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 the 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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We will analyze two types of solving systems of equations:

1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.

In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.

To solve system by term-by-term addition (subtraction) need:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.

The solution of the system is the intersection points of the graphs of the function.

Let us consider in detail the solution of systems using examples.

Example #1:

Let's solve by the substitution method

Solving the system of equations by the substitution method

2x+5y=1 (1 equation)
x-10y=3 (2nd equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y

2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1

3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2

The solution of the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1

It is customary to write points in the first place, we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)

Example #2:

Let's solve by term-by-term addition (subtraction).

Solving a system of equations by the addition method

3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)

1. Select a variable, let's say we select x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.

3x-2y=1 |*2
6x-4y=2

2x-3y=-10 |*3
6x-9y=-30

2. From the first equation, subtract the second to get rid of the variable x. Solve the linear equation.
__6x-4y=2

5y=32 | :5
y=6.4

3. Find x. We substitute the found y in any of the equations, let's say in the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6

The point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)

Do you want to prepare for exams for free? Tutor online for free. No kidding.

In this lesson, we will continue to study the method of solving systems of equations, namely: the method of algebraic addition. First, consider the application of this method on 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 on the application of this method.

Topic: Systems of Equations

Lesson: Algebraic addition method

1. Method of algebraic addition on the example of linear systems

Consider algebraic addition method on the example of linear systems.

Example 1. Solve the system

If we add these two equations, then the y's will cancel each other out, leaving the equation for x.

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

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

2. Algebraic addition method with preliminary adjustment of coefficients

Example 2. Solve the system

The term is present in both equations, so the algebraic addition method is convenient. Subtract the second from the first equation.

Answer: (2; -1).

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

Consider another linear system.

3. Solution of nonlinear systems

Example 3. Solve the system

We want to get rid of y, but the two equations have different coefficients for y. We equalize them, for this we multiply the first equation by 3, the second - by 4.

Example 4. Solve the system

Equalize the coefficients at x

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

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

The method of algebraic addition is also applicable in solving nonlinear systems.

Example 5. Solve the system

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

The same system can be solved by applying the algebraic addition method twice. 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 get rid of the term xy. Multiply the first equation by .

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

Answer:

Example 8. Solve the system

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

Our task was reduced to solving four simple systems.

4. Conclusion

We considered the method of algebraic addition using the example of solving linear and nonlinear systems. In the next lesson, we will consider the method of introducing new variables.

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

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

3. Yu. N. Makarychev, Algebra. Grade 9: textbook. 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. Sh. A. Alimov, Yu. M. Kolyagin, and Yu. V. Sidorov, Algebra. Grade 9 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. Textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.

6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational 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 "SOLVE USE".

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

You need to download the lesson plan on the topic » Algebraic addition method?

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.

1. 1. 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 up 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.