A linear equation in two variables is any equation that has the following form: a*x + b*y =с. Here x and y are two variables, a,b,c are some numbers.
Below are a few examples of linear equations.
1. 10*x + 25*y = 150;
Like equations with one unknown, a linear equation with two variables (unknowns) also has a solution. For example, the linear equation x-y=5, with x=8 and y=3 turns into the correct identity 8-3=5. In this case, the pair of numbers x=8 and y=3 is said to be a solution to the linear equation x-y=5. You can also say that a pair of numbers x=8 and y=3 satisfies the linear equation x-y=5.
Solving a Linear Equation
Thus, the solution to the linear equation a*x + b*y = c is any pair of numbers (x,y) that satisfies this equation, that is, turns the equation with variables x and y into a correct numerical equality. Notice how the pair of numbers x and y are written here. This entry is shorter and more convenient. You just need to remember that the first place in such a record is the value of the variable x, and the second is the value of the variable y.
Please note that the numbers x=11 and y=8, x=205 and y=200 x= 4.5 and y= -0.5 also satisfy the linear equation x-y=5, and therefore are solutions to this linear equation.
Solving a linear equation with two unknowns is not the only one. Every linear equation in two unknowns has infinitely many different solutions. That is, there is infinitely many different two numbers x and y that convert a linear equation into a true identity.
If several equations with two variables have identical solutions, then such equations are called equivalent equations. It should be noted that if equations with two unknowns do not have solutions, then they are also considered equivalent.
Basic properties of linear equations with two unknowns
1. Any of the terms in the equation can be transferred from one part to another, but it is necessary to change its sign to the opposite one. The resulting equation will be equivalent to the original one.
2. Both sides of the equation can be divided by any number that is not zero. As a result, we obtain an equation equivalent to the original one.
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 “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. fold the left side with the left, and the right with the right:
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, written in general form 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 follows in 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 his father, so indicate this in the conditions of the problem.
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 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 ones (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. The more complex the system, the more interesting it is to solve. Most often in secondary school mathematics 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 - a positive and negative value of the 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.
Equality f(x; y) = 0 represents an equation with two variables. The solution to such an equation is a pair of variable values that turns the equation with two variables into a true equality.
If we have an equation with two variables, then, by tradition, we must put x in first place and y in second place.
Consider the equation x – 3y = 10. Pairs (10; 0), (16; 2), (-2; -4) are solutions to the equation under consideration, while pair (1; 5) is not a solution.
To find other pairs of solutions to this equation, it is necessary to express one variable in terms of another - for example, x in terms of y. As a result, we get the equation
x = 10 + 3y. Let's calculate the values of x by choosing arbitrary values of y.
If y = 7, then x = 10 + 3 ∙ 7 = 10 + 21 = 31.
If y = -2, then x = 10 + 3 ∙ (-2) = 10 – 6 = 4.
Thus, pairs (31; 7), (4; -2) are also solutions to the given equation.
If equations with two variables have the same roots, then such equations are called equivalent.
For equations with two variables, theorems on equivalent transformations of equations are valid.
Consider the graph of an equation with two variables.
Let an equation with two variables f(x; y) = 0 be given. All its solutions can be represented by points on the coordinate plane, obtaining a certain set of points on the plane. This set of points on the plane is called the graph of the equation f(x; y) = 0.
Thus, the graph of the equation y – x 2 = 0 is the parabola y = x 2; the graph of the equation y – x = 0 is a straight line; the graph of the equation y – 3 = 0 is a straight line parallel to the x axis, etc.
An equation of the form ax + by = c, where x and y are variables and a, b and c are numbers, is called linear; the numbers a, b are called coefficients of the variables, c is the free term.
The graph of the linear equation ax + by = c is:
Let's plot the equation 2x – 3y = -6.
1. Because none of the coefficients of the variables is equal to zero, then the graph of this equation will be a straight line.
2. To construct a straight line, we need to know at least two of its points. Substitute the x values into the equations and get the y values and vice versa:
if x = 0, then y = 2; (0 ∙ x – 3y = -6);
if y = 0, then x = -3; (2x – 3 ∙ 0 = -6). 
So, we got two points on the graph: (0; 2) and (-3; 0).
3. Let’s draw a straight line through the obtained points and get a graph of the equation
2x – 3y = -6.
If the linear equation ax + by = c has the form 0 ∙ x + 0 ∙ y = c, then we must consider two cases:
1. c = 0. In this case, any pair (x; y) satisfies the equation, and therefore the graph of the equation is the entire coordinate plane;
2. c ≠ 0. In this case, the equation has no solution, which means its graph does not contain a single point.
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Subject:Linear function
Lesson:Linear equation in two variables and its graph
We became familiar with the concepts of a coordinate axis and a coordinate plane. We know that each point on the plane uniquely defines a pair of numbers (x; y), with the first number being the abscissa of the point, and the second being the ordinate.
We will very often encounter a linear equation in two variables, the solution of which is a pair of numbers that can be represented on the coordinate plane.
Equation of the form:
Where a, b, c are numbers, and 
It is called a linear equation with two variables x and y. The solution to such an equation will be any such pair of numbers x and y, substituting which into the equation we will obtain the correct numerical equality.
A pair of numbers will be depicted on the coordinate plane as a point.
For such equations we will see many solutions, that is, many pairs of numbers, and all the corresponding points will lie on the same straight line.
Let's look at an example:
To find solutions to this equation you need to select the corresponding pairs of numbers x and y:
Let , then the original equation turns into an equation with one unknown:
,
That is, the first pair of numbers that is a solution to a given equation (0; 3). We got point A(0; 3)
Let . We get the original equation with one variable: 
, from here, we got point B(3; 0)
Let's put the pairs of numbers in the table:
Let's plot points on the graph and draw a straight line: 
Note that any point on a given line will be a solution to the given equation. Let's check - take a point with a coordinate and use the graph to find its second coordinate. It is obvious that at this point. Let's substitute this pair of numbers into the equation. We get 0=0 - a correct numerical equality, which means a point lying on a line is a solution.
For now, we cannot prove that any point lying on the constructed line is a solution to the equation, so we accept this as true and will prove it later.
Example 2 - graph the equation:
Let's make a table; we only need two points to construct a straight line, but we'll take a third one for control:
In the first column we took a convenient one, we will find it from:
, ,
In the second column we took a convenient one, let's find x:
, , ,
Let's check and find:
, ,
Let's build a graph:
Let's multiply the given equation by two:
From such a transformation, the set of solutions will not change and the graph will remain the same.
Conclusion: we learned to solve equations with two variables and build their graphs, we learned that the graph of such an equation is a straight line and that any point on this line is a solution to the equation
1. Dorofeev G.V., Suvorova S.B., Bunimovich E.A. and others. Algebra 7. 6th edition. M.: Enlightenment. 2010
2. Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7. M.: VENTANA-GRAF
3. Kolyagin Yu.M., Tkacheva M.V., Fedorova N.E. and others. Algebra 7.M.: Enlightenment. 2006
2. Portal for family viewing ().
Task 1: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 960, Art. 210;
Task 2: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 961, Art. 210;
Task 3: Merzlyak A.G., Polonsky V.B., Yakir M.S. Algebra 7, No. 962, Art. 210;
§ 1 Selection of equation roots in real situations
Let's consider this real situation:
The master and apprentice together made 400 custom parts. Moreover, the master worked for 3 days, and the student for 2 days. How many parts did each person make?
Let's create an algebraic model of this situation. Let the master produce parts in 1 day. And the student is at the details. Then the master will make 3 parts in 3 days, and the student will make 2 parts in 2 days. Together they will produce 3 + 2 parts. Since, according to the condition, a total of 400 parts were manufactured, we obtain the equation:
The resulting equation is called a linear equation in two variables. Here we need to find a pair of numbers x and y for which the equation will take the form of a true numerical equality. Note that if x = 90, y = 65, then we get the equality:
3 ∙ 90 + 65 ∙ 2 = 400
Since the correct numerical equality has been obtained, the pair of numbers 90 and 65 will be a solution to this equation. But the solution found is not the only one. If x = 96 and y = 56, then we get the equality:
96 ∙ 3 + 56 ∙ 2 = 400
This is also a true numerical equality, which means that the pair of numbers 96 and 56 is also a solution to this equation. But a pair of numbers x = 73 and y = 23 will not be a solution to this equation. In fact, 3 ∙ 73 + 2 ∙ 23 = 400 will give us the incorrect numerical equality 265 = 400. It should be noted that if we consider the equation in relation to this real situation, then there will be pairs of numbers that, being a solution to this equation, will not be a solution to the problem. For example, a couple of numbers:
x = 200 and y = -100
is a solution to the equation, but the student cannot make -100 parts, and therefore such a pair of numbers cannot be the answer to the question of the problem. Thus, in each specific real situation it is necessary to take a reasonable approach to selecting the roots of the equation.
Let's summarize the first results:
An equation of the form ax + bу + c = 0, where a, b, c are any numbers, is called a linear equation with two variables.
The solution to a linear equation in two variables is a pair of numbers corresponding to x and y, for which the equation turns into a true numerical equality.
§ 2 Graph of a linear equation
The very recording of the pair (x;y) leads us to think about the possibility of depicting it as a point with coordinates xy y on a plane. This means that we can obtain a geometric model of a specific situation. For example, consider the equation:
2x + y - 4 = 0
Let's select several pairs of numbers that will be solutions to this equation and construct points with the found coordinates. Let these be points:
A(0; 4), B(2; 0), C(1; 2), D(-2; 8), E(- 1; 6).
Note that all points lie on the same line. This line is called the graph of a linear equation in two variables. It is a graphical (or geometric) model of a given equation.
If a pair of numbers (x;y) is a solution to the equation
ax + vy + c = 0, then the point M(x;y) belongs to the graph of the equation. We can say the other way around: if the point M(x;y) belongs to the graph of the equation ax + y + c = 0, then the pair of numbers (x;y) is a solution to this equation.
From the geometry course we know:
To construct a straight line, you need 2 points, so to plot a graph of a linear equation with two variables, it is enough to know only 2 pairs of solutions. But guessing the roots is not always a convenient or rational procedure. You can act according to another rule. Since the abscissa of a point (variable x) is an independent variable, you can give it any convenient value. Substituting this number into the equation, we find the value of the variable y.
For example, let the equation be given:
Let x = 0, then we get 0 - y + 1 = 0 or y = 1. This means that if x = 0, then y = 1. A pair of numbers (0;1) is the solution to this equation. Let's set another value for the variable x: x = 2. Then we get 2 - y + 1 = 0 or y = 3. The pair of numbers (2;3) is also a solution to this equation. Using the two points found, it is already possible to construct a graph of the equation x - y + 1 = 0.
You can do this: first assign some specific value to the variable y, and only then calculate the value of x.
§ 3 System of equations
Find two natural numbers whose sum is 11 and difference is 1.
To solve this problem, we first create a mathematical model (namely, an algebraic one). Let the first number be x and the second number y. Then the sum of the numbers x + y = 11 and the difference of the numbers x - y = 1. Since both equations deal with the same numbers, these conditions must be met simultaneously. Usually in such cases a special record is used. The equations are written one below the other and combined with a curly brace.
Such a record is called a system of equations.
Now let’s construct sets of solutions to each equation, i.e. graphs of each of the equations. Let's take the first equation:
If x = 4, then y = 7. If x = 9, then y = 2.
Let's draw a straight line through points (4;7) and (9;2).
Let's take the second equation x - y = 1. If x = 5, then y = 4. If x = 7, then y = 6. We also draw a straight line through the points (5;4) and (7;6). We obtained a geometric model of the problem. The pair of numbers we are interested in (x;y) must be a solution to both equations. In the figure we see a single point that lies on both lines; this is the point of intersection of the lines.
Its coordinates are (6;5). Therefore, the solution to the problem will be: the first required number is 6, the second is 5.
List of used literature:
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 1, Textbook for general education institutions / A.G. Mordkovich. – 10th ed., revised – Moscow, “Mnemosyne”, 2007
- Mordkovich A.G., Algebra 7th grade in 2 parts, Part 2, Problem book for educational institutions / [A.G. Mordkovich and others]; edited by A.G. Mordkovich - 10th edition, revised - Moscow, “Mnemosyne”, 2007
- HER. Tulchinskaya, Algebra 7th grade. Blitz survey: a manual for students of general education institutions, 4th edition, revised and expanded, Moscow, “Mnemosyne”, 2008
- Alexandrova L.A., Algebra 7th grade. Thematic test papers in a new form for students of general education institutions, edited by A.G. Mordkovich, Moscow, “Mnemosyne”, 2011
- Alexandrova L.A. Algebra 7th grade. Independent works for students of general education institutions, edited by A.G. Mordkovich - 6th edition, stereotypical, Moscow, “Mnemosyne”, 2010