System of equations examples. Solving easy problems using addition

The material in this article is intended for a first acquaintance with systems of equations. Here we will introduce the definition of a system of equations and its solutions, and also consider the most common types of systems of equations. As usual, we will give explanatory examples.

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What is a system of equations?

We will approach the definition of the system of equations gradually. First, let’s just say that it is convenient to give it, indicating two points: firstly, the type of recording, and, secondly, the meaning embedded in this recording. Let's look at them in turn, and then generalize the reasoning into the definition of systems of equations.

Let there be several of them in front of us. For example, let's take two equations 2 x+y=−3 and x=5. Let's write them one below the other and combine them on the left with a curly brace:

Records of this type, which are several equations arranged in a column and united on the left by a curly brace, are records of systems of equations.

What do such entries mean? They define the set of all solutions to the equations of the system that are a solution to each equation.

It wouldn't hurt to describe it in other words. Let us assume that some solutions to the first equation are solutions to all other equations of the system. So the system record just means them.

Now we are ready to adequately accept the definition of a system of equations.

Definition.

Systems of equations call records that are equations located one below the other, united on the left by a curly brace, which denote the set of all solutions to equations that are also solutions to each equation of the system.

A similar definition is given in the textbook, however, it is given there not for the general case, but for two rational equations with two variables.

Main types

It is clear that there are an infinite number of different equations. Naturally, there are also an infinite number of systems of equations compiled using them. Therefore, for the convenience of studying and working with systems of equations, it makes sense to divide them into groups according to similar characteristics, and then move on to considering systems of equations of individual types.

The first division suggests itself by the number of equations included in the system. If there are two equations, then we can say that we have a system of two equations, if there are three, then a system of three equations, etc. It is clear that it makes no sense to talk about a system of one equation, since in this case, in essence, we are dealing with the equation itself, and not with the system.

The next division is based on the number of variables involved in writing the equations of the system. If there is one variable, then we are dealing with a system of equations with one variable (they also say with one unknown), if there are two, then with a system of equations with two variables (with two unknowns), etc. For example, is a system of equations with two variables x and y.

This refers to the number of all the different variables involved in the recording. They do not have to all be included in the record of each equation at once; their presence in at least one equation is sufficient. For example, is a system of equations with three variables x, y and z. In the first equation, the variable x is present explicitly, and y and z are implicit (we can assume that these variables have zero), and in the second equation there are x and z, but the variable y is not explicitly presented. In other words, the first equation can be viewed as , and the second – as x+0·y−3·z=0.

The third point in which systems of equations differ is the type of equations themselves.

At school, the study of systems of equations begins with systems of two linear equations in two variables. That is, such systems constitute two linear equations. Here are a couple of examples: And . They learn the basics of working with systems of equations.

When solving more complex problems, you may also encounter systems of three linear equations with three unknowns.

Further in the 9th grade, nonlinear equations are added to systems of two equations with two variables, mostly entire equations of the second degree, less often - higher degrees. These systems are called systems of nonlinear equations; if necessary, the number of equations and unknowns is specified. Let us show examples of such systems of nonlinear equations: And .

And then in systems there are also, for example, . They are usually called simply systems of equations, without specifying which equations. It is worth noting here that most often they simply say “system of equations” about a system of equations, and clarifications are added only if necessary.

In high school, as the material is studied, irrational, trigonometric, logarithmic and exponential equations penetrate into the systems: , , .

If we look even further into the first-year university curriculum, the main emphasis is on the study and solution of systems of linear algebraic equations (SLAEs), that is, equations in which the left-hand sides contain polynomials of the first degree, and the right-hand sides contain certain numbers. But there, unlike at school, they no longer take two linear equations with two variables, but an arbitrary number of equations with an arbitrary number of variables, which often does not coincide with the number of equations.

What is the solution to a system of equations?

The term “solution of a system of equations” directly refers to systems of equations. At school, the definition of solving a system of equations with two variables is given :

Definition.

Solving a system of equations with two variables is called a pair of values ​​of these variables that turns each equation of the system into the correct one, in other words, is a solution to each equation of the system.

For example, a pair of variable values ​​x=5, y=2 (it can be written as (5, 2)) is a solution to a system of equations by definition, since the equations of the system, when x=5, y=2 are substituted into them, turn into correct numerical equalities 5+2=7 and 5−2=3 respectively. But the pair of values ​​x=3, y=0 is not a solution to this system, since when substituting these values ​​into the equations, the first of them will turn into the incorrect equality 3+0=7.

Similar definitions can be formulated for systems with one variable, as well as for systems with three, four, etc. variables.

Definition.

Solving a system of equations with one variable there will be a value of the variable that is the root of all equations of the system, that is, turning all equations into correct numerical equalities.

Let's give an example. Consider a system of equations with one variable t of the form . The number −2 is its solution, since both (−2) 2 =4 and 5·(−2+2)=0 are true numerical equalities. And t=1 is not a solution to the system, since substituting this value will give two incorrect equalities 1 2 =4 and 5·(1+2)=0.

Definition.

Solving a system with three, four, etc. variables called three, four, etc. values ​​of the variables, respectively, turning all equations of the system into true equalities.

So, by definition, a triple of values ​​of the variables x=1, y=2, z=0 is a solution to the system , since 2·1=2, 5·2=10 and 1+2+0=3 are correct numerical equalities. And (1, 0, 5) is not a solution to this system, since when substituting these values ​​of the variables into the equations of the system, the second of them turns into the incorrect equality 5·0=10, and the third too 1+0+5=3.

Note that systems of equations may have no solutions, may have a finite number of solutions, for example, one, two, ..., or may have infinitely many solutions. You will see this as you delve deeper into the topic.

Taking into account the definitions of a system of equations and their solutions, we can conclude that the solution to a system of equations is the intersection of the sets of solutions of all its equations.

To conclude, here are a few related definitions:

Definition.

non-joint, if it has no solutions, otherwise the system is called joint.

Definition.

The system of equations is called uncertain, if it has infinitely many solutions, and certain, if it has a finite number of solutions or does not have them at all.

These terms are introduced, for example, in a textbook, but are used quite rarely at school; they are more often heard in higher educational institutions.

References.

1. Algebra: textbook for 7th grade general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 17th ed. - M.: Education, 2008. - 240 p. : ill. - ISBN 978-5-09-019315-3.
2. Algebra: 9th grade: educational. for general education institutions / [Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, S. B. Suvorova]; edited by S. A. Telyakovsky. - 16th ed. - M.: Education, 2009. - 271 p. : ill. - ISBN 978-5-09-021134-5.
3. Mordkovich A. G. Algebra. 7th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich. - 17th ed., add. - M.: Mnemosyne, 2013. - 175 p.: ill. ISBN 978-5-346-02432-3.
4. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. - 13th ed., erased. - M.: Mnemosyne, 2011. - 222 p.: ill. ISBN 978-5-346-01752-3.
5. Mordkovich A. G. Algebra and the beginnings of mathematical analysis. 11th grade. In 2 hours. Part 1. Textbook for students of general education institutions (profile level) / A. G. Mordkovich, P. V. Semenov. - 2nd ed., erased. - M.: Mnemosyne, 2008. - 287 p.: ill. ISBN 978-5-346-01027-2.
6. Algebra and the beginning of analysis: Proc. for 10-11 grades. general education institutions / A. N. Kolmogorov, A. M. Abramov, Yu. P. Dudnitsyn and others; Ed. A. N. Kolmogorov. - 14th ed. - M.: Education, 2004. - 384 pp.: ill. - ISBN 5-09-013651-3.
7. A. G. Kurosh. Higher algebra course.
8. Ilyin V. A., Poznyak E. G. Analytical geometry: Textbook: For universities. – 5th ed. – M.: Science. Fizmatlit, 1999. – 224 p. – (Course of higher mathematics and mathematical physics). – ISBN 5-02-015234 – X (Issue 3)

In this lesson we will look at methods for solving a system of linear equations. In a course of higher mathematics, systems of linear equations are required to be solved both in the form of separate tasks, for example, “Solve the system using Cramer’s formulas,” and in the course of solving other problems. Systems of linear equations have to be dealt with in almost all branches of higher mathematics.

First, a little theory. What does the mathematical word “linear” mean in this case? This means that the equations of the system All variables included in the first degree: without any fancy stuff like etc., which only participants in mathematical Olympiads are delighted with.

In higher mathematics, not only letters familiar from childhood are used to denote variables.
A fairly popular option is variables with indexes: .
Or the initial letters of the Latin alphabet, small and large:
It is not so rare to find Greek letters: – known to many as “alpha, beta, gamma”. And also a set with indices, say, with the letter “mu”:

The use of one or another set of letters depends on the section of higher mathematics in which we are faced with a system of linear equations. So, for example, in systems of linear equations encountered when solving integrals and differential equations, it is traditional to use the notation

But no matter how the variables are designated, the principles, methods and methods for solving a system of linear equations do not change. Thus, if you come across something scary like , do not rush to close the problem book in fear, after all, you can draw the sun instead, a bird instead, and a face (the teacher) instead. And, funny as it may seem, a system of linear equations with these notations can also be solved.

I have a feeling that the article will turn out to be quite long, so a small table of contents. So, the sequential “debriefing” will be like this:

– Solving a system of linear equations using the substitution method (“school method”);
– Solving the system by term-by-term addition (subtraction) of the system equations;
– Solution of the system using Cramer’s formulas;
– Solving the system using an inverse matrix;
– Solving the system using the Gaussian method.

Everyone is familiar with systems of linear equations from school mathematics courses. Essentially, we start with repetition.

Solving a system of linear equations using the substitution method

This method can also be called the “school method” or the method of eliminating unknowns. Figuratively speaking, it can also be called “an unfinished Gaussian method.”

Example 1

Here we are given a system of two equations with two unknowns. Note that the free terms (numbers 5 and 7) are located on the left side of the equation. Generally speaking, it doesn’t matter where they are, on the left or on the right, it’s just that in problems in higher mathematics they are often located that way. And such a recording should not lead to confusion; if necessary, the system can always be written “as usual”: . Don’t forget that when moving a term from part to part, it needs to change its sign.

What does it mean to solve a system of linear equations? Solving a system of equations means finding many of its solutions. The solution of a system is a set of values ​​of all variables included in it, which turns EVERY equation of the system into a true equality. In addition, the system can be non-joint (have no solutions).Don’t be shy, this is a general definition =) We will have only one “x” value and one “y” value, which satisfy each c-we equation.

There is a graphical method for solving the system, which you can familiarize yourself with in class. The simplest problems with a line. There I talked about geometric sense systems of two linear equations with two unknowns. But now this is the era of algebra, and numbers-numbers, actions-actions.

Let's decide: from the first equation we express:
We substitute the resulting expression into the second equation:

We open the brackets, add similar terms and find the value:

Next, we remember what we danced for:
We already know the value, all that remains is to find:

Answer:

After ANY system of equations has been solved in ANY way, I strongly recommend checking (orally, on a draft or on a calculator). Fortunately, this is done easily and quickly.

1) Substitute the found answer into the first equation:

– the correct equality is obtained.

2) Substitute the found answer into the second equation:

– the correct equality is obtained.

Or, to put it more simply, “everything came together”

The considered solution method is not the only one; from the first equation it was possible to express , and not .
You can do the opposite - express something from the second equation and substitute it into the first equation. By the way, note that the most disadvantageous of the four methods is to express from the second equation:

The result is fractions, but why? There is a more rational solution.

However, in some cases you still cannot do without fractions. In this regard, I would like to draw your attention to HOW I wrote down the expression. Not like this: and in no case like this: .

If in higher mathematics you are dealing with fractional numbers, then try to carry out all calculations in ordinary improper fractions.

Exactly, and not or!

A comma can be used only sometimes, in particular if it is the final answer to some problem, and no further actions need to be performed with this number.

Many readers probably thought “why such a detailed explanation as for a correction class, everything is clear.” Nothing of the kind, it seems like such a simple school example, but there are so many VERY important conclusions! Here's another one:

You should strive to complete any task in the most rational way. If only because it saves time and nerves, and also reduces the likelihood of making a mistake.

If in a problem in higher mathematics you come across a system of two linear equations with two unknowns, then you can always use the substitution method (unless it is indicated that the system needs to be solved by another method). Not a single teacher will think that you are a sucker and will reduce your grade for using the “school method” "
Moreover, in some cases it is advisable to use the substitution method with a larger number of variables.

Example 2

Solve a system of linear equations with three unknowns

A similar system of equations often arises when using the so-called method of indefinite coefficients, when we find the integral of a fractional rational function. The system in question was taken from there by me.

When finding the integral, the goal is fast find the values ​​of the coefficients, rather than using Cramer’s formulas, the inverse matrix method, etc. Therefore, in this case, the substitution method is appropriate.

When any system of equations is given, first of all it is desirable to find out whether it is possible to somehow simplify it IMMEDIATELY? Analyzing the equations of the system, we notice that the second equation of the system can be divided by 2, which is what we do:

Reference: the mathematical sign means “from this follows that” and is often used in problem solving.

Now let's analyze the equations; we need to express some variable in terms of the others. Which equation should I choose? You probably already guessed that the easiest way for this purpose is to take the first equation of the system:

Here, no matter what variable to express, one could just as easily express or .

Next, we substitute the expression for into the second and third equations of the system:

We open the brackets and present similar terms:

Divide the third equation by 2:

From the second equation we express and substitute into the third equation:

Almost everything is ready, from the third equation we find:
From the second equation:
From the first equation:

Check: Substitute the found values ​​of the variables into the left side of each equation of the system:

1)
2)
3)

The corresponding right-hand sides of the equations are obtained, thus the solution is found correctly.

Example 3

Solve a system of linear equations with 4 unknowns

This is an example for you to solve on your own (answer at the end of the lesson).

Solving the system by term-by-term addition (subtraction) of the system equations

When solving systems of linear equations, you should try to use not the “school method”, but the method of term-by-term addition (subtraction) of the equations of the system. Why? This saves time and simplifies calculations, however, now everything will become clearer.

Example 4

Solve a system of linear equations:

I took the same system as in the first example.
Analyzing the system of equations, we notice that the coefficients of the variable are identical in magnitude and opposite in sign (–1 and 1). In such a situation, the equations can be added term by term:

Actions circled in red are performed MENTALLY.
As you can see, as a result of term-by-term addition, we lost the variable. This, in fact, is what the essence of the method is to get rid of one of the variables.

More reliable than the graphical method discussed in the previous paragraph.

Substitution method

We used this method in 7th grade to solve systems of 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 a two-digit number led to a 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 in a simpler form: Let's solve this equation with respect to the variable t:

Both of these values ​​satisfy the condition and therefore are the roots of a 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 simple equations obtained must be considered in turn 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 the given system in a much simpler form, but with respect to the 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 of algebraic addition to solve this system:

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 discussion. You have already gained some experience in solving various equations: linear, quadratic, 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 the graphical solution method.

The method of solving systems of equations graphically involves constructing a graph for each of the specific equations that are included in a given system and are located in the same coordinate plane, as well as 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 common for a graphical system of equations to have either one single correct solution, or an infinite number of solutions, or to have no solutions at all.

Now let’s look at each of these solutions in more detail. And so, a system of equations can have a unique solution if the lines that 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 graph of this 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 plot a graph of 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).

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 decide 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

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, 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 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 using the term-by-term addition (subtraction) method.

3x-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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Methods for solving systems of inequalities

Substitution method

How should you proceed when making a decision?
1. Express one of the variables in terms of another. The variables most often used in equations are x and y. In one of the equations we express one variable in terms of another. Tip: Look at both equations carefully before you start solving, and choose the one where it is easier to express the variable.
2. Substitute the resulting expression into the second equation, instead of the variable that was expressed.
3. Solve the equation that we got.
4. Substitute the resulting solution into the second equation. If there are several solutions, then you need to substitute sequentially so as not to lose a couple of solutions.
5. As a result, you will receive a pair of numbers $(x;y)$, which must be written down as an answer.

Example.
Solve a system with two variables using the substitution method: $\begin(cases)x+y=5, \\xy=6\end(cases)$.

Solution.
Let's take a closer look at our equations. Obviously, expressing y in terms of x in the first equation is much simpler.
$\begin(cases)y=5-x, \\xy=6\end(cases)$.
Let's substitute the first expression into the second equation $\begin(cases)y=5-x, \\x(5-2x)=6\end(cases)$.
Let's solve the second equation separately:
$x(5-x)=6$.
$-x^2+5x-6=0$.
$x^2-5x+6=0$.
$(x-2)(x-3)=0$.
We obtained two solutions to the second equation $x_1=2$ and $x_2=3$.
Substitute sequentially into the second equation.
If $x=2$, then $y=3$. If $x=3$, then $y=2$.
The answer will be two pairs of numbers.
Answer: $(2;3)$ and $(3;2)$.

Algebraic addition method

Example.
Solve the system: $\begin(cases)2x+xy-1=0, \\4y+2xy+6=0\end(cases)$.

Solution.
Let's multiply the first equation by 2.
$\begin(cases)4x+2xy-2=0, \\4y+2xy+6=0\end(cases)$.
Let's subtract the second from the first equation.
$4x+2xy-2-4y-2xy-6=4x-4y-8$.
As you can see, the form of the resulting equation is much simpler than the original one. Now we can use the substitution method.
$\begin(cases)4x-4y-8=0, \\4y+2xy+6=0\end(cases)$.
Let's express x in terms of y in the resulting equation.
$\begin(cases)4x=4y+8, \\4y+2xy+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2(y+2)y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\4y+2y^2+4y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\2y^2+8y+6=0\end(cases)$.
$\begin(cases)x=y+2, \\y^2+4y+3=0\end(cases)$.
$\begin(cases)x=y+2, \\(y+3)(y+1)=0\end(cases)$.
We got $y=-1$ and $y=-3$.
Let's substitute these values ​​sequentially into the first equation. We get two pairs of numbers: $(1;-1)$ and $(-1;-3)$.
Answer: $(1;-1)$ and $(-1;-3)$.

Method for introducing a new variable

Example.
Solve the system: $\begin(cases)\frac(x)(y)+\frac(2y)(x)=3, \\2x^2-y^2=1\end(cases)$.

Solution.
Let us introduce the replacement $t=\frac(x)(y)$.
Let's rewrite the first equation with a new variable: $t+\frac(2)(t)=3$.
Let's solve the resulting equation:
$\frac(t^2-3t+2)(t)=0$.
$\frac((t-2)(t-1))(t)=0$.
We got $t=2$ or $t=1$. Let us introduce the reverse change $t=\frac(x)(y)$.
We got: $x=2y$ and $x=y$.

For each of the expressions, the original system must be solved separately:
$\begin(cases)x=2y, \\2x^2-y^2=1\end(cases)$.    $\begin(cases)x=y, \\2x^2-y^2=1\end(cases)$.
$\begin(cases)x=2y, \\8y^2-y^2=1\end(cases)$.    $\begin(cases)x=y, \\2y^2-y^2=1\end(cases)$.
$\begin(cases)x=2y, \\7y^2=1\end(cases)$.       $\begin(cases)x=2y, \\y^2=1\end(cases)$.
$\begin(cases)x=2y, \\y=±\frac(1)(\sqrt(7))\end(cases)$.      $\begin(cases)x=y, \\y=±1\end(cases)$.
$\begin(cases)x=±\frac(2)(\sqrt(7)), \\y=±\frac(1)(\sqrt(7))\end(cases)$.     $\begin(cases)x=±1, \\y=±1\end(cases)$.
We received four pairs of solutions.
Answer: $(\frac(2)(\sqrt(7));\frac(1)(\sqrt(7)))$; $(-\frac(2)(\sqrt(7));-\frac(1)(\sqrt(7)))$; $(1;1)$; $(-1;-1)$.

Example.
Solve the system: $\begin(cases)\frac(2)(x-3y)+\frac(3)(2x+y)=2, \\\frac(8)(x-3y)-\frac(9 )(2x+y)=1\end(cases)$.

Solution.
Let us introduce the replacement: $z=\frac(2)(x-3y)$ and $t=\frac(3)(2x+y)$.
Let's rewrite the original equations with new variables:
$\begin(cases)z+t=2, \\4z-3t=1\end(cases)$.
Let's use the algebraic addition method:
$\begin(cases)3z+3t=6, \\4z-3t=1\end(cases)$.
$\begin(cases)3z+3t+4z-3t=6+1, \\4z-3t=1\end(cases)$.
$\begin(cases)7z=7, \\4z-3t=1\end(cases)$.
$\begin(cases)z=1, \\-3t=1-4\end(cases)$.
$\begin(cases)z=1, \\t=1\end(cases)$.
Let's introduce the reverse substitution:
$\begin(cases)\frac(2)(x-3y)=1, \\\frac(3)(2x+y)=1\end(cases)$.
$\begin(cases)x-3y=2, \\2x+y=3\end(cases)$.
Let's use the substitution method:
$\begin(cases)x=2+3y, \\4+6y+y=3\end(cases)$.
$\begin(cases)x=2+3y, \\7y=-1\end(cases)$.
$\begin(cases)x=2+3(\frac(-1)(7)), \\y=\frac(-1)(7)\end(cases)$.
$\begin(cases)x=\frac(11)(7), \\x=-\frac(11)(7)\end(cases)$.
Answer: $(\frac(11)(7);-\frac(1)(7))$.

Problems on systems of equations for independent solution