Method of algebraic addition algorithm. Examples of systems of linear equations: solution method
Very often, students find it difficult to choose a way to solve systems of equations.
In this article we will look at one of the ways to solve systems - the substitution method.
If a common solution to two equations is found, then these equations are said to form a system. In a system of equations, each unknown represents the same number in all equations. To show that given equations form a system, they are usually written one below the other and joined by a curly brace, for example
We note that for x = 15 and y = 5, both equations of the system are correct. This pair of numbers is the solution to the system of equations. Each pair of unknown values that simultaneously satisfies both equations of the system is called a solution to the system.
A system can have one solution (as in our example), infinitely many solutions, or no solutions.
How to solve systems using the substitution method? If the coefficients for some unknown in both equations are equal in absolute value(if they are not equal, then we equalize), then by adding both equations (or subtracting one from the other), you can get an equation with one unknown. Then we solve this equation. We determine one unknown. We substitute the resulting value of the unknown into one of the system equations (the first or the second). We find another unknown. Let's look at examples of the application of this method.
Example 1. Solve the system of equations
Here the coefficients for y are equal in absolute value, but opposite in sign. Let's try to add the equations of the system term by term.
We substitute the resulting value x = 4 into some equation of the system (for example, into the first one) and find the value y:
2 *4 +y = 11, y = 11 – 8, y = 3.
Our system has a solution x = 4, y = 3. Or the answer can be written in parentheses as the coordinates of a point, x in the first place, y in the second.
Answer: (4; 3)
Example 2. Solve system of equations
Let's equalize the coefficients for the variable x, to do this we multiply the first equation by 3, and the second by (-2), we get
Be careful when adding equations
Then y = - 2. Substitute the number (-2) instead of y into the first equation, and we get
4x + 3(-2) = - 4. Solve this equation 4x = - 4 + 6, 4x = 2, x = ½.
Answer: (1/2; - 2)
Example 3. Solve the system of equations
Multiply the first equation by (-2)
Solving the system
we get 0 = - 13.
The system has no solutions, since 0 is not equal to (-13).
Answer: there are no solutions.
Example 4. Solve the system of equations
We notice that all coefficients of the second equation are divisible by 3,
let's divide the second equation by three and we get a system that consists of two identical equations.
This system has infinitely many solutions, since the first and second equations are the same (we got only one equation with two variables). How can we imagine the solution to this system? Let's express the variable y from the equation x + y = 5. We get y = 5 – x.
Then answer will be written like this: (x; 5-x), x – any number.
We looked at solving systems of equations using the addition method. If you have any questions or something is not clear, sign up for a lesson and we will solve all the problems with you.
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Using the addition method, the equations of a system are added term by term, and 1 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 the right number, you can multiply the coefficients: .
Next step:
We multiply the 1st equation by ,
We multiply the 3rd equation by ,
System linear equations with two unknowns are two or more linear equations for which it is necessary to find all their common solutions. We will consider systems of two linear equations in two unknowns. General form 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 method
An algorithm for solving a system of linear equations with two unknowns using the addition method.
1. If required, by means of equivalent transformations, 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.
OGBOU "Education Center for Children with Special Educational Needs in Smolensk"
Center distance education
Algebra lesson in 7th grade
Lesson topic: Method algebraic addition.
- 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.
http://zhakulina20090612.blogspot.ru/2011/06/blog-post_25.html
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 training 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.