Solving linear equations with examples. Solving Equations in Two Variables How to Solve Equation 2 1

An equation with one unknown, which, after opening the brackets and reducing similar terms, takes the form

ax + b \u003d 0, where a and b are arbitrary numbers, is called linear equation with one unknown. Today we will figure out how to solve these linear equations.

For example, all equations:

2x + 3 \u003d 7 - 0.5x; 0.3x \u003d 0; x / 2 + 3 \u003d 1/2 (x - 2) - linear.

The value of the unknown that turns the equation into a true equality is called decision or root of the equation .

For example, if in the equation 3x + 7 \u003d 13 instead of the unknown x to substitute the number 2, then we get the correct equality 3 · 2 +7 \u003d 13. Hence, the value x \u003d 2 is the solution or the root of the equation.

And the value x \u003d 3 does not turn the equation 3x + 7 \u003d 13 into a true equality, since 3 · 2 +7 ≠ 13. Hence, the value x \u003d 3 is not a solution or a root of the equation.

The solution of any linear equations is reduced to the solution of equations of the form

ax + b \u003d 0.

We transfer the free term from the left side of the equation to the right, changing the sign in front of b to the opposite, we get

If a ≠ 0, then х \u003d - b / a .

Example 1. Solve the equation 3x + 2 \u003d 11.

Move 2 from the left side of the equation to the right, while changing the sign in front of 2 to the opposite, we get
3x \u003d 11 - 2.

Subtract, then
3x \u003d 9.

To find x, you need to divide the product by a known factor, that is
x \u003d 9: 3.

Hence, the value x \u003d 3 is the solution or the root of the equation.

Answer: x \u003d 3.

If a \u003d 0 and b \u003d 0, then we get the equation 0x \u003d 0. This equation has infinitely many solutions, since when multiplying any number by 0 we get 0, but b is also equal to 0. Any number is a solution to this equation.

Example 2.Solve the equation 5 (x - 3) + 2 \u003d 3 (x - 4) + 2x - 1.

Let's expand the brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.

5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.

Here are similar terms:
0x \u003d 0.

Answer: x is any number.

If a \u003d 0 and b ≠ 0, then we get the equation 0x \u003d - b. This equation has no solutions, since multiplying any number by 0 we get 0, but b ≠ 0.

Example 3.Solve the equation x + 8 \u003d x + 5.

Let us group the members containing unknowns on the left, and free members on the right:
x - x \u003d 5 - 8.

Here are similar terms:
0x \u003d - 3.

Answer: there are no solutions.

On picture 1 shows the scheme for solving the linear equation

Let's draw up a general scheme for solving equations with one variable. Consider the solution to Example 4.

Example 4. Let the equation be solved

1) Multiply all the terms of the equation by the least common multiple of the denominators, equal to 12.

2) After reduction, we get
4 (x - 4) + 32 (x + 1) - 12 \u003d 6 5 (x - 3) + 24x - 2 (11x + 43)

3) To separate the members containing unknown and free members, we expand the brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.

4) Let us group in one part the members containing unknowns, and in the other - free members:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.

5) Here are similar terms:
- 22x \u003d - 154.

6) Divide by - 22, We get
x \u003d 7.

As you can see, the root of the equation is seven.

Generally such equations can be solved according to the following scheme:

a) bring the equation to its whole form;

b) open the brackets;

c) group the terms containing the unknown in one part of the equation, and free terms in the other;

d) bring similar members;

e) solve an equation of the form ax \u003d b, which was obtained after bringing similar terms.

However, this scheme is not required for every equation. When solving many simpler equations, one has to start not with the first, but with the second ( Example. 2), third ( Example. thirteen) and even from the fifth stage, as in example 5.

Example 5.Solve the equation 2x \u003d 1/4.

Find the unknown x \u003d 1/4: 2,
x \u003d 1/8 .

Consider the solution of some linear equations found in the main state exam.

Example 6.Solve the equation 2 (x + 3) \u003d 5 - 6x.

2x + 6 \u003d 5 - 6x

2x + 6x \u003d 5 - 6

Answer: - 0, 125

Example 7.Solve the equation - 6 (5 - 3x) \u003d 8x - 7.

- 30 + 18x \u003d 8x - 7

18x - 8x \u003d - 7 +30

Answer: 2.3

Example 8. Solve the equation

3 (3x - 4) \u003d 4.7x + 24

9x - 12 \u003d 28x + 24

9x - 28x \u003d 24 + 12

Example 9.Find f (6) if f (x + 2) \u003d 3 7th

Decision

Since we need to find f (6), and we know f (x + 2),
then x + 2 \u003d 6.

Solve the linear equation x + 2 \u003d 6,
we get x \u003d 6 - 2, x \u003d 4.

If x \u003d 4, then
f (6) \u003d 3 7-4 \u003d 3 3 \u003d 27

Answer: 27.

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Let us consider two types of solutions to 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. Express one variable from any equation.
2. Substitute. We substitute the obtained value into another equation instead of the expressed variable.
3. Solve the resulting equation in one variable. We find a solution to the system.

To solve system by term-by-term addition (subtraction) need to:
1.Choose a variable for which we will make the same coefficients.
2. We add or subtract equations, in the end we get an equation with one variable.
3. 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's consider in detail the solution of systems using examples.

Example # 1:

Let's solve by substitution method

Solving a System of Equations by the Substitution Method

2x + 5y \u003d 1 (1 equation)
x-10y \u003d 3 (2 equation)

1. We express
It can be seen that in the second equation there is a variable x with a coefficient of 1, from which it turns out that it is easiest to express the variable x from the second equation.
x \u003d 3 + 10y

2. After we have expressed, we substitute 3 + 10y in the first equation instead of the variable x.
2 (3 + 10y) + 5y \u003d 1

3. Solve the resulting equation in one variable.
2 (3 + 10y) + 5y \u003d 1 (expand brackets)
6 + 20y + 5y \u003d 1
25y \u003d 1-6
25y \u003d -5 |: (25)
y \u003d -5: 25
y \u003d -0.2

The solution to 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. Find x, in the first paragraph where we expressed there we substitute y.
x \u003d 3 + 10y
x \u003d 3 + 10 * (- 0.2) \u003d 1

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

Example # 2:

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

Solving a system of equations by the addition method

3x-2y \u003d 1 (1 equation)
2x-3y \u003d -10 (2 equation)

1.Choose a variable, say, choose x. In the first equation the variable x has a coefficient of 3, in the second 2. It is necessary to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. The first equation is multiplied by 2, and the second by 3, and we get a total factor of 6.

3x-2y \u003d 1 | * 2
6x-4y \u003d 2

2x-3y \u003d -10 | * 3
6x-9y \u003d -30

2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y \u003d 2

5y \u003d 32 | :five
y \u003d 6.4

3. Find x. Substitute the found y into any of the equations, let's say in the first equation.
3x-2y \u003d 1
3x-2 * 6.4 \u003d 1
3x-12.8 \u003d 1
3x \u003d 1 + 12.8
3x \u003d 13.8 |: 3
x \u003d 4.6

The intersection point will be x \u003d 4.6; y \u003d 6.4
Answer: (4.6; 6.4)

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In the 7th grade mathematics course, they first meet with equations in two variables, but they are studied only in the context of systems of equations with two unknowns. That is why a whole series of problems drop out of sight, in which some conditions are introduced on the coefficients of the equation that limit them. In addition, methods for solving problems such as "Solve an equation in natural numbers or integers" are also left without attention, although problems of this kind are found more and more often in the USE materials and on entrance exams.

Which equation will be called a two-variable equation?

So, for example, the equations 5x + 2y \u003d 10, x 2 + y 2 \u003d 20, or xy \u003d 12 are equations in two variables.

Consider the equation 2x - y \u003d 1. It turns into a true equality for x \u003d 2 and y \u003d 3, so this pair of values \u200b\u200bof the variables is a solution to the equation under consideration.

Thus, the solution to any equation with two variables is the set of ordered pairs (x; y), the values \u200b\u200bof the variables that this equation turns into a true numerical equality.

An equation with two unknowns can:

and) have one solution. For example, the equation x 2 + 5y 2 \u003d 0 has a unique solution (0; 0);

b) have multiple solutions. For example, (5 - | x |) 2 + (| y | - 2) 2 \u003d 0 has 4 solutions: (5; 2), (-5; 2), (5; -2), (-5; - 2);

in) have no solutions. For example, the equation x 2 + y 2 + 1 \u003d 0 has no solutions;

d) have infinitely many solutions. For example, x + y \u003d 3. The solutions of this equation will be numbers, the sum of which is 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 evaluative methods. The equation is usually transformed into a form from which a system for finding unknowns can be obtained.

Factorization

Example 1.

Solve the equation: xy - 2 \u003d 2x - y.

Decision.

We group the terms for the purpose of factoring:

(xy + y) - (2x + 2) \u003d 0. Move out the common factor from each parenthesis:

y (x + 1) - 2 (x + 1) \u003d 0;

(x + 1) (y - 2) \u003d 0. We have:

y \u003d 2, x is any real number or x \u003d -1, y is any real number.

Thus, the answer is all pairs of the form (x; 2), x € R and (-1; y), y € R.

Equality to zero of non-negative numbers

Example 2.

Solve the equation: 9x 2 + 4y 2 + 13 \u003d 12 (x + y).

Decision.

We group:

(9x 2 - 12x + 4) + (4y 2 - 12y + 9) \u003d 0. Now each parenthesis can be folded using the squared difference formula.

(3x - 2) 2 + (2y - 3) 2 \u003d 0.

The sum of two non-negative expressions is zero only if 3x - 2 \u003d 0 and 2y - 3 \u003d 0.

This means that x \u003d 2/3 and y \u003d 3/2.

Answer: (2/3; 3/2).

Evaluation method

Example 3.

Solve the equation: (x 2 + 2x + 2) (y 2 - 4y + 6) \u003d 2.

Decision.

In each bracket, select a full square:

((x + 1) 2 + 1) ((y - 2) 2 + 2) \u003d 2. 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 \u003d 1 and (y - 2) 2 + 2 \u003d 2, which means x \u003d -1, y \u003d 2.

Answer: (-1; 2).

Let's get acquainted with another method for solving equations with two variables of the second degree. This method is that the equation is considered as square with respect to any variable.

Example 4.

Solve the equation: x 2 - 6x + y - 4√y + 13 \u003d 0.

Decision.

Solve the equation as square with respect to x. Let's find the discriminant:

D \u003d 36 - 4 (y - 4√y + 13) \u003d -4y + 16√y - 16 \u003d -4 (√y - 2) 2. The equation will have a solution only when D \u003d 0, that is, if y \u003d 4. Substitute the value of y into the original equation and find that x \u003d 3.

Answer: (3; 4).

Often in equations with two unknowns they indicate constraints on variables.

Example 5.

Solve the whole equation: x 2 + 5y 2 \u003d 20x + 2.

Decision.

Rewrite the equation as x 2 \u003d -5y 2 + 20x + 2. The right side of the resulting equation when divided by 5 gives remainder 2. Therefore, x 2 is not divisible by 5. But the square of a number that is not divisible by 5 gives remainder 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) \u003d 3.

Decision.

Select the complete squares in each bracket:

((| x | - 2) 2 + 1) ((y + 3) 2 + 3) \u003d 3. The left side of the equation is always greater than or equal to 3. Equality is possible provided | x | - 2 \u003d 0 and y + 3 \u003d 0. Thus, x \u003d ± 2, y \u003d -3.

Answer: (2; -3) and (-2; -3).

Example 7.

For each pair of negative integers (x; y) satisfying the equation
x 2 - 2xy + 2y 2 + 4y \u003d 33, calculate the sum (x + y). In the answer, indicate the smallest of the amounts.

Decision.

Let's select complete squares:

(x 2 - 2xy + y 2) + (y 2 + 4y + 4) \u003d 37;

(x - y) 2 + (y + 2) 2 \u003d 37. Since x and y are integers, their squares are also integers. The sum of the squares of two integers, equal to 37, is obtained if we add 1 + 36. Therefore:

(x - y) 2 \u003d 36 and (y + 2) 2 \u003d 1