Usually the equations of the system are written in a column one below the other and combined with a curly brace
A system of equations of this type, where a, b, c- numbers, and x, y- variables are called system of linear equations.
When solving a system of equations, properties that are valid for solving equations are used.
Solving a system of linear equations using the substitution method
Let's look at an example 
1) Express the variable in one of the equations. For example, let's express y in the first equation, we get the system:
2) Substitute into the second equation of the system instead of y expression 3x-7:
3) Solve the resulting second equation:
4) We substitute the resulting solution into the first equation of the system:
A 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 addition
Let's solve the system of equations from the previous example addition method.
1) Transform the system so that the coefficients for one of the variables become opposite. Let's multiply the first equation of the system by "3".
2) Add the equations of the system term by term. We rewrite the second equation of the system (any) without changes.
3) We substitute the resulting solution 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 comes down to finding the coordinates of the common points of the graphs of the equations.
The graph of a linear function is a straight line. Two lines on a plane can intersect at one point, be parallel, or coincide. Accordingly, a system of equations can: a) have a unique solution; b) have no solutions; c) have an infinite number of solutions.
2) The solution to 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
Changing variables can lead to solving a simpler system of equations than the original one.
Consider the solution of the system
Let's introduce the replacement , then
Let's move on to the initial variables
Special cases
Without solving a system of linear equations, one can determine the number of its solutions from the coefficients of the corresponding variables.
In this case, it is convenient to express x in terms of y from the second equation of the system and substitute the resulting expression instead of x in the first equation:
The first equation is an equation with one variable y. Let's solve it:
5(7-3y)-2y = -16
We substitute the resulting y value into the expression for x:
Answer: (-2; 3).
In this system, it is easier to express y in terms of x from the first equation and substitute the resulting expression instead of y in the second equation:
The second equation is an equation with one variable x. Let's solve it:
3x-4(-1.5-3.5x)=23
In the expression for y, instead of x, we substitute x=1 and find y:
Answer: (1; -5).
Here it is more convenient to express y in terms of x from the second equation (since dividing by 10 is easier than dividing by 4, -9 or 3):
Let's solve the first equation:
4x-9(1.6-0.3x)= -1
4x-14.4+2.7x= -1
Substitute x=2 and find y:
Answer: (2; 1).
Before applying the substitution method, this system should be simplified. Both sides of the first equation can be multiplied by the lowest common denominator, in the second equation we open the brackets and present similar terms:
We obtained a system of linear equations with two variables. Now let's apply the substitution. It is convenient to express a through b from the second equation:
We solve the first equation of the system:
3(21.5 + 2.5b) – 7b = 63
It remains to find the value of a:
According to the formatting rules, we write the answer in parentheses separated by a semicolon in alphabetical order.
Answer: (14; -3).
When expressing one variable through another, it is sometimes more convenient to leave it with a certain coefficient.
Let us analyze two types of solutions to systems of equations:
1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.
In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.
The solution to the system is the intersection points of the function graphs.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by substitution method
Solving a system of equations using 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, which means that it is easiest to express the variable x from the second equation.
x=3+10y
2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1
3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution to the equation system 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 point where we expressed it we substitute y.
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 using the term-by-term addition (subtraction) method.
Solving a system of equations using the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. We choose a variable, let’s say we choose 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. Subtract the second from the first equation 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 into any of the equations, let’s say into 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 intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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Place of the lesson in the lesson system: third lesson on the topic “Systems of two linear equations with two variables”
Lesson type: learning new knowledge
Educational technology: developing critical thinking through reading and writing
Teaching Method: study
Lesson objectives: master another way to solve systems of linear equations with two variables - the addition method
Tasks:
- subject: formation of practical skills in solving systems of linear equations using the method of substitution;
- meta-subject: develop thinking, conscious perception of educational material;
- personal: fostering cognitive activity, a culture of communication and instilling interest in the subject.
As a result, the student:
- Knows the definition of a system of linear equations with two variables;
- Knows what it means to solve a system of linear equations in two variables;
- Able to write a system of linear equations with two variables;
- Understands how many solutions a system of linear equations with two variables can have;
- Can determine whether a system has solutions, and if so, how many;
- Knows the algorithm for solving systems of linear equations using substitution, algebraic addition, and graphical methods.
Problematic question:“How to solve a system of linear equations in two variables?”
Key questions: How and why do we use equations in life?
Equipment: presentation; multimedia projector; screen; computer, algebra workbook: 7th grade: to the textbook by A.G. Mordkovich et al. “Algebra – 7” 2012
Resources (where information on the topic comes from: books, textbooks, Internet, etc.): textbook “Algebra – 7” 2012, A.G. Mordkovich
Forms of organizing students' educational activities (group, pair-group, frontal, etc.): individual, partly frontal, partly steam
Evaluation criteria:
- A – knowledge and understanding +
- B – application and reasoning
- C – message +
- D – reflection and evaluation
Areas of interaction:
- ATL - Be able to use time effectively, plan your activities in accordance with your goals and objectives, and determine the most rational sequence of activities. Ability to answer questions, give reasons, argue. Be able to analyze and evaluate your own educational and cognitive activities, find ways to solve problems.
- HI students explore the consequences of human activities
During the classes
I. Lesson organization
II. Self-preparation check
a) No. 12.2(b, c).
Answer:(5; 3). Answer:(2; 3).
Answer: (4;2)
Express one variable in terms of another:
- p = p /(g * h) – liquid density
- p = g * p * h - liquid pressure at the bottom of the vessel
- h = p /(g * p) – height
- p = m / V - density
- m = V * p -mass
- p = m / V – density
Algorithm for solving a system of two equations with two variables using the substitution method:
- Express y in terms of x from the first (or second) equation of the system.
- Substitute the expression obtained in the first step instead of y into the second (first) equation of the system.
- Solve the equation obtained in the second step for x.
- Substitute the value x found in the third step into the expression y in terms of x obtained in the first step.
- Write the answer as a pair of values (x; y), which were found in the third and fourth steps, respectively.
Independent work:
In the workbook pp. 46 – 47.
- to “3” No. 6(a);
- to “4” No. 6(b);
- to “5” No. 7.
III. Updating of reference knowledge
What is a system of linear equations in two variables?
A system of equations is two or more equations for which it is necessary to find all their common solutions.
What is the solution to a system of equations in two variables?
A solution to a system of two equations with 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.
How many solutions can a system of linear equations in two variables have?
If the slopes are equal, then the lines are parallel and there are no roots.
If the angular coefficients are not equal, then the lines intersect, one root (coordinates of the intersection point).
If the slopes are equal, then the lines coincide and the root is infinitely large.
IV. Learning new material
Fill in the blanks: Appendix 1 (followed by self-test on the slides)
V. Work on the topic of the lesson
In class: No. 13.2(a, d), 13.3(a, d).
VI. Homework
Paragraph 13 - textbook; dictionary; No. 13.2(b, c), 13.3(b, c).
VII. Lesson summary
- Hooray!!! I understand everything!
- There are some things I need to work on!
- There were failures, but I will overcome everything!
VIII. Solving problems with a military component
Main battle tank T-80.
Adopted into service in 1976. The world's first production tank with a main power plant based on a gas turbine engine.
Basic tactical and technical data (TTD):
Weight, t – 46
Speed, km/h – 70
Cruising range, km – 335-370
Armament: 125 mm smoothbore gun (40 pieces of ammunition);
12.7 mm machine gun (300 pieces of ammunition);
7.62-mm PKT machine gun (ammunition 2000 pcs.)
How long can a T-80 tank remain in motion without refueling?
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. The substitution method allows you to easily solve systems of linear equations of any complexity. The essence of the method is that, using the first expression of the system, we express “y”, and then we substitute the resulting expression into the second equation of the system instead of “y”. Since the equation already contains not two unknowns, but only one, we can easily find the value of this variable, and then use it to determine the value of the second.
Suppose we are given a system of linear equations of the following form:
\[\left\(\begin(matrix) 3x-y-10=0\\ x+4y-12=0 \end(matrix)\right.\]
Let's express \
\[\left\(\begin(matrix) 3x-10=y\\ x+4y-12=0 \end(matrix)\right.\]
Let us substitute the resulting expression into equation 2:
\[\left\(\begin(matrix) y=3x-10\\ x+4(3x-10)-12=0 \end(matrix)\right.\]
Let's find the value \
Let's simplify and solve the equation by opening brackets and taking into account the rules for transferring terms:
Now we know the value \ Let's use this to find the value \
Answer: \[(4;2).\]
Where can I solve a system of equations online using the substitution method?
You can solve the system of equations on our website. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also find out how to solve the equation on our website. And if you still have questions, you can ask them in our VKontakte group.
