How to solve a system of equations with one unknown. Online calculator. Solving a system of two linear equations in two variables. Substitution and addition method

The article introduces the concept of defining a system of equations and its solution. Frequently encountered cases of system solutions will be considered. The examples provided will help explain the solution in detail.

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Definition of a system of equations

To move on to defining a system of equations, you need to pay attention to two points: the type of record and its meaning. To understand this, we need to dwell in detail on each of the types, then we can come to the definition of systems of equations.

For example, let’s take two equations 2 x + y = − 3 and x = 5, and then combine them with a curly brace like this:

2 x + y = - 3, x = 5.

Equations joined by curly braces are considered to be records of systems of equations. They define sets of solutions to the equations of a given system. Every decision must be everyone's decision given equations.

In other words, this means that any solutions to the first equation will be solutions to all equations combined by the system.

Definition 1

Systems of equations- this is a certain number of equations, united by a curly brace, having many solutions to the equations, which are simultaneously solutions for the entire system.

Main types of systems of equations

There are quite a lot of types of equations, like systems of equations. In order to make it convenient to solve and study, they are divided into groups according to certain characteristics. This will help in considering systems of equations of individual types.

To begin with, equations are classified by the number of equations. If there is one equation, then it is an ordinary equation; if there are more, then we are dealing with a system consisting of two or more equations.

Another classification concerns the number of variables. When the number of variables is 1, we say that we are dealing with a system of equations with one unknown, when 2 – with two variables. Let's look at an example

x + y = 5, 2 x - 3 y = 1

Obviously, the system of equations includes two variables x and y.

When writing such equations, the number of all variables present in the record is counted. Their presence in each equation is not necessary. At least one equation must have one variable. Let's consider an example of a system of equations

2 x = 11, x - 3 z 2 = 0, 2 7 x + y - z = - 3

This system has 3 variables x, y, z. The first equation has an explicit x and implicit y and z. Implicit variables are variables that have a 0 in the coefficient. The second equation has x and z, and y is an implicit variable. Otherwise it can be written like this

2 x + 0 y + 0 z = 11

And the other equation is x + 0 · y − 3 · z = 0.

The third classification of equations is type. They take place at school simple equations and systems of equations, starting with systems of two linear equations with two variables . This means that the system includes 2 linear equations. For example, consider

2 x - y = 1, x + 2 y = - 1 and - 3 x + y = 0. 5 , x + 2 2 3 y = 0

These are the basic simplest linear equations. Next, you may encounter systems containing 3 or more unknowns.

In 9th grade they solve equations with two variables and nonlinear ones. In whole equations, the degree is increased to increase complexity. Such systems are called systems of nonlinear equations with a certain number of equations and unknowns. Let's look at examples of such systems

x 2 - 4 x y = 1, x - y = 2 and x = y 3 x y = - 5

Both systems have two variables and both are nonlinear.

When solving one may encounter fractional rational equations. For example

x + y = 3, 1 x + 1 y = 2 5

They can simply call it a system of equations without specifying which ones. The type of system itself is rarely specified.

Senior grades move on to the study of irrational, trigonometric and exponential equations. For example,

x + y - x · y = 5 , 2 · x · y = 3 , x + y = 5 · π 2 , sin x + cos 2 y = - 1 , y - log 3 x = 1 , x y = 3 12 .

Higher education institutions study and research solutions to linear systems algebraic equations(SLAU). Left side of such equations contains polynomials with the first degree, and the right hand contains some numbers. The difference from school ones is that the number of variables and the number of equations can be arbitrary, most often not matching.

Solving systems of equations

Definition 2

Solving a system of equations with two variables is a pair of variables that, when substituted, turns each equation into a correct numerical inequality, that is, it is a solution for each equation of a given system.

For example, a pair of values ​​x = 5 and y = 2 are a solution to the system of equations x + y = 7, x - y = 3. Because when substituting, the equations turn into the correct numerical inequalities 5 + 2 = 7 and 5 − 2 = 3. If we substitute the pair x = 3 and y = 0, then the system will not be solved, since the substitution will not give the correct equation, namely, we get 3 + 0 = 7.

Let us formulate a definition for systems containing one or more variables.

Definition 3

Solving a system of equations with one variable– this is the value of the variable, which is the root of the equations of the system, which means that all equations will be converted into correct numerical equalities.

Let us consider the example of a system of equations with one variable t

t 2 = 4, 5 (t + 2) = 0

The number - 2 is a solution to the equation, since (− 2) · 2 = 4, and 5 · (− 2 + 2) = 0 are true numerical equalities. At t = 1, the system is not solved, since upon substitution we obtain two incorrect equalities 12 = 4 and 5 · (1 + 2) = 0.

Definition 4

Solving a system with three or more variables they call three, four, and further values, respectively, that turn all the equations of the system into correct equalities.

If we have the values ​​of the variables x = 1, y = 2, z = 0, then substituting them into the system of equations 2 · x = 2, 5 · y = 10, x + y + z = 3, we get 2 · 1 = 2, 5 · 2 = 10 and 1 + 2 + 0 = 3. This means that these numerical inequalities are correct. And the values ​​(1, 0, 5) will not be a solution, since, having substituted the values, the second of them will be incorrect, as well as the third: 5 0 = 10, 1 + 0 + 5 = 3.

Systems of equations may have no solutions at all or may have infinite set. This can be verified by in-depth study of this topic. We can come to the conclusion that a system of equations is the intersection of sets of solutions to all its equations. Let's expand on a few definitions:

Definition 5

Incompatible a system of equations is called when it has no solutions, otherwise it is called joint.

Definition 6

Uncertain a system is called when it has an infinite number of solutions, and certain with a finite number of solutions or in their absence.

Such terms are rarely used in school, as they are intended for higher education programs. educational institutions. Familiarity with equivalent systems will deepen your existing knowledge of solving systems of equations.

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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 terms, 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. Algebraic addition method: 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.

Lesson and presentation on the topic: "Systems of equations. Substitution method, addition method, method of introducing a new variable"

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Methods for solving systems of inequalities

Guys, we have studied systems of equations and learned how to solve them using graphs. Now let's see what other ways to solve systems exist?
Almost all the methods for solving them are no different from those we studied in 7th grade. Now we need to make some adjustments according to the equations that we have learned to solve.
The essence of all the methods described in this lesson is to replace the system with an equivalent system with more simple view and the method of solution. Guys, remember what an equivalent system is.

Substitution method

The first way to solve systems of equations with two variables is well known to us - this is the substitution method. We used this method to solve linear equations. Now let's see how to solve equations in the general case?

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 close 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

We also studied this method in 7th grade.
It is known that rational equation from two variables we can multiply by any number, not forgetting to multiply both sides of the equation. We multiplied one of the equations by a certain number so that when adding the resulting equation to the second equation of the system, one of the variables was destroyed. Then the equation was solved for the remaining variable.
This method still works, although it is not always possible to destroy one of the variables. But it allows you to significantly simplify the form of one of the equations.

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

We also studied this method, but let's look at it again.

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

Solve systems:
1. $\begin(cases)2x-2y=6,\\xy =-2\end(cases)$.
2. $\begin(cases)x+y^2=3, \\xy^2=4\end(cases)$.
3. $\begin(cases)xy+y^2=3,\\y^2-xy=5\end(cases)$.
4. $\begin(cases)\frac(2)(x)+\frac(1)(y)=4, \\\frac(1)(x)+\frac(3)(y)=9\ end(cases)$.
5. $\begin(cases)\frac(5)(x^2-xy)+\frac(4)(y^2-xy)=-\frac(1)(6), \\\frac(7 )(x^2-xy)-\frac(3)(y^2-xy)=\frac(6)(5)\end(cases)$.
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 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.

Solving a system of equations using the addition 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 overall coefficient 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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Solve the system with two unknowns - this means finding all pairs of variable values ​​that satisfy each of the given equations. Each such pair is called system solution.

Example:
The pair of values ​​\(x=3\);\(y=-1\) is a solution to the first system, because when substituting these threes and minus ones into the system instead of \(x\) and \(y\), both equations will become into the correct equalities \(\begin(cases)3-2\cdot (-1)=5 \\3 \cdot 3+2 \cdot (-1)=7 \end(cases)\)

But \(x=1\); \(y=-2\) - is not a solution to the first system, because after substitution the second equation “does not converge” \(\begin(cases)1-2\cdot(-2)=5 \\3\cdot1+2 \cdot(-2)≠7 \end(cases)\)

Note that such pairs are often written shorter: instead of "\(x=3\); \(y=-1\)" they write like this: \((3;-1)\).

How to solve a system of linear equations?

There are three main ways to solve systems of linear equations:

1. Substitution method.

\(\begin(cases)x-2y=5\\3x+2y=7 \end(cases)\)\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3x+2y= 7\end(cases)\)\(\Leftrightarrow\)

Substitute the resulting expression instead of this variable into another equation of the system.

\(\Leftrightarrow\) \(\begin(cases)x=5+2y\\3(5+2y)+2y=7\end(cases)\)\(\Leftrightarrow\)

1. \(\begin(cases)13x+9y=17\\12x-2y=26\end(cases)\)

   In the second equation, each term is even, so we simplify the equation by dividing it by \(2\).
   

   \(\begin(cases)13x+9y=17\\6x-y=13\end(cases)\)

   This system can be solved in any of the following ways, but it seems to me that the substitution method is the most convenient here. Let's express y from the second equation.
   

   \(\begin(cases)13x+9y=17\\y=6x-13\end(cases)\)

   Let's substitute \(6x-13\) instead of \(y\) into the first equation.
   

   \(\begin(cases)13x+9(6x-13)=17\\y=6x-13\end(cases)\)

   The first equation turned into an ordinary one. Let's solve it.

   First, let's open the brackets.
   

   \(\begin(cases)13x+54x-117=17\\y=6x-13\end(cases)\)

   Let's move \(117\) to the right and present similar terms.

   \(\begin(cases)67x=134\\y=6x-13\end(cases)\)

   Let's divide both sides of the first equation by \(67\).

   \(\begin(cases)x=2\\y=6x-13\end(cases)\)

   Hurray, we found \(x\)! Let's substitute its value into the second equation and find \(y\).

   \(\begin(cases)x=2\\y=12-13\end(cases)\)\(\Leftrightarrow\)\(\begin(cases)x=2\\y=-1\end(cases )\)

   Let's write down the answer.