Instruction

Addition method.
You need to write two strictly under 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 of this system of equations: x=116, y=11.

Graphic way.
It consists in the practical finding of the coordinates of the point at which the lines are mathematically written in the system of equations. You should draw graphs of both lines separately in the same coordinate system. General view: - y \u003d kx + 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 \u003d 4

Y \u003d -3x + 1.
A straight line is built according to the first one, for convenience it needs to be written down: y \u003d 2x-4. Come up with (easier) values ​​for x, substituting it into the equation, solving it, find y. Two points are obtained, along which a straight line is built. (see pic.)
x 0 1

y -4 -2
A straight line is constructed according to the second equation: y \u003d -3x + 1.
Also build a line. (see pic.)

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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Helpful advice

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

Sources:

• Algebra Grade 8
• 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 certain number of variables. There are several ways to solve them.

You will need

• -Ruler and pencil;
• -calculator.

Instruction

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 members. When applying this method, each system is the coordinates of the points corresponding to each equation. First, in each case, express one variable in terms of the other. Then set the x variable to any number of values. Two is enough. Plug into the equation and find y. Build a coordinate system, mark the obtained points on it and draw a straight 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 inconsistent if they are parallel to each other. And it has infinitely many solutions when the lines merge with each other.

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

Any solution to a system of equations is worth checking. To do this, substitute the obtained values ​​instead of the variables. You can also find its solution in several ways. 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 do a certain set of actions with these numbers.

You will need

• - paper;
• - Pen or pencil.

Instruction

Imagine that you have 8 rabbits in front of you, and you only have 5 carrots. Think you need to buy more carrots so that each rabbit gets a carrot.

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

When you substituted a number for x, you were doing the same operation as subtracting 5 from 8. Thus, to find unknown term, subtract the known term from the sum.

Let's say you have 20 rabbits and only 5 carrots. Let's compose . An equation is an equality that holds only for certain values ​​of the letters included in it. The letters whose values ​​you want to find are called. Write an equation with one unknown, call it x. When solving our problem about rabbits, the following equation is obtained: 5 + x = 20.

Let's find the difference between 20 and 5. When subtracting, the number from which it is subtracted is 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 rabbits.

Make a check: 5 + 15 = 20. The equation is correct. Of course, when it comes to such simple , the check is not necessary. However, when it comes to equations with three-digit, four-digit, and so on, it is imperative to perform a check in order 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, it is necessary 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 the Cramer method. Cramer's method, in addition to solving the system, allows one to evaluate whether the system is solvable before finding the values ​​of the unknowns.

Instruction

The substitution method consists in sequentially one unknown through two others and substituting the result obtained 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 get a linear expression for z through the coefficients of the equations of the system. Now go "back": plug z into the second equation and find y, then plug z and y into the first equation and find x. The process is generally shown in the figure until z is found. Further, the record in general form will be too cumbersome, in practice, substituting , you can quite easily find all three unknowns.

Cramer's method consists in compiling the matrix of the system and calculating the determinant of this matrix, as well as three more auxiliary matrices. The matrix of the system is composed of the coefficients at the unknown terms of the equations. The column containing the numbers on the right side of the equations, the column of the right side. 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 supply 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.

Instruction

If two of the three systems have only two of the three unknowns, try expressing some variables in terms of the others and plugging them into the equation with three unknown. Your goal with this is to turn it into a normal the equation with the unknown. 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 by or a variable so that two unknowns are reduced at once. If there is such an opportunity, use it, most likely, the subsequent decision will not be difficult. Do not forget that when multiplying by a number, you must multiply both the left side and the right side. Similarly, when subtracting equations, remember that the right hand side must also be subtracted.

If the previous methods did not help, use the general method for solving any equations with three unknown. To do this, rewrite the equations in the form a11x1 + a12x2 + a13x3 \u003d b1, a21x1 + a22x2 + a23x3 \u003d b2, a31x1 + a32x2 + a33x3 \u003d b3. Now make a matrix of coefficients at x (A), a matrix of unknowns (X) and a matrix of free ones (B). Pay attention, multiplying the matrix of coefficients by the matrix of unknowns, you will get a matrix, a matrix of free members, that is, A * X \u003d B.

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

You can also find a solution to a system of three equations using the Cramer method. To do this, find the third-order determinant ∆ corresponding to the matrix of the system. Then sequentially 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 of equations with three unknowns

Starting to solve a system of equations, figure out what these equations are. The methods of solving linear equations are well studied. Nonlinear equations are most often not solved. There are only one special cases, each of which is practically individual. Therefore, the study of methods of solution should begin with linear equations. Such equations can be solved even purely algorithmically.

Instruction

Begin the 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. So the coefficient a21 emphasizes the fact that it is written in the second equation in the first place. In the generally accepted notation, the system is written by equations located one below the other, jointly denoted by a curly bracket on the right or left (for more details, see Fig. 1a).

The numbering of the equations is arbitrary. Choose the simplest one, such as one in which one of the variables is preceded by a factor of 1, or at least an integer. If this is equation (1), then further express, say, the unknown Y in terms of X (the case of eliminating Y). To do this, transform (1) to the form a12*Y=b1-a11*X (or a11*X=b1-a12*Y if X is excluded)) 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 relationship between Y and X, finally get the second unknown Y=(a11* b2-a21*b1)/(a11*a22-a12*a21).

If the system were given with specific numerical coefficients, then the calculations would be less cumbersome. On the other hand, the general solution makes it possible to consider the fact that, for the unknowns found, they are exactly the same. Yes, and the numerators are visible some patterns of 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 should learn the general system of equations of n equations.

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

A system is a union of two or more equalities, each of which has two or more unknowns. There are two main ways to solve systems of linear equations that are used in the school curriculum. One of them is called the method, the other is the addition method.

Standard form of a system of two equations

In standard form, the first equation is a1*x+b1*y=c1, the second equation is a2*x+b2*y=c2, and so on. For example, in the case of two parts of the system in both given a1, a2, b1, b2, c1, c2 are some numerical coefficients presented in specific equations. In turn, x and y are unknowns whose values ​​need to be determined. The desired values ​​turn both equations simultaneously into true equalities.

Solution of the system by 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 a few simple steps. The first of these is to transform any of the equations in such a way that the numerical coefficients for the variable x or y in both equations coincide in absolute value, but differ in sign.

For example, let a system consisting of two equations be 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 factor of -2, which will lead it to the form -12x-4y=-12. The correct choice of the coefficient is one of the key tasks in the process of solving the system by the addition method, 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 equal in value but opposite in sign coefficients will lead it to the form -10x=-4. After that, it is necessary to solve this simple equation, from which it unambiguously follows that x=0.4.

The last step in the solution process is the substitution of the found value of one of the variables into any of the initial 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 system shown in the example.

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, an equality of the form 0.4 * 6 + 1.8 * 2 = 6 is obtained, which is correct.

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A linear equation with two variables has the general form ax + by + c = 0. In it, a, b and c are coefficients - some numbers; and x and y are variables - unknown numbers to be found.

The solution of a linear equation with two variables is a pair of numbers x and y, for which ax + by + c = 0 is a true equality.

A particular linear equation with two variables (for example, 3x + 2y - 1 = 0) has a set of solutions, that is, a set of pairs of numbers for which the equation is true. A linear equation with two variables is transformed into a linear function of the form y = kx + m, which is a straight line on the coordinate plane. The coordinates of all points lying on this line are solutions of a linear equation in two variables.

If two linear equations of the form ax + by + c = 0 are given and it is required to find such values ​​of x and y for which both of them will have solutions, then they say that it is necessary solve the system of equations. The system of equations is written under a common curly bracket. Example:

A system of equations can have no solution if the lines that are the graphs of the corresponding linear functions do not intersect (that is, they are parallel to each other). To conclude that there is no solution, it is enough to transform both linear equations with two variables to the form y = kx + m. If k is the same number in both equations, then the system has no solutions.

If a system of equations turns out to consist of two identical equations (which may not be obvious immediately, but after transformations), then it has an infinite number of solutions. In this case, we are talking about uncertainty.

In all other cases, the system has one solution. This conclusion can be drawn from the fact that any two non-parallel lines can intersect at only one point. It is this point of intersection that will lie both the first line and the second, that is, it will be the solution of both the first equation and the second. Therefore, to be a solution to a system of equations. However, it is necessary to stipulate the situations when certain restrictions are imposed on the values ​​of x and y (usually by the condition of the problem). For example, x > 0, y > 0. In this case, even if the system of equations has a solution, but it does not satisfy the condition, then it is concluded that the system of equations has no solutions under the given conditions.

There are three ways to solve a system of equations:

1. selection method. Most of the time this is very difficult to do.
2. Graphic method. When two lines are drawn on the coordinate plane (graphs of the functions of the corresponding equations) and their intersection point is found. This method may give inaccurate results if the coordinates of the intersection point are fractional numbers.
3. Algebraic methods. They are versatile and reliable.

We are already familiar with the concept of a linear equation in two unknowns. Equations can be present in one problem both individually and several equations at once. In such cases, the equations are combined into a system of equations.

What is a system of linear equations

System of equations are two or more equations for which it is necessary to find all their common solutions. Usually, to write a system of equations, they are written in a column and draw one common curly bracket. The system of linear equations is written below.

(4x + 3y = 6
( 2x + y = 4

This record means that a system of two equations is given, with two variables. If there were three equations in the system, then it would be a system of three equations. And so for any number of equations.

If all the equations present in the system are linear, then they say that a system of linear equations is given. In the example above, a system of two linear equations is just presented. As noted above, the system can have general solutions. We will discuss the term "general solution" below.

What is the solution?

A solution to a system of two equations with two unknowns is a pair of numbers (x, y) such that if these numbers are substituted into the equations of the system, then each of the equations of the system turns into a true equality.

For example, we have a system of two linear equations. The solution to the first equation will be all pairs of numbers that satisfy this equation.

For the second equation, the solution will be pairs of numbers that satisfy this equation. If there is such a pair of numbers that satisfies both the first and the second equation, then this pair of numbers will be the solution to the system of two linear equations with two unknowns.

Graphic solution

Graphically, the solution of a linear equation are all points of some line on the plane.

For a system of linear equations, we will have several lines (according to the number of equations). And the solution to the system of equations will be the point at which ALL lines intersect. If there is no such point, then the system will have no solutions. The point at which all lines intersect belongs to each of these lines, so the solution is called general.

By the way, plotting the equations of the system and finding their common point is one of the ways to solve the system of equations. This method is called graphic.

Other Ways to Solve Linear Equations

There are other ways to solve systems of linear equations with two variables. Basic methods for solving systems of linear equations with two unknowns.

We will analyze two types of solving systems of equations:

1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.

In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.

To solve system by term-by-term addition (subtraction) need:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.

The solution of the system is the intersection points of the graphs of the function.

Let us consider in detail the solution of systems using examples.

Example #1:

Let's solve by the substitution method

Solving the system of equations by the substitution method

2x+5y=1 (1 equation)
x-10y=3 (2nd equation)

1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y

2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1

3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2

The solution of the system of equations is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first paragraph where we expressed we substitute y there.
x=3+10y
x=3+10*(-0.2)=1

It is customary to write points in the first place, we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)

Example #2:

Let's solve by term-by-term addition (subtraction).

Solving a system of equations by the addition method

3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)

1. Select a variable, let's say we select x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.

3x-2y=1 |*2
6x-4y=2

2x-3y=-10 |*3
6x-9y=-30

2. From the first equation, subtract the second to get rid of the variable x. Solve the linear equation.
__6x-4y=2

5y=32 | :5
y=6.4

3. Find x. We substitute the found y in any of the equations, let's say in the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6

The point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)

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

More reliable than the graphical method discussed in the previous paragraph.

Substitution method

We used this method in the 7th grade to solve systems of 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 denoted 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 by the substitution method (see example 1 from § 4).

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 at the third step instead of x into the expression y through x obtained at the first step.
5. Write down the answer in the form of pairs of values ​​(x; y), which were found, respectively, in the third and fourth steps.

4) Substitute in turn each of the found values ​​of y into the formula x \u003d 5 - Zy. If then
5) Pairs (2; 1) and solutions of 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. We recall the essence of the method in the following example.

Example 2 Solve a system of equations

We multiply all the 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 algebraic addition of two equations of the original system, an equation was obtained that is 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 by the substitution method. From the second equation we find Substituting this expression instead of y into the first equation of the system, we obtain

It remains to substitute the found values ​​\u200b\u200bof x into the formula

If x = 2 then

Thus, we have found two solutions to the system:

Method for introducing new variables

You got acquainted with 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 a system of equations

Let's introduce a new variable Then the first equation of the system can be rewritten in a simpler form: Let's solve this equation with respect to the variable t:

Both of these values ​​satisfy the condition , and therefore are the roots of a rational equation with the variable t. But that means either from where we find that x = 2y, or
Thus, using the method of introducing a new variable, we managed, as it were, to “stratify” the first equation of the system, which is 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 \u003d 3, which we have not yet remembered. In other words, the problem is reduced to solving two systems of equations:

It is necessary to find solutions for 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: we substitute the expression 2y instead of x into the second equation of the system. Get

Since x \u003d 2y, we find x 1 \u003d 2, x 2 \u003d 2, respectively. Thus, two solutions to 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: we substitute the expression 2x instead of y in the second equation of the system. Get

This equation has no roots, which means that the system of equations has no solutions. Thus, only the solutions of the first system should be included in the answer.

Answer: (2; 1); (-2;-1).

The method of introducing new variables in 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. The 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 a system of equations

Let's introduce two new variables:

We learn that then

This will allow us to rewrite the given system in a much simpler form, but with respect to the new variables a and b:

Since a \u003d 1, then from the equation a + 6 \u003d 2 we find: 1 + 6 \u003d 2; 6=1. Thus, for the variables a and b, we got one solution:

Returning to the variables x and y, we obtain the system of equations

We apply the algebraic addition method to solve this system:

Since then from the equation 2x + y = 3 we find:
Thus, for the variables x and y, we got one solution:

Let us conclude this section with a brief but rather serious theoretical discussion. You have already gained some experience in solving various 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 section, we introduced the notion 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 said to be equivalent if they have the same solutions or if both systems have no solutions.

All three methods (substitution, algebraic addition, and introduction of new variables) that we have 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. And 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 this system and are in the same coordinate plane, and also where it is required to find the intersection 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 common for a graphical system of equations to have either one single correct solution, or an infinite number of solutions, or not to have solutions at all.

Now let's take a closer look at each of these solutions. And so, the system of equations can have a unique solution if the lines, which are the graphs of the equations of the system, intersect. If these lines are parallel, then such a system of equations has absolutely no solutions. In the case of the coincidence of the direct graphs of the equations of the system, then such a system allows you to find many solutions.

Well, now let's take a look at the algorithm for solving a system of two equations with 2 unknowns using a graphical method:

First, at first we build a graph of the 1st equation;
The second step will be to plot 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 with an example. We are given a system of equations to be solved:

Solving Equations

1. First, we will build a graph of this equation: x2+y2=9.

But it should be noted that this graph of equations will be a circle centered at the origin, and its radius will be equal to three.

2. Our next step will be to plot an equation such as: y = x - 3.

In this case, we must build a line and find the points (0;−3) and (3;0).

3. Let's see what we got. We see that the 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 at the intersection of a straight line with a circle are precisely the solutions of both equations of the system. And from this it follows that these numbers are also solutions of this system of equations.

That is, the answer of this solution is the numbers: (3;0) and (0;−3).