Instructions

Addition method.
You need to write two strictly below each other:

549+45y+4y=-7, 45y+4y=549-7, 49y=542, y=542:49, y≈11.
In an arbitrarily chosen (from the system) equation, insert the number 11 instead of the already found “game” and calculate the second unknown:

X=61+5*11, x=61+55, x=116.
The answer to this system of equations is x=116, y=11.

Graphic method.
It consists of practically finding the coordinates of the point at which the lines are mathematically written in a system of equations. The graphs of both lines should be drawn separately in the same coordinate system. General view: – y=khx+b. To construct a straight line, it is enough to find the coordinates of two points, and x is chosen arbitrarily.
Let the system be given: 2x – y=4

Y=-3x+1.
A straight line is constructed using the first one, for convenience it should be written down: y=2x-4. Come up with (easier) values ​​for x, substituting it into the equation, solving it, and finding y. We get two points along which a straight line is constructed. (see picture)
x 0 1

y -4 -2
A straight line is constructed using the second equation: y=-3x+1.
Also construct a straight line. (see picture)

y 1 -5
Find the coordinates of the intersection point of two constructed lines on the graph (if the lines do not intersect, then the system of equations does not have - so).

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Useful advice

If the same system of equations is solved by three in different ways, the answer will be the same (if the solution is correct).

Sources:

• 8th grade algebra
• solve an equation with two unknowns online
• Examples of system solutions linear equations with two

System equations represents a collection mathematical notations, each of which contains a number of variables. There are several ways to solve them.

You will need

• -ruler and pencil;
• -calculator.

Instructions

Let's consider the sequence of solving the system, which consists of linear equations having the form: a1x + b1y = c1 and a2x + b2y = c2. Where x and y are unknown variables, and b,c are free terms. When applying this method, each system represents the coordinates of points corresponding to each equation. To begin, in each case, express one variable in terms of another. Then set the variable x to any number of values. Two is enough. Substitute into the equation and find y. Construct a coordinate system, mark the resulting points on it and draw a line through them. Similar calculations must be carried out for other parts of the system.

The system has the only solution, if the constructed lines intersect and one common point. It is incompatible if parallel to each other. And it has infinitely many solutions when the lines merge with each other.

This method considered very visual. The main disadvantage is that the calculated unknowns have approximate values. A more accurate result is given by the so-called algebraic methods.

Any solution to a system of equations is worth checking. To do this, substitute the resulting values ​​instead of the variables. You can also find its solution using several methods. If the solution of the system is correct, then everyone should turn out the same.

Often there are equations in which one of the terms is unknown. To solve an equation, you need to remember and perform a certain set of actions with these numbers.

You will need

• - a sheet of paper;
• - pen or pencil.

Instructions

Imagine that there are 8 rabbits in front of you, and you only have 5 carrots. Think about it, you still need to buy more carrots so that each rabbit gets one.

Let's present this problem in the form of an equation: 5 + x = 8. Let's substitute the number 3 in place of x. Indeed, 5 + 3 = 8.

When you substituted a number for x, you did the same thing as when you subtracted 5 from 8. So, to find unknown term, subtract the known term from the sum.

Let's say you have 20 rabbits and only 5 carrots. Let's make it up. An equation is an equality that holds only for certain values ​​of the letters included in it. The letters whose meanings need to be found are called . Write an equation with one unknown, call it x. When solving our rabbit problem, we get the following equation: 5 + x = 20.

Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is the one being reduced. The number that is being subtracted is called , and end result called difference. So, x = 20 – 5; x = 15. You need to buy 15 carrots for the rabbits.

Check: 5 + 15 = 20. The equation is solved correctly. Of course, when we're talking about about such simple ones, it is not necessary to perform a check. However, when you have equations with three-digit, four-digit, etc. numbers, you definitely need to check to be absolutely sure of the result of your work.

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Useful advice

To find the unknown minuend, you need to add the subtrahend to the difference.

To find the unknown subtrahend, you need to subtract the difference from the minuend.

Tip 4: How to solve a system of three equations with three unknowns

A system of three equations with three unknowns may not have solutions, despite sufficient quantity equations. You can try to solve it using the substitution method or using Cramer's method. Cramer's method, in addition to solving the system, allows you to evaluate whether the system is solvable before finding the values ​​of the unknowns.

Instructions

The substitution method consists of sequentially sequentially one unknown through two others and substituting the resulting result into the equations of the system. Let a system of three equations be given in general view:

a1x + b1y + c1z = d1

a2x + b2y + c2z = d2

a3x + b3y + c3z = d3

Express x from the first equation: x = (d1 - b1y - c1z)/a1 - and substitute into the second and third equations, then express y from the second equation and substitute into the third. You will obtain a linear expression for z through the coefficients of the system equations. Now go “backward”: substitute z into the second equation and solve for y, and then substitute z and y into the first and solve for x. The process is generally shown in the figure before finding z. Further writing in general form will be too cumbersome; in practice, by substituting , you can quite easily find all three unknowns.

Cramer's method consists of constructing a system matrix and calculating the determinant of this matrix, as well as three more auxiliary matrices. The system matrix is ​​composed of coefficients for the unknown terms of the equations. A column containing the numbers on the right-hand sides of equations, a column of right-hand sides. It is not used in the system, but is used when solving the system.

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Please note

All equations in the system must provide additional information independent of other equations. Otherwise, the system will be underdetermined and it will not be possible to find an unambiguous solution.

Useful advice

After solving the system of equations, substitute the found values ​​into the original system and check that they satisfy all the equations.

By itself equation with three unknown has many solutions, so most often it is supplemented by two more equations or conditions. Depending on what the initial data are, the course of the decision will largely depend.

You will need

• - a system of three equations with three unknowns.

Instructions

If two of the three systems have only two of the three unknowns, try to express some variables in terms of the others and substitute them into equation with three unknown. Your goal in this case is to turn it into normal equation with an unknown person. If this is , the further solution is quite simple - substitute the found value into other equations and find all the other unknowns.

Some systems of equations can be subtracted from one equation by another. See if it is possible to multiply one of or a variable so that two unknowns are canceled at once. If there is such an opportunity, take advantage of it; most likely, the subsequent solution will not be difficult. Don't forget that when multiplying by a number you must multiply as left side, and the right one. Likewise, when subtracting equations, it is necessary to remember that right side must also be deducted.

If the previous methods did not help, use in a general way solutions to any equations with three unknown. To do this, rewrite the equations in the form a11x1+a12x2+a13x3=b1, a21x1+a22x2+a23x3=b2, a31x1+a32x2+a33x3=b3. Now create a matrix of coefficients for x (A), a matrix of unknowns (X) and a matrix of free ones (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you get the matrix, matrix free members, that is, A*X=B.

Find matrix A to the power (-1) by first finding , note that it should not be equal to zero. After this, multiply the resulting matrix by matrix B, as a result you will receive the desired matrix X, indicating all the values.

You can also find a solution to a system of three equations using Cramer's method. To do this, find the third-order determinant ∆ corresponding to the system matrix. Then successively find three more determinants ∆1, ∆2 and ∆3, substituting the values ​​of the free terms instead of the values ​​of the corresponding columns. Now find x: x1=∆1/∆, x2=∆2/∆, x3=∆3/∆.

Sources:

• solutions to equations with three unknowns

When starting to solve a system of equations, figure out what kind of equations they are. Methods for solving linear equations have been studied quite well. Nonlinear equations most often they do not dare. There are only one special cases, each of which is practically individual. Therefore, the study of solution techniques should begin with linear equations. Such equations can even be solved purely algorithmically.

Instructions

Begin your learning process by learning how to solve a system of two linear equations with two unknowns X and Y by elimination. a11*X+a12*Y=b1 (1); a21*X+a22*Y=b2 (2). The coefficients of the equations are indicated by indices indicating their locations. Thus, the coefficient a21 emphasizes the fact that it is written in the first place in the second equation. In generally accepted notation, the system is written by equations located one below the other, collectively designated curly brace on the right or left (for more details, see Fig. 1a).

The numbering of equations is arbitrary. Choose the simplest one, such as one in which one of the variables is preceded by a coefficient of 1 or at least an integer. If this is equation (1), then further express, say, the unknown Y in terms of X (the case of excluding Y). To do this, transform (1) to the form a12*Y=b1-a11*X (or a11*X=b1-a12*Y when excluding X)), and then Y=(b1-a11*X)/a12. Substituting the latter into equation (2) write a21*X+a22*(b1-a11*X)/a12=b2. Solve this equation for X.
a21*X+a22*b1/a12-a11*a22*X/a12=b2; (a21-a11*a22/a12)*X=b2-a22*b1/a12;
X=(a12* b2-a22*b1)/(a12*a21-a11*a22) or X=(a22* b1-a12*b2)/(a11*a22-a12*a21).
Using the found connection between Y and X, you will finally obtain the second unknown Y=(a11* b2-a21*b1)/(a11*a22-a12*a21).

If the system had been specified with specific numerical coefficients, then the calculations would have been less cumbersome. But general solution makes it possible to consider the fact that the unknowns found are exactly the same. Yes, and the numerators show some patterns in their construction. If the dimension of the system of equations were greater than two, then the elimination method would lead to very cumbersome calculations. To avoid them, they are designed purely algorithmic methods solutions. The simplest of them is Cramer's algorithm (Cramer's formulas). For you should find out general system equations from n equations.

System n linear algebraic equations with n unknowns has the form (see Fig. 1a). In it, aij are the coefficients of the system,
xj – unknowns, bi – free terms (i=1, 2, ... , n; j=1, 2, ... , n). Such a system can be written compactly in matrix form AX=B. Here A is the matrix of system coefficients, X is the column matrix of unknowns, B is the column matrix of free terms (see Figure 1b). According to Cramer's method, each unknown xi =∆i/∆ (i=1,2…,n). The determinant ∆ of the coefficient matrix is ​​called the main one, and ∆i the auxiliary one. For each unknown, the auxiliary determinant is found by replacing the i-th column of the main determinant with a column of free terms. The Cramer method for the case of second- and third-order systems is presented in detail in Fig. 2.

The system is a combination of two or more equalities, each of which contains two or more unknowns. There are two main ways to solve systems of linear equations that are used within school curriculum. One of them is called the method, the other - the addition method.

Standard form of a system of two equations

Solving the system using the addition method

For example, suppose a system consisting of two equations is given. The first of them has the form 2x+4y=8, the second has the form 6x+2y=6. One of the options for completing the task is to multiply the second equation by a coefficient of -2, which will lead it to the form -12x-4y=-12. The right choice coefficient is one of the key tasks in the process of solving a system by addition, since it determines the entire further course of the procedure for finding unknowns.

Now it is necessary to add the two equations of the system. Obviously, the mutual destruction of variables with coefficients equal in value but opposite in sign will lead to the form -10x=-4. After this, it is necessary to solve this simple equation, from which it clearly follows that x = 0.4.

The last step in the solution process is to substitute the found value of one of the variables into any of the original equalities available in the system. For example, substituting x=0.4 into the first equation, you can get the expression 2*0.4+4y=8, from which y=1.8. Thus, x=0.4 and y=1.8 are the roots of the example system.

In order to make sure that the roots were found correctly, it is useful to check by substituting the found values ​​into the second equation of the system. For example, in in this case we get an equality of the form 0.4*6+1.8*2=6, which is true.

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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 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)

Do you want to prepare for exams for free? Tutor online for free. No joke.

Using this math program you can solve a system of two linear equations with two variable method substitution and 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

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

Rules for entering decimal fractions.
Whole and fractional part 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 for 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. 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.