More reliable than the graphical method discussed in the previous paragraph.
Substitution method
We used this method in 7th grade to solve systems linear equations. The algorithm that was developed in the 7th grade is quite suitable for solving systems of any two equations (not necessarily linear) with two variables x and y (of course, the variables can be designated by other letters, which does not matter). In fact, we used this algorithm in the previous paragraph, when the problem of double digit number led to mathematical model, which is a system of equations. We solved this system of equations above using the substitution method (see example 1 from § 4).
An algorithm for using the substitution method when solving a system of two equations with two variables x, y.
1. Express y in terms of x from one equation of the system.
2. Substitute the resulting expression instead of y into another equation of the system.
3. Solve the resulting equation for x.
4. Substitute in turn each of the roots of the equation found in the third step instead of x into the expression y through x obtained in the first step.
5. Write the answer in the form of pairs of values (x; y), which were found in the third and fourth steps, respectively.
4) Substitute one by one each of the found values of y into the formula x = 5 - 3. If then 
5) Pairs (2; 1) and solutions to a given system of equations.
Answer: (2; 1);
Algebraic addition method
This method, like the substitution method, is familiar to you from the 7th grade algebra course, where it was used to solve systems of linear equations. Let us recall the essence of the method using the following example.
Example 2. Solve system of equations
Let's multiply all terms of the first equation of the system by 3, and leave the second equation unchanged: 
Subtract the second equation of the system from its first equation:
As a result of the algebraic addition of two equations of the original system, an equation was obtained that was simpler than the first and second equations of the given system. With this simpler equation we have the right to replace any equation of a given system, for example the second one. Then the given system of equations will be replaced by a simpler system:
This system can be solved using the substitution method. From the second equation we find. Substituting this expression instead of y into the first equation of the system, we get
It remains to substitute the found values of x into the formula
If x = 2 then
Thus, we found two solutions to the system: 
Method for introducing new variables
You were introduced to the method of introducing a new variable when solving rational equations with one variable in the 8th grade algebra course. The essence of this method for solving systems of equations is the same, but from a technical point of view there are some features that we will discuss in the following examples.
Example 3. Solve system of equations
Let's introduce a new variable. Then the first equation of the system can be rewritten into a more in simple form: Let's solve this equation for the variable t:
Both of these values satisfy the condition and therefore are roots rational equation with variable t. But that means either where we find that x = 2y, or
Thus, using the method of introducing a new variable, we managed to “stratify” the first equation of the system, which was quite complex in appearance, into two simpler equations:
x = 2 y; y - 2x.
What's next? And then each of the two received simple equations need to be considered one by one in a system with the equation x 2 - y 2 = 3, which we have not yet remembered. In other words, the problem comes down to solving two systems of equations:
We need to find solutions to the first system, the second system and include all the resulting pairs of values in the answer. Let's solve the first system of equations:
Let's use the substitution method, especially since everything is ready for it here: let's substitute the expression 2y instead of x into the second equation of the system. We get
Since x = 2y, we find, respectively, x 1 = 2, x 2 = 2. Thus, two solutions of the given system are obtained: (2; 1) and (-2; -1). Let's solve the second system of equations:
Let's use the substitution method again: substitute the expression 2x instead of y into the second equation of the system. We get
This equation has no roots, which means the system of equations has no solutions. Thus, only the solutions of the first system need to be included in the answer.
Answer: (2; 1); (-2;-1).
The method of introducing new variables when solving systems of two equations with two variables is used in two versions. First option: one new variable is introduced and used in only one equation of the system. This is exactly what happened in example 3. Second option: two new variables are introduced and used simultaneously in both equations of the system. This will be the case in example 4.
Example 4. Solve system of equations
Let's introduce two new variables:
Let's take into account that then
This will allow you to rewrite this system in a much simpler form, but relatively new variables a and b:
Since a = 1, then from the equation a + 6 = 2 we find: 1 + 6 = 2; 6=1. Thus, regarding the variables a and b, we got one solution:
Returning to the variables x and y, we obtain a system of equations
Let us apply the method to solve this system algebraic addition:
Since then from the equation 2x + y = 3 we find:
Thus, regarding the variables x and y, we got one solution:
Let us conclude this paragraph with a brief but rather serious theoretical conversation. You have already gained some experience in solving different equations: linear, square, rational, irrational. You know that the main idea of solving an equation is to gradually move from one equation to another, simpler, but equivalent to the given one. In the previous paragraph we introduced the concept of equivalence for equations with two variables. This concept is also used for systems of equations.
Definition.
Two systems of equations with variables x and y are called equivalent if they have the same solutions or if both systems have no solutions.
All three methods (substitution, algebraic addition and introducing new variables) that we discussed in this section are absolutely correct from the point of view of equivalence. In other words, using these methods, we replace one system of equations with another, simpler, but equivalent to the original system.
Graphical method for solving systems of equations
We have already learned how to solve systems of equations in such common and reliable ways as the method of substitution, algebraic addition and the introduction of new variables. Now let's remember the method that you already studied in the previous lesson. That is, let's repeat what you know about graphical method solutions.
The method of solving systems of equations graphically is the construction of a graph for each of the specific equations that are included in a given system and are in one coordinate plane, and also where it is necessary to find the intersections of the points of these graphs. To solve this system of equations are the coordinates of this point (x; y).
It should be remembered that it is typical for a graphical system of equations to have either one single the right decision, or infinite set solutions, or have no solutions at all.
Now let’s look at each of these solutions in more detail. And so, the system of equations can have only decision in case the straight lines, which are the graphs of the system’s equations, intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. If the direct graphs of the equations of the system coincide, then such a system allows one to find many solutions.
Well, now let’s look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:
Firstly, first we build a graph of the 1st equation;
The second step will be to construct a graph that relates to the second equation;
Thirdly, we need to find the intersection points of the graphs.
And as a result, we get the coordinates of each intersection point, which will be the solution to the system of equations.
Let's look at this method in more detail using an example. We are given a system of equations that needs to be solved:
Solving equations
1. First, we will build a schedule given equation: x2+y2=9.
But it should be noted that this graph of the equations will be a circle with a center at the origin, and its radius will be equal to three.
2. Our next step will be to graph an equation such as: y = x – 3.
In this case, we must construct a straight line and find the points (0;−3) and (3;0).
3. Let's see what we got. We see that the straight line intersects the circle at two of its points A and B.
Now we are looking for the coordinates of these points. We see that the coordinates (3;0) correspond to point A, and the coordinates (0;−3) correspond to point B.
And what do we get as a result?
The numbers (3;0) and (0;−3) obtained when the line intersects the circle are precisely the solutions to both equations of the system. And from this it follows that these numbers are also solutions to this system of equations.
That is, the answer to this solution is the numbers: (3;0) and (0;−3).
Let us first recall the definition of a solution to a system of equations with two variables.
Definition 1
A pair of numbers is called a solution to a system of equations in two variables if substituting them into the equation results in a true equality.
In the future we will consider systems of two equations with two variables.
Exist four basic ways to solve systems of equations: substitution method, addition method, graphic method, a way to maintain new variables. Let's look at these methods specific examples. To describe the principle of using the first three methods, we will consider a system of two linear equations with two unknowns:
Substitution method
The substitution method is as follows: take any of these equations and express $y$ in terms of $x$, then $y$ is substituted into the system equation, from where the variable $x is found.$ After this, we can easily calculate the variable $y.$
Example 1
Let us express $y$ from the second equation in terms of $x$:
Let's substitute into the first equation and find $x$:
\ \ \
Let's find $y$:
Answer: $(-2,\ 3)$
Addition method.
Let's look at this method using an example:
Example 2
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Multiplying the second equation by 3, we get:
\[\left\( \begin(array)(c) (2x+3y=5) \\ (9x-3y=-27) \end(array) \right.\]
Now let's add both equations together:
\ \ \
Let's find $y$ from the second equation:
\[-6-y=-9\] \
Answer: $(-2,\ 3)$
Note 1
Note that in this method it is necessary to multiply one or both equations by such numbers that during addition one of the variables “disappears”.
Graphic method
The graphical method is as follows: both equations of the system are depicted on the coordinate plane and the point of their intersection is found.
Example 3
\[\left\( \begin(array)(c) (2x+3y=5) \\ (3x-y=-9) \end(array) \right.\]
Let us express $y$ from both equations in terms of $x$:
\[\left\( \begin(array)(c) (y=\frac(5-2x)(3)) \\ (y=3x+9) \end(array) \right.\]
Let's depict both graphs on the same plane:
Picture 1.
Answer: $(-2,\ 3)$
Method for introducing new variables
Let's look at this method using the following example:
Example 4
\[\left\( \begin(array)(c) (2^(x+1)-3^y=-1) \\ (3^y-2^x=2) \end(array) \right .\]
Solution.
This system is equivalent to the system
\[\left\( \begin(array)(c) ((2\cdot 2)^x-3^y=-1) \\ (3^y-2^x=2) \end(array) \ right.\]
Let $2^x=u\ (u>0)$, and $3^y=v\ (v>0)$, we get:
\[\left\( \begin(array)(c) (2u-v=-1) \\ (v-u=2) \end(array) \right.\]
Let us solve the resulting system using the addition method. Let's add up the equations:
\ \
Then from the second equation, we get that
Returning to the replacement, we get new system exponential equations:
\[\left\( \begin(array)(c) (2^x=1) \\ (3^y=3) \end(array) \right.\]
We get:
\[\left\( \begin(array)(c) (x=0) \\ (y=1) \end(array) \right.\]
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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 paragraph 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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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 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 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 true equality, therefore, we have found the right solution.