In this lesson we will continue to study the method of solving systems of equations, namely the method of algebraic addition. First, let's look at the application of this method using an example linear equations and its essence. Let's also remember how to equalize coefficients in equations. And we will solve a number of problems using this method.
Topic: Systems of equations
Lesson: Algebraic addition method
1. Method of algebraic addition using linear systems as an example
Let's consider algebraic addition method using the example of linear systems.
Example 1. Solve the system
If we add these two equations, then y cancels out, leaving an equation for x.
If we subtract the second from the first equation, the x's cancel each other out, and we get an equation for y. This is the meaning of the algebraic addition method.
We solved the system and remembered the method of algebraic addition. Let's repeat its essence: we can add and subtract equations, but we must ensure that we get an equation with only one unknown.
2. Method of algebraic addition with preliminary equalization of coefficients
Example 2. Solve the system
The term is present in both equations, so the algebraic addition method is convenient. Let's subtract the second from the first equation.
Answer: (2; -1).
Thus, after analyzing the system of equations, you can see that it is convenient for the method of algebraic addition, and apply it.
Let's consider another linear system.
3. Solution of nonlinear systems
Example 3. Solve the system
We want to get rid of y, but the coefficients of y are different in the two equations. Let's equalize them; to do this, multiply the first equation by 3, the second by 4.
Example 4. Solve the system 
Let's equalize the coefficients for x
You can do it differently - equalize the coefficients for y.
We solved the system by applying the algebraic addition method twice.
The algebraic addition method is also applicable to solving nonlinear systems.
Example 5. Solve the system
Let's add these equations together and we'll get rid of y.
The same system can be solved by applying the algebraic addition method twice. Let's add and subtract from one equation another.
Example 6. Solve the system 
Answer: 
Example 7. Solve the system 
Using the method of algebraic addition we will get rid of the xy term. Let's multiply the first equation by .
The first equation remains unchanged, instead of the second we write the algebraic sum.
Answer: 
Example 8. Solve the system
Multiply the second equation by 2 to isolate a perfect square.
Our task was reduced to solving four simple systems.
4. Conclusion
We examined the method of algebraic addition using the example of solving linear and nonlinear systems. In the next lesson we will look at the method of introducing new variables.
1. Mordkovich A.G. et al. Algebra 9th grade: Textbook. For general education Institutions.- 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.
2. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill.
3. Makarychev Yu. N. Algebra. 9th grade: educational. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. — 7th ed., rev. and additional - M.: Mnemosyne, 2008.
4. Alimov Sh. A., Kolyagin Yu. M., Sidorov Yu. V. Algebra. 9th grade. 16th ed. - M., 2011. - 287 p.
5. Mordkovich A. G. Algebra. 9th grade. In 2 hours. Part 1. Textbook for students of general education institutions / A. G. Mordkovich, P. V. Semenov. — 12th ed., erased. - M.: 2010. - 224 p.: ill.
6. Algebra. 9th grade. In 2 parts. Part 2. Problem book for students of general education institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. — 12th ed., rev. - M.: 2010.-223 p.: ill.
1. College section. ru in mathematics.
2. Internet project “Tasks”.
3. Educational portal“I WILL SOLVE THE USE.”
1. Mordkovich A.G. et al. Algebra 9th grade: Problem book for students of general education institutions / A.G. Mordkovich, T.N. Mishustina et al. - 4th ed. - M.: Mnemosyne, 2002.-143 p.: ill. No. 125 - 127.
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A system of linear equations with two unknowns is two or more linear equations for which it is necessary to find all their common 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, use equivalent transformations to equalize the coefficients of one of the unknown variables in both equations.
2. By adding or subtracting the resulting equations, obtain a linear equation with one unknown
3. Solve the resulting equation with one unknown and find one of the variables.
4. Substitute the resulting expression into any of the two equations of the system and solve this equation, thus obtaining the second variable.
5. Check the solution.
An example of a solution using the addition method
For greater clarity, let us solve the following system of linear equations with two unknowns using the addition method:
(3*x + 2*y = 10;
(5*x + 3*y = 12;
Since none of the variables have identical coefficients, we equalize the coefficients of the variable y. To do this, multiply the first equation by three, and the second equation by two.
(3*x+2*y=10 |*3
(5*x + 3*y = 12 |*2
We get the following system of equations:
(9*x+6*y = 30;
(10*x+6*y=24;
Now we subtract the first from the second equation. We present similar terms and solve the resulting linear equation.
10*x+6*y - (9*x+6*y) = 24-30; x=-6;
We substitute the resulting value into the first equation from our original system and solve the resulting equation.
(3*(-6) + 2*y =10;
(2*y=28; y =14;
The result is a pair of numbers x=6 and y=14. We are checking. Let's make a substitution.
(3*x + 2*y = 10;
(5*x + 3*y = 12;
{3*(-6) + 2*(14) = 10;
{5*(-6) + 3*(14) = 12;
{10 = 10;
{12=12;
As you can see, we got two correct equalities, therefore, we found the correct solution.
Systems of equations are widely used in the economic sector in mathematical modeling various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.
Systems of equations are used not only in mathematics, but also in physics, chemistry and biology, when solving problems of finding population size.
A system of linear equations is two or more equations with several variables for which it is necessary to find a common solution. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.
Linear equation
Equations of the form ax+by=c are called linear. The designations x, y are the unknowns whose value must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving an equation by plotting it will look like a straight line, all points of which are solutions to the polynomial.
Types of systems of linear equations
The simplest examples are considered to be systems of linear equations with two variables X and Y.
F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.
Solve system of equations - this means finding values (x, y) at which the system turns into a true equality or establishing that suitable values of x and y do not exist.
A pair of values (x, y), written as the coordinates of a point, is called a solution to a system of linear equations.
If systems have one common solution or no solution exists, they are called equivalent.
Homogeneous systems of linear equations are systems whose right-hand side is equal to zero. If the right part after the equal sign has a value or is expressed by a function, such a system is heterogeneous.
The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three or more variables.
When faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not the case. The number of equations in the system does not depend on the variables; there can be as many of them as desired.
Simple and complex methods for solving systems of equations
There is no general analytical method for solving such systems; all methods are based on numerical solutions. IN school course Mathematics describes in detail such methods as permutation, algebraic addition, substitution, as well as graphical and matrix methods, solution by the Gaussian method.
The main task when teaching solution methods is to teach how to correctly analyze the system and find the optimal solution algorithm for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of using a particular method
Solving examples of systems of linear equations in the 7th grade general education curriculum is quite simple and explained in great detail. In any mathematics textbook, this section is given enough attention. Solving examples of systems of linear equations using the Gauss and Cramer method is studied in more detail in the first years of higher education.
Solving systems using the substitution method
The actions of the substitution method are aimed at expressing the value of one variable in terms of the second. The expression is substituted into the remaining equation, then it is reduced to a form with one variable. The action is repeated depending on the number of unknowns in the system
Let us give a solution to an example of a system of linear equations of class 7 using the substitution method:
As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Solution this example does not cause difficulties and allows you to obtain the Y value. The last step is to check the obtained values.
It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and expressing the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, solving by substitution is also inappropriate.
Solution of an example of a system of linear inhomogeneous equations:
Solution using algebraic addition
When searching for solutions to systems using the addition method, they perform term-by-term addition and multiplication of equations by different numbers. The ultimate goal of mathematical operations is an equation in one variable.
Application of this method requires practice and observation. Solving a system of linear equations using the addition method when there are 3 or more variables is not easy. Algebraic addition is convenient to use when equations contain fractions and decimals.
Solution algorithm:
- Multiply both sides of the equation by a certain number. As a result of the arithmetic operation, one of the coefficients of the variable should become equal to 1.
- Add the resulting expression term by term and find one of the unknowns.
- Substitute the resulting value into the 2nd equation of the system to find the remaining variable.
Method of solution by introducing a new variable
A new variable can be introduced if the system requires finding a solution for no more than two equations; the number of unknowns should also be no more than two.
The method is used to simplify one of the equations by introducing a new variable. The new equation is solved for the introduced unknown, and the resulting value is used to determine the original variable.
The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to the standard one quadratic trinomial. You can solve a polynomial by finding the discriminant.
It is necessary to find the value of the discriminant using the well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the factors of the polynomial. In the given example, a=1, b=16, c=39, therefore D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is one solution: x = -b / 2*a.
The solution for the resulting systems is found by the addition method.
Visual method for solving systems
Suitable for 3 equation systems. The method is to build on coordinate axis graphs of each equation included in the system. The coordinates of the points of intersection of the curves and will be general decision systems.
The graphical method has a number of nuances. Let's look at several examples of solving systems of linear equations in a visual way.
As can be seen from the example, for each line two points were constructed, the values of the variable x were chosen arbitrarily: 0 and 3. Based on the values of x, the values for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.
The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.
IN following example you need to find a graphical solution to a system of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.
As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.
The systems from examples 2 and 3 are similar, but when constructed it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether a system has a solution or not; it is always necessary to construct a graph.
The matrix and its varieties
Matrices are used to concisely write a system of linear equations. A matrix is a table special type filled with numbers. n*m has n - rows and m - columns.
A matrix is square when the number of columns and rows are equal. A matrix-vector is a matrix of one column with an infinitely possible number of rows. A matrix with ones along one of the diagonals and other zero elements is called identity.
An inverse matrix is a matrix, when multiplied by which the original one turns into a unit matrix; such a matrix exists only for the original square one.
Rules for converting a system of equations into a matrix
In relation to systems of equations, the coefficients and free terms of the equations are written as matrix numbers; one equation is one row of the matrix.
A matrix row is said to be nonzero if at least one element of the row is not zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.
The matrix columns must strictly correspond to the variables. This means that the coefficients of the variable x can be written only in one column, for example the first, the coefficient of the unknown y - only in the second.
When multiplying a matrix, all elements of the matrix are sequentially multiplied by a number.
Options for finding the inverse matrix
The formula for finding the inverse matrix is quite simple: K -1 = 1 / |K|, where K -1 is the inverse matrix, and |K| is the determinant of the matrix. |K| must not be equal to zero, then the system has a solution.
The determinant is easily calculated for a two-by-two matrix; you just need to multiply the diagonal elements by each other. For the “three by three” option there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the numbers of columns and rows of elements are not repeated in the work.
Solving examples of systems of linear equations using the matrix method
The matrix method of finding a solution allows you to reduce cumbersome entries when solving systems with a large number of variables and equations.
In the example, a nm are the coefficients of the equations, the matrix is a vector x n are variables, and b n are free terms.
Solving systems using the Gaussian method
In higher mathematics, the Gaussian method is studied together with the Cramer method, and the process of finding solutions to systems is called the Gauss-Cramer solution method. These methods are used to find variable systems with a large number of linear equations.
The Gauss method is very similar to solutions by substitution and algebraic addition, but is more systematic. In the school course, the solution by the Gaussian method is used for systems of 3 and 4 equations. The purpose of the method is to reduce the system to the form of an inverted trapezoid. By algebraic transformations and substitutions, the value of one variable is found in one of the equations of the system. The second equation is an expression with 2 unknowns, while 3 and 4 are, respectively, with 3 and 4 variables.
After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.
In school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:
As can be seen from the example, at step (3) two equations were obtained: 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. Solving any of the equations will allow you to find out one of the variables x n.
Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.
The Gaussian method is difficult for students to understand high school, but is one of the most interesting ways to develop the ingenuity of children enrolled in advanced study programs in mathematics and physics classes.
For ease of recording, calculations are usually done as follows:
The coefficients of the equations and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right. Roman numerals indicate the numbers of equations in the system.
First, write down the matrix to be worked with, then all the actions carried out with one of the rows. The resulting matrix is written after the "arrow" sign and the necessary algebraic operations are continued until the result is achieved.
The result should be a matrix in which one of the diagonals is equal to 1, and all other coefficients are equal to zero, that is, the matrix is reduced to a unit form. We must not forget to perform calculations with numbers on both sides of the equation.
This recording method is less cumbersome and allows you not to be distracted by listing numerous unknowns.
The free use of any solution method will require care and some experience. Not all methods are of an applied nature. Some methods of finding solutions are more preferable in a particular area of human activity, while others exist for educational purposes.
Using this mathematical program, you can solve a system of two linear equations with 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 in 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.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems 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 whole numbers, but also fractions in the form of decimals and ordinary fractions.
Rules for entering decimal fractions.
Integer and fractional parts in 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 a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Examples.
-1&2/3y + 5/3x = 55
2.1p + 55 = -2/7(3.5p - 2&1/8q)
Solve system of equations
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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.
Books (textbooks) Abstracts of the Unified State Examination and the Unified State Examination tests online Games, puzzles Plotting graphs of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary educational institutions of Russia Catalog of Russian universities List of tasksWith 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:
- Look at the system and choose a variable that has identical (or opposite) coefficients in each equation;
- Perform algebraic subtraction (for opposite numbers - addition) of equations from each other, and then bring similar terms;
- 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:
- Addition method;
- 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 certainly be among the roots of the 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:
- 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.
- We substitute the found variable into any of the system equations to find the second one.
- 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 $y$.
- 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, an additional technique is used, 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 No. 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, however, 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:
- We look at the system and analyze each equation.
- 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.
- We find one variable.
- We substitute the found variable into one of the two equations of the system and find the second.
- 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 No. 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 coefficients, but for none of the variables do the coefficients fit into each other 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 obtained 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!