More reliable than the graphical method discussed in the previous paragraph.
Substitution method
We used this method in 7th grade to solve systems linear equations. The algorithm that was developed in the 7th grade is quite suitable for solving systems of any two equations (not necessarily linear) with two variables x and y (of course, the variables can be designated by other letters, which does not matter). In fact, we used this algorithm in the previous paragraph, when the problem of a two-digit number led to a mathematical model, which is a system of equations. We solved this system of equations above using the substitution method (see example 1 from § 4).
An algorithm for using the substitution method when solving a system of two equations with two variables x, y.
1. Express y in terms of x from one equation of the system.
2. Substitute the resulting expression instead of y into another equation of the system.
3. Solve the resulting equation for x.
4. Substitute in turn each of the roots of the equation found in the third step instead of x into the expression y through x obtained in the first step.
5. Write the answer in the form of pairs of values (x; y), which were found in the third and fourth steps, respectively.
4) Substitute one by one each of the found values of y into the formula x = 5 - 3. If then 
5) Pairs (2; 1) and solutions to a given system of equations.
Answer: (2; 1);
Algebraic addition method
This method, like the substitution method, is familiar to you from the 7th grade algebra course, where it was used to solve systems of linear equations. Let us recall the essence of the method using the following example.
Example 2. Solve system of equations
Let's multiply all terms of the first equation of the system by 3, and leave the second equation unchanged: 
Subtract the second equation of the system from its first equation:
As a result of the algebraic addition of two equations of the original system, an equation was obtained that was simpler than the first and second equations of the given system. With this simpler equation we have the right to replace any equation of a given system, for example the second one. Then the given system of equations will be replaced by a simpler system:
This system can be solved using the substitution method. From the second equation we find. Substituting this expression instead of y into the first equation of the system, we get
It remains to substitute the found values of x into the formula
If x = 2 then
Thus, we found two solutions to the system: 
Method for introducing new variables
You were introduced to the method of introducing a new variable when solving rational equations with one variable in the 8th grade algebra course. The essence of this method for solving systems of equations is the same, but from a technical point of view there are some features that we will discuss in the following examples.
Example 3. Solve system of equations
Let's introduce a new variable. Then the first equation of the system can be rewritten 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 “stratify” the first equation of the system, which was quite complex in appearance, into two simpler equations:
x = 2 y; y - 2x.
What's next? And then each of the two simple equations obtained must be considered in turn in a system with the equation x 2 - y 2 = 3, which we have not yet remembered. In other words, the problem comes down to solving two systems of equations:
We need to find solutions to the first system, the second system and include all the resulting pairs of values in the answer. Let's solve the first system of equations:
Let's use the substitution method, especially since everything is ready for it here: let's substitute the expression 2y instead of x into the second equation of the system. We get
Since x = 2y, we find, respectively, x 1 = 2, x 2 = 2. Thus, two solutions of the given system are obtained: (2; 1) and (-2; -1). Let's solve the second system of equations:
Let's use the substitution method again: substitute the expression 2x instead of y into the second equation of the system. We get
This equation has no roots, which means the system of equations has no solutions. Thus, only the solutions of the first system need to be included in the answer.
Answer: (2; 1); (-2;-1).
The method of introducing new variables when solving systems of two equations with two variables is used in two versions. First option: one new variable is introduced and used in only one equation of the system. This is exactly what happened in example 3. Second option: two new variables are introduced and used simultaneously in both equations of the system. This will be the case in example 4.
Example 4. Solve system of equations
Let's introduce two new variables:
Let's take into account that then
This will allow you to rewrite this 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 the graphical solution method.
The method of solving systems of equations graphically is the construction of a graph for each of the specific equations that are included in a given 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 it is typical for a graphical system of equations to have either one single the right decision, either an infinite number of solutions, or no solutions at all.
Now let’s look at each of these solutions in more detail. And so, a system of equations can have a unique solution if the lines that are the graphs of the system’s equations intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. If the direct graphs of the equations of the system coincide, then such a system allows one to find many solutions.
Well, now let’s look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:
Firstly, first we build a graph of the 1st equation;
The second step will be to construct a graph that relates to the second equation;
Thirdly, we need to find the intersection points of the graphs.
And as a result, we get the coordinates of each intersection point, which will be the solution to the system of equations.
Let's look at this method in more detail using an example. We are given a system of equations that needs to be solved:
Solving equations
1. First, we will build a schedule given 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).
Instructions
Substitution MethodExpress one variable and substitute it into another equation. You can express any variable at your discretion. For example, express y from the second equation:
x-y=2 => y=x-2Then substitute everything into the first equation:
2x+(x-2)=10 Move everything without the “x” to the right side and calculate:
2x+x=10+2
3x=12 Next, to get x, divide both sides of the equation by 3:
x=4. So, you found “x. Find "y. To do this, substitute “x” into the equation from which you expressed “y”:
y=x-2=4-2=2
y=2.
Do a check. To do this, substitute the resulting values into the equations:
2*4+2=10
4-2=2
The unknowns have been found correctly!
A way to add or subtract equations Get rid of any variable right away. In our case, this is easier to do with “y.
Since in “y” there is a “+” sign, and in the second one “-”, then you can perform the addition operation, i.e. left side add it to the left one and add the right one to the right one:
2x+y+(x-y)=10+2Convert:
2x+y+x-y=10+2
3x=12
x=4Substitute “x” into any equation and find “y”:
2*4+y=10
8+y=10
y=10-8
y=2By the 1st method you can see that they were found correctly.
If there are no clearly defined variables, then it is necessary to slightly transform the equations.
In the first equation we have “2x”, and in the second we simply have “x”. In order for x to be reduced during addition, multiply the second equation by 2:
x-y=2
2x-2y=4Then subtract the second from the first equation:
2x+y-(2x-2y)=10-4 Note that if there is a minus before the bracket, then after opening, change it to the opposite:
2x+y-2x+2y=6
3у=6
find y=2x by expressing from any equation, i.e.
x=4
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Tip 2: How to solve a linear equation in two variables
The equation, V general view written ax+bу+c=0 is called a linear equation with two variables. Such an equation itself contains an infinite number of solutions, so in problems it is always supplemented with something - another equation or limiting conditions. Depending on the conditions provided by the problem, solve a linear equation with two variables should different ways.
You will need
- - linear equation with two variables;
- - second equation or additional conditions.
Instructions
Given a system of two linear equations, solve it as follows. Choose one of the equations in which the coefficients are variables smaller and express one of the variables, for example, x. Then substitute this value containing y into the second equation. In the resulting equation there will be only one variable y, move all parts with y to the left side, and free ones to the right. Find y and substitute into any of the original equations to find x.
There is another way to solve a system of two equations. Multiply one of the equations by a number so that the coefficient of one of the variables, such as x, is the same in both equations. Then subtract one of the equations from the other (if the right-hand side is not equal to 0, remember to subtract the right-hand sides in the same way). You will see that the x variable has disappeared and only one y variable remains. Solve the resulting equation, and substitute the found value of y into any of the original equalities. Find x.
The third way to solve a system of two linear equations is graphical. Draw a coordinate system and graph two straight lines whose equations are given in your system. To do this, substitute any two x values into the equation and find the corresponding y - these will be the coordinates of the points belonging to the line. The most convenient way to find the intersection with the coordinate axes is to simply substitute the values x=0 and y=0. The coordinates of the point of intersection of these two lines will be the tasks.
If there is only one linear equation in the problem conditions, then you have been given additional conditions through which you can find a solution. Read the problem carefully to find these conditions. If variables x and y indicate distance, speed, weight - feel free to set the limit x≥0 and y≥0. It is quite possible that x or y hides the number of apples, etc. – then the values can only be . If x is the son's age, it is clear that he cannot be older than father, so indicate this in the task conditions.
Sources:
- how to solve an equation with one variable
By itself the 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 the equation with three unknown. Your goal in this case is to turn it into normal the 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. Remember that when multiplying by a number, you must multiply both the left side and the right side. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.
If previous methods didn't help, use it 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 variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, 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
Solving a system of equations is challenging and exciting. How more complex system, the more interesting it is to solve it. Most often in mathematics high school There are systems of equations with two unknowns, but in higher mathematics there may be more variables. Systems can be solved using several methods.
Instructions
The most common method for solving a system of equations is substitution. To do this, you need to express one variable in terms of another and substitute it into the second the equation systems, thus leading the equation to one variable. For example, given the following equations: 2x-3y-1=0;x+y-3=0.
From the second expression it is convenient to express one of the variables, moving everything else to the right side of the expression, not forgetting to change the sign of the coefficient: x = 3-y.
Open the brackets: 6-2y-3y-1=0;-5y+5=0;y=1. We substitute the resulting value y into the expression: x=3-y;x=3-1;x=2.
In the first expression, all terms are 2, you can take 2 out of the bracket to the distributive property of multiplication: 2*(2x-y-3)=0. Now both parts of the expression can be reduced by this number, and then expressed as y, since the modulus coefficient for it is equal to one: -y = 3-2x or y = 2x-3.
Just as in the first case, we substitute this expression into the second the equation and we get: 3x+2*(2x-3)-8=0;3x+4x-6-8=0;7x-14=0;7x=14;x=2. Substitute the resulting value into the expression: y=2x -3;y=4-3=1.
We see that the coefficient for y is the same in value, but different in sign, therefore, if we add these equations, we will completely get rid of y: 4x+3x-2y+2y-6-8=0; 7x-14=0; x=2. Substitute the value of x into any of the two equations of the system and get y=1.
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Biquadratic the equation represents the equation fourth degree, the general form of which is represented by the expression ax^4 + bx^2 + c = 0. Its solution is based on the use of the method of substitution of unknowns. In this case, x^2 is replaced by another variable. Thus, the result is an ordinary square the equation, which needs to be solved.
Instructions
Solve the quadratic the equation, resulting from the replacement. To do this, first calculate the value in accordance with the formula: D = b^2? 4ac. In this case, the variables a, b, c are the coefficients of our equation.
Find the roots of the biquadratic equation. To do this, take the square root of the solutions obtained. If there was one solution, then there will be two - positive and negative meaning square root. If there were two solutions, the biquadratic equation will have four roots.
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One of the classical methods for solving systems of linear equations is the Gauss method. It consists in the sequential elimination of variables, when a system of equations using simple transformations is transformed into a stepwise system, from which all variables are sequentially found, starting with the last ones.
Instructions
First, bring the system of equations into a form where all the unknowns are in a strictly defined order. For example, all unknown X's will appear first on each line, all Y's will come after X's, all Z's will come after Y's, and so on. There should be no unknowns on the right side of each equation. Mentally determine the coefficients in front of each unknown, as well as the coefficients on the right side of each equation.
In the 7th grade mathematics course, we encounter for the first time equations with two variables, but they are studied only in the context of systems of equations with two unknowns. That is why it falls out of sight whole line problems in which certain conditions are introduced on the coefficients of the equation that limit them. In addition, methods for solving problems like “Solve an equation in natural or integer numbers” are also ignored, although in Unified State Exam materials And in entrance exams, problems of this kind are encountered more and more often.
Which equation will be called an equation with two variables?
So, for example, the equations 5x + 2y = 10, x 2 + y 2 = 20, or xy = 12 are equations in two variables.
Consider the equation 2x – y = 1. It becomes true when x = 2 and y = 3, so this pair of variable values is a solution to the equation in question.
Thus, the solution to any equation with two variables is a set of ordered pairs (x; y), values of the variables that turn this equation into a true numerical equality.
An equation with two unknowns can:
A) have one solution. For example, the equation x 2 + 5y 2 = 0 has a unique solution (0; 0);
b) have multiple solutions. For example, (5 -|x|) 2 + (|y| – 2) 2 = 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);
V) have no solutions. For example, the equation x 2 + y 2 + 1 = 0 has no solutions;
G) have infinitely many solutions. For example, x + y = 3. The solutions to this equation will be numbers whose sum is equal to 3. The set of solutions to this equation can be written in the form (k; 3 – k), where k is any real number.
The main methods for solving equations with two variables are methods based on factoring expressions, isolating a complete square, using the properties of a quadratic equation, limited expressions, and estimation methods. The equation is usually transformed into a form from which a system for finding the unknowns can be obtained.
Factorization
Example 1.
Solve the equation: xy – 2 = 2x – y.
Solution.
We group the terms for the purpose of factorization:
(xy + y) – (2x + 2) = 0. From each bracket we take out a common factor:
y(x + 1) – 2(x + 1) = 0;
(x + 1)(y – 2) = 0. We have:
y = 2, x – any real number or x = -1, y – any real number.
Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.
Equal to zero is not negative numbers
Example 2.
Solve the equation: 9x 2 + 4y 2 + 13 = 12(x + y).
Solution.
Grouping:
(9x 2 – 12x + 4) + (4y 2 – 12y + 9) = 0. Now each bracket can be folded using the squared difference formula.
(3x – 2) 2 + (2y – 3) 2 = 0.
The sum of two non-negative expressions is zero only if 3x – 2 = 0 and 2y – 3 = 0.
This means x = 2/3 and y = 3/2.
Answer: (2/3; 3/2).
Estimation method
Example 3.
Solve the equation: (x 2 + 2x + 2)(y 2 – 4y + 6) = 2.
Solution.
In each bracket we select a complete square:
((x + 1) 2 + 1)((y – 2) 2 + 2) = 2. Let’s estimate 
the meaning of the expressions in parentheses.
(x + 1) 2 + 1 ≥ 1 and (y – 2) 2 + 2 ≥ 2, then the left side of the equation is always at least 2. Equality is possible if:
(x + 1) 2 + 1 = 1 and (y – 2) 2 + 2 = 2, which means x = -1, y = 2.
Answer: (-1; 2).
Let's get acquainted with another method for solving equations with two variables of the second degree. This method consists of treating the equation as square with respect to some variable.
Example 4.
Solve the equation: x 2 – 6x + y – 4√y + 13 = 0.
Solution.
Let's solve the equation as a quadratic equation for x. Let's find the discriminant:
D = 36 – 4(y – 4√y + 13) = -4y + 16√y – 16 = -4(√y – 2) 2 . The equation will have a solution only when D = 0, that is, if y = 4. We substitute the value of y into the original equation and find that x = 3.
Answer: (3; 4).
Often in equations with two unknowns they indicate restrictions on variables.
Example 5.
Solve the equation in whole numbers: x 2 + 5y 2 = 20x + 2.
Solution.
Let's rewrite the equation in the form x 2 = -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives a remainder of 2. Therefore, x 2 is not divisible by 5. But the square of a number not divisible by 5 gives a remainder of 1 or 4. Thus, equality is impossible and there are no solutions.
Answer: no roots.
Example 6.
Solve the equation: (x 2 – 4|x| + 5)(y 2 + 6y + 12) = 3.
Solution.
Let's highlight perfect squares in each bracket:
((|x| – 2) 2 + 1)((y + 3) 2 + 3) = 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided |x| – 2 = 0 and y + 3 = 0. Thus, x = ± 2, y = -3.
Answer: (2; -3) and (-2; -3).
Example 7.
For every pair of negative integers (x;y) satisfying the equation
x 2 – 2xy + 2y 2 + 4y = 33, calculate the sum (x + y). Please indicate the smallest amount in your answer.
Solution.
Let's select complete squares:
(x 2 – 2xy + y 2) + (y 2 + 4y + 4) = 37;
(x – y) 2 + (y + 2) 2 = 37. Since x and y are integers, their squares are also integers. We get the sum of the squares of two integers equal to 37 if we add 1 + 36. Therefore:
(x – y) 2 = 36 and (y + 2) 2 = 1 
(x – y) 2 = 1 and (y + 2) 2 = 36.
Solving these systems and taking into account that x and y are negative, we find solutions: (-7; -1), (-9; -3), (-7; -8), (-9; -8).
Answer: -17.
Don't despair if you have difficulty solving equations with two unknowns. With a little practice, you can handle any equation.
Still have questions? Don't know how to solve equations in two variables?
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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
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.
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: /
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)
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 for 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.
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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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If you solve the same system of equations in three 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 solving systems of linear equations with two
System equations is a collection of mathematical records, 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 a unique solution if the constructed lines intersect and have 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. More exact result give 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
- - 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 subtracted is called , and the final result is called the 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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Helpful 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 a sufficient number of 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 form:
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 find 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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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.
Helpful 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 the 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 the equation with three unknown. Your goal in this case is to turn it into normal the 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. Remember that when multiplying by a number, you must multiply both the left side and the right side. Likewise, when subtracting equations, you must remember that the right-hand side must also be subtracted.
If the previous methods did not help, use the general method of solving 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 variables (B). Please note that by multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix of free terms, 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 and jointly denoted by a curly bracket 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 next 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 were specified with specific numerical coefficients, then the calculations would be less cumbersome. But common decision 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, purely algorithmic solutions have been developed. The simplest of them is Cramer's algorithm (Cramer's formulas). For you should find out general system equations from n equations.
A system of 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 determinant, and ∆i the auxiliary one. For every unknown auxiliary qualifier found by replacing the i-th column of the main determinant with the 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
At standard form the first equation has the form a1*x+b1*y=c1, the second equation has the form a2*x+b2*y=c2 and so on. For example, in the case of two parts of the system, both given a1, a2, b1, b2, c1, c2 are some numerical coefficients represented in specific equations. In turn, x and y represent unknowns whose values need to be determined. The required values turn both equations simultaneously into true equalities.Solving the system using the addition method
In order to solve the system, that is, to find those values of x and y that will turn them into true equalities, you need to take several simple steps. The first of them is to transform either equation so that the numerical coefficients for the variable x or y in both equations are the same in magnitude, but different in sign.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 this case we get an equality of the form 0.4*6+1.8*2=6, which is correct.
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