An example of solving a system of equations by the substitution method. System of equations
system linear equations with two unknowns - these are 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 with two unknowns. General form a system of two linear equations with two unknowns is shown 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 with two unknowns is a pair of numbers (x, y) such that if these numbers are substituted into the equations of the system, then each of the equations of the system turns into a true equality. Consider one of the ways to solve a system of linear equations, namely the substitution method.
Algorithm for solving by substitution method
Algorithm for solving a system of linear equations by substitution method:
1. Choose one equation (it is better to choose the one where the numbers are smaller) and express one variable from it through another, for example, x through y. (you can also y through x).
2. Substitute the resulting expression instead of the corresponding variable in another equation. Thus, we get a linear equation with one unknown.
3. We solve the resulting linear equation and get the solution.
4. We substitute the obtained solution into the expression obtained in the first paragraph, we obtain the second unknown from the solution.
5. Verify the resulting solution.
Example
To make it more clear, let's solve a small example.
Example 1 Solve the system of equations:
(x+2*y=12
(2*x-3*y=-18
Solution:
1. From the first equation of this system, we express the variable x. We have x= (12 -2*y);
2. Substitute this expression into the second equation, we get 2*x-3*y=-18; 2*(12 -2*y) - 3*y = -18; 24 - 4y - 3*y = -18;
3. We solve the resulting linear equation: 24 - 4y - 3*y = -18; 24-7*y=-18; -7*y = -42; y=6;
4. We substitute the result obtained into the expression obtained in the first paragraph. x= (12 -2*y); x=12-2*6 = 0; x=0;
5. We check the obtained solution, for this we substitute the numbers found in the original system.
(x+2*y=12;
(2*x-3*y=-18;
{0+2*6 =12;
{2*0-3*6=-18;
{12 =12;
{-18=-18;
Got true equalities, therefore, we correctly found the solution.
Let's figure it out how to solve systems of equations by substitution?
1) Express the unknown from the first or second equation of the system X or at(as we prefer);
2) Substitute in another equation (into the one from which the unknown was not expressed) instead of the unknown X or at(if expressed X, substitute instead X; if expressed at, substitute instead at) the resulting expression;
3) We solve the equation that we received. We find X or y;
4) We substitute the obtained value of the unknown and find the second unknown.
The rule is written down. Now let's try to apply it when solving a system of equations.
Example 1.
Let's take a closer look at the system of equations. We notice that from the first equation it is easier to express at.
We express at:
-2y \u003d 11 - 3x
y \u003d (11 - 3x) / (-2)
y \u003d -5.5 + 1.5x
Now carefully substitute into the second equation instead of at expression -5.5 + 1.5x.
We get: 4x - 5 (-5.5 + 1.5x) \u003d 3
We solve this equation:
4x + 27.5 - 7.5x \u003d 3
-3.5x \u003d 3 - 27.5
-3.5x = -24.5
x \u003d -24.5 / (-3.5)
We substitute into the expression y \u003d - 5.5 + 1.5x instead of X the value we found. We get:
y \u003d - 5.5 + 1.5 7 \u003d -5.5 + 10.5 \u003d 5.
Answer: (7; 5)
Interestingly, if we express from the first equation not at, a X, will the answer change?
Let's try to express X from the first equation.
x = (11 + 2y)/3
Substitute instead of X into the second equation, the expression (11 + 2y) / 3, we get an equation with one unknown and solve it.
4(11 + 2y)/3 - 5y = 3, multiply both sides of the equation by 3, we get
4(11 + 2y) - 15y=9
44 + 8y - 15y \u003d 9
–7y = 9 – 44
y = -35/(-7)
We find the variable x by substituting 5 into the expression x = (11 + 2y)/3.
x \u003d (11 + 2 5) / 3 \u003d (11 + 10) / 3 \u003d 21/3 \u003d 7
Answer: (7; 5)
As you can see the answer is the same. If you are careful and careful, then no matter what variable you express - X or at you will get the correct answer.
Quite often students ask: Are there other ways to solve systems besides addition and substitution?»
There is some modification of the substitution method - way to compare unknowns .
1) It is necessary to express the same unknown from each equation of the system through the second.
2) The obtained unknowns are compared, an equation with one unknown is obtained.
3) Find the value of one unknown.
4) Substitute the obtained value of the unknown and find the second unknown.
Example 2. Solve a system of equations
From the two equations, we express the variable X through at.
We obtain from the first equation x = (13 - 6y) / 5, and from the second x = (-1 - 18y) / 7.
Comparing these expressions, we obtain an equation with one unknown and solve it:
(13 - 6y) / 5 = (-1 - 18y) / 7
7 (13 - 6y) \u003d 5 (-1 - 18y)
91 - 42y \u003d -5 - 90y
-42y + 90y \u003d -5 - 91
y \u003d - 96 / 48
Unknown X find by substituting the value at into one of the expressions for X.
(13 – 6(– 2)) / 5= (13+12) / 5 = 25/5 = 5
Answer: (5; -2).
I think that you will succeed too. If you have any questions, please come to my class.
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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 method2x+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 method3x-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)
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1 . FULL NAME. teachers: ____Tkachuk Natalya Petrovna _________________________________________________________________________________________________
2. Class: _8 Date: .11.03________ Subject_-mathematics, No. 71 of the lesson according to the schedule:
3. Lesson topic Solving systems by the substitution method 4 . The place and role of the lesson in the topic under study :. Knowledge consolidation lesson. The purpose of the lesson :
Educational: to develop knowledge of solving systems of equations by substitution. Know/Understand: if the graphs have common points, then the system has solutions; if the graphs do not have common points, then the system of solutions does not; algorithm for solving systems of equations.Be able to solve systems by substitution method To promote the development of skills to apply the acquired knowledge in non-standard (typical) conditionsDeveloping: To promote the development of students' skills to summarize the knowledge gained, to analyze, synthesize, compare, draw the necessary conclusions. To promote the development of skills to apply the acquired knowledge in non-standard and typical conditions.Educational: To promote the development of a creative attitude to learning activities
Characteristics of the stages of the lesson
Activitystudents
Self-determination.
Activate cognitive activity
Solve the system
verbal
Frontal
Greeting students. holding. Creating a situation of readiness for the lesson, success in the upcoming lesson.
Check readiness for the lesson.
2.Updating knowledge.
To identify the quality and level of mastery of knowledge and skills acquired at previous lessons on this topic
Find out if a pair of numbers is a solution to the system. x=5 y=9
What operations can be performed with equations?
(multiply both sides of the equation by the same number, divide by a number not equal to zero ....)
Group work
Frontal. Group-analysis of algorithms for solving problems;
Asks leading questions if necessary.
They answer the questions asked.
3. Statement of the educational task, the objectives of the lesson.
Formation
and skill development
define and formulate
problem, goal and topic
to study lines
How the system of equations is solved by the method of addition, by the method of substitution.
What is the best way to use the solution. this system?
Group work.
Individual.
Frontal.
What steps did we take to find out the cost of the purchase?
What topic will we study?
They speak out.
4. The stage of updating knowledge on the topic
To promote the development of skills to distinguish and compare lines. Provide conditions for the development of skills to competently, clearly and accurately express their thoughts.
№ 621
To find out mutual arrangement direct
2x+0.5y=1.2 and x-4y=0
Is it possible to determine if the lines intersect or not by their coefficients?
2. Make equations of lines that are parallel to each other.
Working with the textbook
Work in pairs with self-examination
Frontal, individual. problem solving workshop
Asks leading questions if necessary. Draws a parallel with previously studied material.
Provides motivation to complete the proposed tasks.
Leads students to the conclusion about the existence of formulas.
Solve problems, answer the teacher's questions if necessary. Perform the exercise in the notebook.
They take turns commenting, analyzing, determining the causes and solutions.
5.Work by yourself
application of acquired knowledge. Updating knowledge and skills in problem solving.
Formation and development of skills in reading numbers. Planning your activities to solve the problem, monitoring the result, correcting the result, self-regulation
1 var -
2 var
Independent work. Neighbor check.
« brainstorm»,
Controls the execution of work.
Carries out: individual control; selective control.
Encourages you to express your opinion.
They solve problems. Carry out: self-assessment; mutual check; give a preliminary assessment.
6. Lesson assessment, self-assessment.
Formation and development of the ability to analyze and comprehend their achievements.
The ability to determine the level of mastery of educational material.
Evaluation of intermediate results and self-regulation to increase the motivation of learning activities
Evaluation at every stage
1. Can you plot linear equations?
2. Whether you know how to determine whether they intersect or not.
3. Do you know the algorithm for solving systems of equations?
4. What methods do you know for solving systems of equations?
Group work.
Group and individual..
Encourages you to express your opinion.
Carry out: self-assessment and assessment of a friend.
7. The results of the lesson. Homework.
The ability to correlate the goals and results of one's own activity. Maintaining a healthy spirit of competition to maintain motivation for learning activities; participation in a group discussion of problems.
p. 4.4 No. 623
Group work.
Frontal-Isolation and formulation of a cognitive goal - reflection of the methods and conditions of action
Analysis and synthesis of objects
Encourages you to express your opinion.
Gives a comment on homework; task to search for features in the text ...
Children participate in the discussion, analyze, speak. Reflect and record their achievements.
Today in class I learned...
Today in class I learned...
Usually, the equations of the system are written in a column one below the other and combined with a curly bracket
A system of equations of this kind, where a, b, c- numbers, and x, y- variables, called system of linear equations.
When solving a system of equations, properties are used that are valid for solving equations.
Solving a system of linear equations by the substitution method
Consider an example
1) Express a variable in one of the equations. For example, let's express y in the first equation, we get the system:
2) We substitute into the second equation of the system instead of y expression 3x-7:
3) We solve the resulting second equation:
4) The resulting solution is substituted into the first equation of the system:
The system of equations has a unique solution: a pair of numbers x=1, y=-4. Answer: (1; -4) , written in brackets, in the first position the value x, On the second - y.
Solving a system of linear equations by adding
Let's solve the system of equations from the previous example addition method.
1) Transform the system in such a way that the coefficients for one of the variables become opposite. Multiply the first equation of the system by "3".
2) We add term by term the equations of the system. The second equation of the system (any) is rewritten without changes.
3) The resulting solution is substituted into the first equation of the system:
Solving a system of linear equations graphically
The graphical solution of a system of equations with two variables is reduced to finding the coordinates of the common points of the equation graphs.
The graph of a linear function is a straight line. Two straight lines in a plane can intersect at one point, be parallel or coincide. Accordingly, the system of equations can: a) have a unique solution; b) have no decisions; c) have an infinite number of solutions.
2) The solution of the system of equations is the point (if the equations are linear) of the intersection of the graphs.
Graphic solution of the system
Method for introducing new variables
The change of variables can lead to the solution of a simpler system of equations than the original one.
Consider the solution of the system
We introduce a replacement, then
Going to the original variables
Special cases
Without solving the system of linear equations, one can determine the number of its solutions by the coefficients of the corresponding variables.