the first equation, we find y = -1.

Answer: (3; - 1).

Sample design:

Solve a system of equations using the algebraic addition method

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 the variable x has opposite coefficients. This means that we will do the same as in the first example: we will leave one of the equations unchanged. For example, 2y + 2x = 10. And we get the second by addition.

Now we have a system of equations:

We easily find from the second equation y = 1, and then from the first equation x = 4.

Sample design:

Let's summarize:

We learned 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: graphical, variable replacement method and 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:

  1. Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for educational institutions/ A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007.
  2. Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007.
  3. HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, Mnemosyne, 2008.
  4. Alexandrova L.A., Algebra 7th grade. Thematic testing work V new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011.
  5. Alexandrova L.A. Algebra 7th grade. Independent work for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010.

OGBOU "Education Center for Children with Special Educational Needs in Smolensk"

Center distance education

Algebra lesson in 7th grade

Lesson topic: Method of algebraic addition.

      1. Lesson type: Lesson of initial presentation of new knowledge.

Purpose of the lesson: control the level of acquisition of knowledge and skills in solving systems of equations using the method of substitution; developing skills and abilities to solve systems of equations using 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, giving reasons for it. Personal UUD: f to form a positive motivation for learning, to create a positive emotional attitude of the student towards the lesson and the subject.

Form of work: individual

Lesson steps:

1) Organizational stage.

organize the student’s work on the topic through creating an attitude towards integrity of thinking and understanding of this topic.

2. Questioning the student on the material assigned for homework, updating knowledge.

Purpose: to test the student’s knowledge acquired during the implementation homework, identify errors, do work on errors. Review the material from the previous lesson.

3. Studying new material.

1). develop the ability to solve systems of linear equations using the addition method;

2). develop and improve existing knowledge in new situations;

3). cultivate control and self-control skills, develop independence.

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Goal: preserve vision, relieve eye fatigue while working in class.

5. Consolidation of the studied material

Purpose: to test the knowledge, skills and abilities acquired in the lesson

6. Lesson summary, information about homework, reflection.

Lesson progress (working in an electronic Google document):

1. Today I wanted to start the lesson with Walter’s philosophical riddle.

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 remember the basic concepts on the topic:

Before us is a system of two equations.

Let's remember how we solved 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 equations of a system with two variables. We can solve equations with only one variable.

Only by obtaining an equation with one variable were we able to solve the system of equations.

3. We proceed to solve the following system:

Let's choose an equation in which it is convenient to express one variable through another.

There is no such equation.

Those. In this situation, the previously studied method is not suitable for 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 using a new method.

What do we need to do to learn how to solve systems using a new method?

know the rules (algorithm) for solving a system of equations, complete practical tasks

Let's start developing a new method.

Pay attention to the conclusion we made after solving the first system. It was possible to solve the system only after we obtained a linear equation with one variable.

Look at the system of equations and think about how to get one equation with one variable from two given equations.

Add up the equations.

What does it mean to add equations?

Separately compose the sum of the left sides, the sum of the right sides of the equations and equate the resulting sums.

Let's try. We work together with me.

13x+14x+17y-17y=43+11

We have obtained a linear equation with one variable.

Have you solved the system of equations?

The solution to the system is a pair of numbers.

How to find y?

Substitute the found value of x into the system equation.

Does it matter which equation we substitute the value of x into?

This means that 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.

While solving the system, we discussed the algorithm for solving the system using 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.

Let's consider the problem:

The farm has chickens and sheep. How many of both are there if they together have 19 heads and 46 legs?

Knowing that there are 19 chickens and sheep in total, let’s create the first equation: x + y = 19

4x - the number of legs of sheep

2у - number of legs in chickens

Knowing that there are only 46 legs, let’s create the second equation: 4x + 2y = 46

Let's create a system of equations:

Let's solve the system of equations using the solution algorithm using the addition method.

Problem! The coefficients in front of x and y are not equal and not 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 complete the problem using the algebraic addition method, fixing new material and find out how many chickens and sheep there were on the farm.

Additional tasks:

6.

Reflection.

I give a grade for my work in class -...

6. Internet resources used:

Google services for education

Mathematics teacher Sokolova N. N.

A system of linear equations with two unknowns is two or more linear equations for which it is necessary to find all of them general solutions. We will consider systems of two linear equations in two unknowns. General view a system of two linear equations with two unknowns is presented in the figure below:

( a1*x + b1*y = c1,
( a2*x + b2*y = c2

Here x and y are unknown variables, a1,a2,b1,b2,c1,c2 are some real numbers. A solution to a system of two linear equations in two unknowns is a pair of numbers (x,y) such that if we substitute these numbers into the equations of the system, then each of the equations of the system turns into a true equality. There are several ways to solve a system of linear equations. Let's consider one of the ways to solve a system of linear equations, namely the addition method.

Algorithm for solving by addition

An algorithm for solving a system of linear equations with two unknowns using the addition method.

1. If required, use equivalent transformations to equalize the coefficients of one of the unknown variables in both equations.

2. By adding or subtracting the resulting equations, obtain a linear equation with one unknown

3. Solve the resulting equation with one unknown and find one of the variables.

4. Substitute the resulting expression into any of the two equations of the system and solve this equation, thus obtaining the second variable.

5. Check the solution.

An example of a solution using the addition method

For greater clarity, let us solve the following system of linear equations with two unknowns using the addition method:

(3*x + 2*y = 10;
(5*x + 3*y = 12;

Since none of the variables have identical coefficients, we equalize the coefficients of the variable y. To do this, multiply the first equation by three, and the second equation by two.

(3*x+2*y=10 |*3
(5*x + 3*y = 12 |*2

We get the following system of equations:

(9*x+6*y = 30;
(10*x+6*y=24;

Now we subtract the first from the second equation. We present similar terms and solve the resulting linear equation.

10*x+6*y - (9*x+6*y) = 24-30; x=-6;

We substitute the resulting value into the first equation from our original system and solve the resulting equation.

(3*(-6) + 2*y =10;
(2*y=28; y =14;

The result is a pair of numbers x=6 and y=14. We are checking. Let's make a substitution.

(3*x + 2*y = 10;
(5*x + 3*y = 12;

{3*(-6) + 2*(14) = 10;
{5*(-6) + 3*(14) = 12;

{10 = 10;
{12=12;

As you can see, we got two correct equalities, therefore, we found the correct solution.

In this lesson we will continue to study the method of solving systems of equations, namely the method of algebraic addition. First, let's look at the application of this method using the example of linear equations and its essence. Let's also remember how to equalize coefficients in equations. And we will solve a number of problems using this method.

Topic: Systems of equations

Lesson: Algebraic addition method

1. Method of algebraic addition using linear systems as an example

Let's consider algebraic addition method using the example of linear systems.

Example 1. Solve the system

If we add these two equations, then y cancels out, leaving an equation for x.

If we subtract the second from the first equation, the x's cancel each other out, and we get an equation for y. This is the meaning of the algebraic addition method.

We solved the system and remembered the method of algebraic addition. Let's repeat its essence: we can add and subtract equations, but we must ensure that we get an equation with only one unknown.

2. Method of algebraic addition with preliminary equalization of coefficients

Example 2. Solve the system

The term is present in both equations, so the algebraic addition method is convenient. Let's subtract the second from the first equation.

Answer: (2; -1).

Thus, after analyzing the system of equations, you can see that it is convenient for the method of algebraic addition, and apply it.

Let's consider another linear system.

3. Solution of nonlinear systems

Example 3. Solve the system

We want to get rid of y, but the coefficients of y are different in the two equations. Let's equalize them; to do this, multiply the first equation by 3, the second by 4.

Example 4. Solve the system

Let's equalize the coefficients for x

You can do it differently - equalize the coefficients for y.

We solved the system by applying the algebraic addition method twice.

The algebraic addition method is also applicable when solving nonlinear systems.

Example 5. Solve the system

Let's add these equations together and we'll get rid of y.

The same system can be solved by applying the algebraic addition method twice. Let's add and subtract from one equation another.

Example 6. Solve the system

Answer:

Example 7. Solve the system

Using the method of algebraic addition we will get rid of the xy term. Let's multiply the first equation by .

The first equation remains unchanged, instead of the second we write the algebraic sum.

Answer:

Example 8. Solve the system

Multiply the second equation by 2 to isolate a perfect square.

Our task was reduced to solving four simple systems.

4. Conclusion

We examined the method of algebraic addition using the example of solving linear and nonlinear systems. In the next lesson we will look at the method of introducing new variables.

1. Mordkovich A.G. et al. Algebra 9th grade: Textbook. For general education Institutions.- 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.

2. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.

3. Makarychev Yu. N. Algebra. 9th grade: educational. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. — 7th ed., rev. and additional - M.: Mnemosyne, 2008.

4. Alimov Sh. A., Kolyagin Yu. M., Sidorov Yu. V. Algebra. 9th grade. 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.

6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. — 12th ed., rev. - M.: 2010.-223 p.: ill.

1. College section. ru in mathematics.

2. Internet project “Tasks”.

3. Educational portal“I WILL SOLVE THE USE.”

1. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 125 - 127.

Need to download lesson plan on topic » Algebraic addition method?

Using the addition method, the equations of a system are added term by term, and one or both (several) equations can be multiplied by any number. As a result, they come to an equivalent SLE, where in one of the equations there is only one variable.

To solve the system method of term-by-term addition (subtraction) follow these steps:

1. 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- these are the intersection points of the function graphs.

Let's look at examples.

Example 1.

Given system:

Having analyzed this system, you can notice that the coefficients of the variable are equal in magnitude and different in sign (-1 and 1). In this case, the equations can be easily added term by term:

We perform the actions circled in red in our minds.

The result of term-by-term addition was the disappearance of the variable y. This is precisely the meaning of the method - to get rid of one of the variables.

-4 - y + 5 = 0 → y = 1,

In system form, the solution looks something 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 disadvantage - when you express any variable from any equation, you will get a solution in ordinary fractions. But solving fractions takes a lot of time and the likelihood 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:

You need to find a number that can be divided by 3 and on 4 , and it is necessary that this number be the minimum possible. This least common multiple. If it’s hard for you to find a suitable number, you can multiply the coefficients: .

Next step:

We multiply the 1st equation by ,

We multiply the 3rd equation by ,