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 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’s solve using the addition method the following system linear equations with two unknowns:

(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 equalities, therefore, we found the right decision.

Using this math program You can solve a system of two linear equations in two variables using the substitution method and the addition method.

The program not only gives the answer to the problem, but also gives detailed solution with explanations of the solution steps in two ways: the substitution method and the addition method.

This program may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.

This way you can conduct your own training and/or training of yours. younger brothers or sisters, while the level of education in the field of problems being solved increases.

Rules for entering equations

Any Latin letter can act as a variable.
For example: \(x, y, z, a, b, c, o, p, q\), etc.

When entering equations you can use parentheses. In this case, the equations are first simplified. The equations after simplifications must be linear, i.e. of the form ax+by+c=0 with the accuracy of the order of elements.
For example: 6x+1 = 5(x+y)+2

In equations you can use not only integers, but also fractional numbers in the form of decimals and ordinary fractions.

Rules for entering decimal fractions.
Whole and fraction V decimals can be separated by either a dot or a comma.
For example: 2.1n + 3.5m = 55

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.
The denominator cannot be negative.
When entering numerical fraction The numerator is separated from the denominator by a division sign: /
Whole part separated from the fraction by an ampersand: &

Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)

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A little theory.

Solving systems of linear equations. Substitution method

The sequence of actions when solving a system of linear equations using the substitution method:
1) express one variable from some equation of the system in terms of another;
2) substitute the resulting expression into another equation of the system instead of this variable;

$$ \left\( \begin(array)(l) 3x+y=7 \\ -5x+2y=3 \end(array) \right. $$

Let's express y in terms of x from the first equation: y = 7-3x. Substituting the expression 7-3x into the second equation instead of y, we obtain the system:
$$ \left\( \begin(array)(l) y = 7-3x \\ -5x+2(7-3x)=3 \end(array) \right. $$

It is easy to show that the first and second systems have the same solutions. In the second system, the second equation contains only one variable. Let's solve this equation:
$$ -5x+2(7-3x)=3 \Rightarrow -5x+14-6x=3 \Rightarrow -11x=-11 \Rightarrow x=1 $$

Substituting the number 1 instead of x into the equality y=7-3x, we find the corresponding value of y:
$$ y=7-3 \cdot 1 \Rightarrow y=4 $$

Pair (1;4) - solution of the system

Systems of equations in two variables that have the same solutions are called equivalent. Systems that do not have solutions are also considered equivalent.

Solving systems of linear equations by addition

Let's consider another way to solve systems of linear equations - the addition method. When solving systems in this way, as well as when solving by substitution, we move from this system to another, equivalent system, in which one of the equations contains only one variable.

The sequence of actions when solving a system of linear equations using the addition method:
1) multiply the equations of the system term by term, selecting factors so that the coefficients of one of the variables become opposite numbers;
2) add the left and right sides of the system equations term by term;
3) solve the resulting equation with one variable;
4) find the corresponding value of the second variable.

Example. Let's solve the system of equations:
$$ \left\( \begin(array)(l) 2x+3y=-5 \\ x-3y=38 \end(array) \right. $$

In the equations of this system, the coefficients of y are opposite numbers. By adding the left and right sides of the equations term by term, we obtain an equation with one variable 3x=33. Let's replace one of the equations of the system, for example the first one, with the equation 3x=33. Let's get the system
$$ \left\( \begin(array)(l) 3x=33 \\ x-3y=38 \end(array) \right. $$

From the equation 3x=33 we find that x=11. Substituting this x value into the equation \(x-3y=38\) we get an equation with the variable y: \(11-3y=38\). Let's solve this equation:
\(-3y=27 \Rightarrow y=-9 \)

Thus, we found the solution to the system of equations by addition: \(x=11; y=-9\) or \((11;-9)\)

Taking advantage of the fact that in the equations of the system the coefficients of y are opposite numbers, we reduced its solution to the solution of an equivalent system (by summing both sides of each of the equations of the original system), in which one of the equations contains only one variable.

1. Substitution method: from any equation of the system we express one unknown through another and substitute it into the second equation of the system.

Task. Solve the system of equations:

Solution. From the first equation of the system we express at through X and substitute it into the second equation of the system. Let's get the system equivalent to the original one.

After bringing similar members the system will take the form:

From the second equation we find: . Substituting this value into the equation at = 2 - 2X, we get at= 3. Therefore, the solution to this system is a pair of numbers.

2. Method algebraic addition : By adding two equations, you get an equation with one variable.

Task. Solve the system equation:

Solution. Multiplying both sides of the second equation by 2, we obtain the system equivalent to the original one. Adding the two equations of this system, we arrive at the system

After bringing similar terms, this system will take the form: From the second equation we find . Substituting this value into equation 3 X + 4at= 5, we get , where . Therefore, the solution to this system is a pair of numbers.

3. Method for introducing new variables: we are looking for some repeating expressions in the system, which we will denote by new variables, thereby simplifying the appearance of the system.

Task. Solve the system of equations:

Solution. Let's write it down this system otherwise:

Let x + y = u, xy = v. Then we get the system

Let's solve it using the substitution method. From the first equation of the system we express u through v and substitute it into the second equation of the system. Let's get the system those.

From the second equation of the system we find v 1 = 2, v 2 = 3.

Substituting these values ​​into the equation u = 5 - v, we get u 1 = 3,
u 2 = 2. Then we have two systems

Solving the first system, we get two pairs of numbers (1; 2), (2; 1). The second system has no solutions.

Exercises for independent work

1. Solve systems of equations using the substitution method.

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 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 sort of “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 given 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 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 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.

Method for solving systems of equations graphically represents the construction of a graph for each of specific equations, which are included in this 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 for graphics system equations tend to have either one unique correct solution, 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 lines that are graphs of the equations of the system 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 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).

With this video I begin a series of lessons dedicated to systems of equations. Today we will talk about solving systems of linear equations addition method- this is one of the most simple ways, but at the same time one of the most effective.

The addition method consists of three simple steps:

1. Look at the system and choose a variable that has identical (or opposite) coefficients in each equation;
2. Execute algebraic subtraction(For opposite numbers- addition) of equations from each other, and then bring similar terms;
3. Solve the new equation obtained after the second step.

If everything is done correctly, then at the output we will get a single equation with one variable- it won’t be difficult to solve it. Then all that remains is to substitute the found root into the original system and get the final answer.

However, in practice everything is not so simple. There are several reasons for this:

• Solving equations using the addition method implies that all lines must contain variables with equal/opposite coefficients. What to do if this requirement is not met?
• Not always, after adding/subtracting equations in the indicated way, we get a beautiful construction that can be easily solved. Is it possible to somehow simplify the calculations and speed up the calculations?

To get the answer to these questions, and at the same time understand a few additional subtleties that many students fail at, watch my video lesson:

With this lesson we begin a series of lectures devoted to systems of equations. And we will start from the simplest of them, namely those that contain two equations and two variables. Each of them will be linear.

Systems is 7th grade material, but this lesson will also be useful for high school students who want to brush up on their knowledge of this topic.

In general, there are two methods for solving such systems:

1. Addition method;
2. A method of expressing one variable in terms of another.

Today we will deal with the first method - we will use the method of subtraction and addition. But to do this, you need to understand the following fact: once you have two or more equations, you can take any two of them and add them to each other. They are added member by member, i.e. “X’s” are added to “X’s” and similar ones are given, “Y’s” with “Y’s” are similar again, and what is to the right of the equal sign is also added to each other, and similar ones are also given there.

The results of such machinations will be a new equation, which, if it has roots, they will definitely be among the roots original equation. Therefore, our task is to do the subtraction or addition in such a way that either $x$ or $y$ disappears.

How to achieve this and what tool to use for this - we’ll talk about this now.

Solving easy problems using addition

So, we learn to use the addition method using the example of two simple expressions.

Task No. 1

\[\left\( \begin(align)& 5x-4y=22 \\& 7x+4y=2 \\\end(align) \right.\]

Note that $y$ has a coefficient of $-4$ in the first equation, and $+4$ in the second. They are mutually opposite, so it is logical to assume that if we add them up, then in the resulting sum the “games” will be mutually destroyed. Add it up and get:

Let's solve the simplest construction:

Great, we found the "x". What should we do with it now? We have the right to substitute it into any of the equations. Let's substitute in the first:

\[-4y=12\left| :\left(-4 \right) \right.\]

Answer: $\left(2;-3 \right)$.

Problem No. 2

\[\left\( \begin(align)& -6x+y=21 \\& 6x-11y=-51 \\\end(align) \right.\]

The situation here is completely similar, only with “X’s”. Let's add them up:

We have the simplest linear equation, let's solve it:

Now let's find $x$:

Answer: $\left(-3;3 \right)$.

Important points

So, we have just solved two simple systems of linear equations using the addition method. Key points again:

1. If there are opposite coefficients for one of the variables, then it is necessary to add all the variables in the equation. In this case, one of them will be destroyed.
2. We substitute the found variable into any of the system equations to find the second one.
3. The final response record can be presented in different ways. For example, like this - $x=...,y=...$, or in the form of coordinates of points - $\left(...;... \right)$. The second option is preferable. The main thing to remember is that the first coordinate is $x$, and the second is $y$.
4. The rule of writing the answer in the form of point coordinates is not always applicable. For example, it cannot be used when the variables are not $x$ and $y$, but, for example, $a$ and $b$.

In the following problems we will consider the technique of subtraction when the coefficients are not opposite.

Solving easy problems using the subtraction method

Task No. 1

\[\left\( \begin(align)& 10x-3y=5 \\& -6x-3y=-27 \\\end(align) \right.\]

Note that there are no opposite coefficients here, but there are identical ones. Therefore, we subtract the second from the first equation:

Now we substitute the value $x$ into any of the system equations. Let's go first:

Answer: $\left(2;5\right)$.

Problem No. 2

\[\left\( \begin(align)& 5x+4y=-22 \\& 5x-2y=-4 \\\end(align) \right.\]

We again see the same coefficient of $5$ for $x$ in the first and second equation. Therefore, it is logical to assume that you need to subtract the second from the first equation:

We have calculated one variable. Now let's find the second one, for example, by substituting the value $y$ into the second construction:

Answer: $\left(-3;-2 \right)$.

Nuances of the solution

So what do we see? Essentially, the scheme is no different from the solution of previous systems. The only difference is that we do not add equations, but subtract them. We are doing algebraic subtraction.

In other words, as soon as you see a system consisting of two equations in two unknowns, the first thing you need to look at is the coefficients. If they are the same anywhere, the equations are subtracted, and if they are opposite, the addition method is used. This is always done so that one of them disappears, and in the final equation, which remains after subtraction, only one variable remains.

Of course, that's not all. Now we will consider systems in which the equations are generally inconsistent. Those. There are no variables in them that are either the same or opposite. In this case, to solve such systems, we use additional dose, namely, multiplying each of the equations by a special coefficient. How to find it and how to solve such systems in general, we’ll talk about this now.

Solving problems by multiplying by a coefficient

Example #1

\[\left\( \begin(align)& 5x-9y=38 \\& 3x+2y=8 \\\end(align) \right.\]

We see that neither for $x$ nor for $y$ the coefficients are not only mutually opposite, but also in no way correlated with the other equation. These coefficients will not disappear in any way, even if we add or subtract the equations from each other. Therefore, it is necessary to apply multiplication. Let's try to get rid of the $y$ variable. To do this, we multiply the first equation by the coefficient of $y$ from the second equation, and the second equation by the coefficient of $y$ from the first equation, without touching the sign. We multiply and get a new system:

\[\left\( \begin(align)& 10x-18y=76 \\& 27x+18y=72 \\\end(align) \right.\]

Let's look at it: at $y$ the coefficients are opposite. In such a situation, it is necessary to use the addition method. Let's add:

Now we need to find $y$. To do this, substitute $x$ into the first expression:

\[-9y=18\left| :\left(-9 \right) \right.\]

Answer: $\left(4;-2 \right)$.

Example No. 2

\[\left\( \begin(align)& 11x+4y=-18 \\& 13x-6y=-32 \\\end(align) \right.\]

Again, the coefficients for none of the variables are consistent. Let's multiply by the coefficients of $y$:

\[\left\( \begin(align)& 11x+4y=-18\left| 6 \right. \\& 13x-6y=-32\left| 4 \right. \\\end(align) \right .\]

\[\left\( \begin(align)& 66x+24y=-108 \\& 52x-24y=-128 \\\end(align) \right.\]

Our new system is equivalent to the previous one, but the coefficients of $y$ are mutually opposite, and therefore it is easy to apply the addition method here:

Now let's find $y$ by substituting $x$ into the first equation:

Answer: $\left(-2;1 \right)$.

Nuances of the solution

The key rule here is the following: we always multiply only by positive numbers- this will save you from stupid and offensive mistakes associated with changing signs. In general, the solution scheme is quite simple:

1. We look at the system and analyze each equation.
2. If we see that neither $y$ nor $x$ the coefficients are consistent, i.e. they are neither equal nor opposite, then we do the following: we select the variable that we need to get rid of, and then we look at the coefficients of these equations. If we multiply the first equation by the coefficient from the second, and the second, correspondingly, multiply by the coefficient from the first, then in the end we will get a system that is completely equivalent to the previous one, and the coefficients of $y$ will be consistent. All our actions or transformations are aimed only at getting one variable in one equation.
3. We find one variable.
4. We substitute the found variable into one of the two equations of the system and find the second.
5. We write the answer in the form of coordinates of points if we have variables $x$ and $y$.

But even such a simple algorithm has its own subtleties, for example, the coefficients of $x$ or $y$ can be fractions and other “ugly” numbers. We will now consider these cases separately, because in them you can act somewhat differently than according to the standard algorithm.

Solving problems with fractions

Example #1

\[\left\( \begin(align)& 4m-3n=32 \\& 0.8m+2.5n=-6 \\\end(align) \right.\]

First, notice that the second equation contains fractions. But note that you can divide $4$ by $0.8$. We will receive $5$. Let's multiply the second equation by $5$:

\[\left\( \begin(align)& 4m-3n=32 \\& 4m+12.5m=-30 \\\end(align) \right.\]

We subtract the equations from each other:

We found $n$, now let's count $m$:

Answer: $n=-4;m=5$

Example No. 2

\[\left\( \begin(align)& 2.5p+1.5k=-13\left| 4 \right. \\& 2p-5k=2\left| 5 \right. \\\end(align )\right.\]

Here, as in the previous system, there are fractional odds, however, for none of the variable coefficients do not fit into each other by an integer number of times. Therefore, we use the standard algorithm. Get rid of $p$:

\[\left\( \begin(align)& 5p+3k=-26 \\& 5p-12.5k=5 \\\end(align) \right.\]

We use the subtraction method:

Let's find $p$ by substituting $k$ into the second construction:

Answer: $p=-4;k=-2$.

Nuances of the solution

That's all optimization. In the first equation, we did not multiply by anything at all, but multiplied the second equation by $5$. As a result, we received a consistent and even identical equation for the first variable. In the second system we followed a standard algorithm.

But how do you find the numbers by which to multiply equations? After all, if we multiply by fractions, we get new fractions. Therefore, the fractions must be multiplied by a number that would give a new integer, and after that the variables must be multiplied by coefficients, following the standard algorithm.

In conclusion, I would like to draw your attention to the format for recording the response. As I already said, since here we have not $x$ and $y$, but other values, we use a non-standard notation of the form:

Solving complex systems of equations

As a final note to today's video tutorial, let's look at a couple of really complex systems. Their complexity will consist in the fact that they will have variables on both the left and right. Therefore, to solve them we will have to apply preprocessing.

System No. 1

\[\left\( \begin(align)& 3\left(2x-y \right)+5=-2\left(x+3y ​​\right)+4 \\& 6\left(y+1 \right )-1=5\left(2x-1 \right)+8 \\\end(align) \right.\]

Each equation carries a certain complexity. Therefore, let's treat each expression as with a regular linear construction.

In total, we get the final system, which is equivalent to the original one:

\[\left\( \begin(align)& 8x+3y=-1 \\& -10x+6y=-2 \\\end(align) \right.\]

Let's look at the coefficients of $y$: $3$ fits into $6$ twice, so let's multiply the first equation by $2$:

\[\left\( \begin(align)& 16x+6y=-2 \\& -10+6y=-2 \\\end(align) \right.\]

The coefficients of $y$ are now equal, so we subtract the second from the first equation: $$

Now let's find $y$:

Answer: $\left(0;-\frac(1)(3) \right)$

System No. 2

\[\left\( \begin(align)& 4\left(a-3b \right)-2a=3\left(b+4 \right)-11 \\& -3\left(b-2a \right )-12=2\left(a-5 \right)+b \\\end(align) \right.\]

Let's transform the first expression:

Let's deal with the second one:

\[-3\left(b-2a \right)-12=2\left(a-5 \right)+b\]

\[-3b+6a-12=2a-10+b\]

\[-3b+6a-2a-b=-10+12\]

In total, our initial system will take the following form:

\[\left\( \begin(align)& 2a-15b=1 \\& 4a-4b=2 \\\end(align) \right.\]

Looking at the coefficients of $a$, we see that the first equation needs to be multiplied by $2$:

\[\left\( \begin(align)& 4a-30b=2 \\& 4a-4b=2 \\\end(align) \right.\]

Subtract the second from the first construction:

Now let's find $a$:

Answer: $\left(a=\frac(1)(2);b=0 \right)$.

That's all. I hope this video tutorial will help you understand this difficult topic, namely solving systems of simple linear equations. There will be many more lessons on this topic: we will look at more complex examples, where there will be more variables, and the equations themselves will already be nonlinear. See you again!