Solution X 5. Solution of linear equations with examples. How to solve equations: addition or subtraction
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The equation is called equality in which there is an unknown member - x. Its value and it is necessary to find.
An unknown value is called the root of the equation. Solve the equation means finding its root, and for this you need to know the properties of equations. Equations for grade 5 are simple, but if you learn how to solve them correctly, you will not have problems with them in the future.
The main property of equations
With the change of both parts of the equation to the same value, it continues to remain the same equation with the same root. Let's decide a few examples to better understand this rule.
How to solve equations: addition or subtraction
Suppose we have a type equation:
- a + x \u003d b - here a and b - numbers, and X is an unknown member of the equation.
If we add to both parts of the equation (or subtract out of them) the value of C, it will not change:
- a + x + c \u003d b + with
- a + X - C \u003d B - s.
Example 1.
We use this property to solve the equation:
- 37 + x \u003d 51
Subscribe from both parts number 37:
- 37 + x-37 \u003d 51-37
we get:
- x \u003d 51-37.
The root of the equation x \u003d 14.
If we look closely at the last equation, we will see that it is the same as the first. We simply moved the term 37 of one part of the equation to another, replacing a plus for minus.
It turns out that any number can be transferred from one part of the equation to another with the opposite sign.
Example 2.
- 37 + x \u003d 37 + 22
We carry out the same action, we move the number 37 from the left part of the equation to the right:
- x \u003d 37-37 + 22
Since 37-37 \u003d 0, then we just cut and get it:
- x \u003d 22.
The same members of the equation with one sign in different parts Equations can be reduced (verify).
Multiplication and division of equations
Both parts of equality can also be multiplied or divided into one and the same number:
If equality a \u003d b is divided or multiplied by C, it will not change:
- a / s \u003d b / s,
- aC \u003d BC.
Example 3.
- 5x \u003d 20.
We divide both parts of the 5 equation:
- 5x / 5 \u003d 20/5.
Since 5/5 \u003d 1, then these multiplier and divider in the left side of the equation reducing and get:
- x \u003d 20/5, x \u003d 4
Example 4.
- 5x \u003d 5A.
If both parts of the equation are divided by 5, we get:
- 5x / 5 \u003d 5a / 5.
5 in the numerator and denominator of the left and right part are reduced, it turns out x \u003d a. So, the same multipliers in the left and right of the equations are reduced.
Sure one more example:
- 13 + 2x \u003d 21
We transfer the term 13 from the left side of the equation to the right with the opposite sign:
- 2x \u003d 21 - 13
- 2x \u003d 8.
We divide both parts of the equation by 2, we get:
- x \u003d 4.
Equation with one unknown, which, after disclosing brackets and bringing similar members to take the form
ah + B \u003d 0where a and b arbitrary numbers are called linear equation with one unknown. Today we will describe how these linear equations are solved.
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 an unknown, the equation to the right equality is called by decision or the root of the equation .
For example, if in the equation 3x + 7 \u003d 13 instead of unknown x substitute the number 2, then we obtain the right equality 3 · 2 +7 \u003d 13. So the value x \u003d 2 is a solution or the root of the equation.
And the value x \u003d 3 does not draw the equation 3x + 7 \u003d 13 to the correct equality, since 3 · 2 +7 ≠ 13. So the value x \u003d 3 is not a solution or the root of the equation.
The solution of any linear equations is reduced to solving the equations of the form
ah + B \u003d 0.
We transfer a free member from the left side of the equation to the right, changing the sign before b to the opposite, we get
If a ≠ 0, then x \u003d - b / a .
Example 1. Decide equation 3x + 2 \u003d 11.
We transfer 2 of the left part of the equation to the right, changing the sign before 2 to the opposite, we get
3x \u003d 11 - 2.
Perform subtraction, then
3x \u003d 9.
To find x need to divide the work on the famous multiplier, that is
x \u003d 9: 3.
So, 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, I get the equation 0x \u003d 0. This equation has infinitely many solutions, since with multiplying any number by 0 we obtain 0, but b is also 0. The solution of this equation is any number.
Example 2.Decide equation 5 (x - 3) + 2 \u003d 3 (x - 4) + 2x - 1.
Recall brackets:
5x - 15 + 2 \u003d 3x - 12 + 2x - 1.
5x - 3x - 2x \u003d - 12 - 1 + 15 - 2.
We give similar members:
0x \u003d 0.
Answer: X - any number.
If a \u003d 0 and b ≠ 0, I get the equation 0x \u003d - b. This equation of solutions does not have, since when multiplying any number by 0 we obtain 0, but b ≠ 0.
Example 3.Decide the equation x + 8 \u003d x + 5.
We grouped on the left part of the members containing unknown, and in the right - free members:
x - x \u003d 5 - 8.
We give similar members:
0x \u003d - 3.
Answer: No solutions.
On the figure 1. The scheme of solving a linear equation is depicted
Make a general scheme for solving equations with one variable. Consider the solution of Example 4.
Example 4. Let it be necessary to solve the equation
1) I will multiply all members of the equation for the smallest general multiple denominator, equal to 12.
2) after the reduction we get
4 (x - 4) + 3 · 2 (x + 1) - 12 \u003d 6 · 5 (x - 3) + 24x - 2 (11x + 43)
3) To separate the members containing unknown and free members, open brackets:
4x - 16 + 6x + 6 - 12 \u003d 30x - 90 + 24x - 22x - 86.
4) group members containing unknowns in one part containing unknowns, and in the other - free members:
4x + 6x - 30x - 24x + 22x \u003d - 90 - 86 + 16 - 6 + 12.
5) We give similar members:
- 22x \u003d - 154.
6) we divide on - 22, we get
x \u003d 7.
As we see, the root of the equation is equal to seven.
In general, such equations can be solved according to the following scheme:
a) bring the equation to a whole mind;
b) reveal brackets;
c) group members containing an unknown, in one part of the equation, and free members in another;
d) lead similar members;
e) solve the equation of the form Ah \u003d B, which was obtained after bringing such members.
However, this scheme is not mandatory for any equation. When solving many more simple equations We have to start not from the first, but from the second ( Example. 2.), the third ( Example. 13) And even from the fifth stage, as in Example 5.
Example 5.Decide equation 2x \u003d 1/4.
We find an unknown x \u003d 1/4: 2,
x \u003d 1/8. .
Consider the solution of some linear equations found at the main state exam.
Example 6.Decide equation 2 (x + 3) \u003d 5 - 6x.
2x + 6 \u003d 5 - 6x
2x + 6x \u003d 5 - 6
Answer: - 0, 125
Example 7.Decide equation - 6 (5 - 3x) \u003d 8x - 7.
- 30 + 18x \u003d 8x - 7
18x - 8x \u003d - 7 +30
Answer: 2,3
Example 8. Decide equation
3 (3 - 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 7
Decision
Since it is necessary to find f (6), and we know f (x + 2),
that 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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Multipling the system of normal NTTXT1 + BT1 \u003d 0 equations to the inverse matrix N-1
get:
(34)
(35)
The solution of normal equations by the method of circulation.
A-priory reverse matrix, N-1N \u003d E. This equality is used to substantiate the method for determining the elements of the return matrix. Let T \u003d 2.
This implies:
- 1st system of weighing normal equations.
- 2nd system of weight normal equations.
In general, as a result of such actions, T systems of weight normal equations are obtained by t equations in each system. These systems have the same coefficient matrix, as well as the main one, with unknown ΔХJ and differ from it only by columns of free members. In the j-Ohm equation of the J-th system, the free member is -1, the rest are zero. Weighting normal equation systems are solved in parallel with the main system, in the general scheme, using additional columns for free members of these systems (Table 9). To control the calculated values \u200b\u200bof the elements of the inverse matrix qij are substituted into the total equations drawn up for weight systems. For example, for t \u003d 2, these equations will look at:
(+ [RAB]) Q11 + (+) Q12 - 1 \u003d 0;
(+) Q21 + (+) Q22 - 1 \u003d 0.
For pre-controls, the equality qij \u003d Qji (I ≠ J) is served.
The elements of the return matrix Qij are called weight coefficients.
Table 9.
Determination of the elements of the return matrix in the Gauss scheme
3.6. Evaluation of accuracy based on equalization materials
The average quadratic error of the parameter function is determined by the formula:
where
(36)
The average quadratic error of the weight unit;
(37)
Reverse weight of parameter function or in matrix form:
(38)
The reverse weight of the parameter equal to the diagonal element of the reverse matrix.
3.7. Block diagram of the parameter method of adjustment
1. Analyzes the totality of measurements yi, determine T - the number of required measurements. Install the system of measurements PI (i \u003d 1, 2, ..., n).
2. Select independent parameters x1, x2, ..., xt, the number of which is t.
3. Compile parametric equations. The equalized values \u200b\u200bof all measured values \u200b\u200bare expressed as the functions of the selected parameters.
4. Find the approximate values \u200b\u200bof the parameters x0j.
5. Parametric communication equations lead to a linear form, calculate coefficients and free members of parametric equations of amendments.
6. Compose a function of parameters to assess its accuracy. The weight function is linearized.
7. Compile normal equations, Calculate the coefficients and free members of normal equations.
8. Solve normal equations, calculate the corrections to the approximate values \u200b\u200bof the parameters and control them.
9. Calculate the amendments VI to the measurement results, and control νi and.
10. Calculate the parameters equalized by measurement results and control the adjustment.
11. Calculate the inverse weights of the parameters and parameter functions.
12. Perform an estimate of the accuracy of measurement results, calculate the average quadratic error of the weight unit.
13. Calculate the average quadratic errors of equalized values.
One of the most important skills with admission to Grade 5 is the ability to solve the simplest equations. Since the 5th class is not so far from elementary school, then the types of equations that the student can solve is not so much. We will introduce you to all the main types of equations that you need to be able to decide if you want enlighten into a physical and mathematical school.
1 Type: "Lukovichny"
These are equations that are almost likely to meet you when admission to any school or grace 5 class as a separate task. They are easy to distinguish from others: the variable is present only 1 time. For example, or.
They are solved very simply: you just need to "get" to an unknown, gradually "removing" everything is too much that surrounds her - as if to clean the bulb - hence the same name. To solve, it suffices to remember several rules from the second class. List them all:
Addition
- the term1 + terms2 \u003d amount
- cONDUCTION 1 \u003d Amount - Speed2
- cONDUCTION2 \u003d Amount - Speed1
Subtraction
- reduced - subtractable \u003d difference
- reduced \u003d subtractable + difference
- substant \u003d Reduced - Difference
Multiplication
- multiplier1 * multiplier2 \u003d work
- multiplier1 \u003d Work: Multiplier2
- multiplier2 \u003d Work: Multiplier1
Division
- delimi: Divider \u003d Private
- delimi \u003d divider * Private
- divider \u003d Delimi: Private
We will analyze on the example, how to apply these rules.
Note that we divide 
On and get. In this situation, we know the divider and private. To find a divide, you need a divider to multiply by private:
We became a little closer to myself. Now we see that 
Adjasted and turns out. It means that it is necessary to find one of the terms, you need to subtract the well-known alignment amount:
And one more "layer" is removed from an unknown! Now we see the situation with the known value of the work () and one famous factor ().
Now the situation "diminished - subtractable \u003d difference"
And the last step is a famous work () and one of the multipliers () 
2 Type: Equations with brackets
The equations of this type are most often found in tasks - it is precisely 90% of all tasks for them for admission to Grade 5. Unlike "Lukovichny equations" The variable here can meet several times, so it is impossible to solve it from the previous item. Typical equations: or
The main difficulty is to reveal brackets correctly. After it managed to do this, it should be given such terms (numbers to numbers, variable variables), and after that we get the simplest "Lukovichny equation"Who can solve. But first things first.
Disclosure of brackets. We give several rules that should be used in this case. But, as practice shows, the student starts to completely disclose the student only after 70-80 proceeded tasks. The main rule is: any multiplier behind the brackets must be multiplied by each group inside the brackets. A minus, standing in front of the bracket, changes the sign of all expressions that they stand inside. So, the basic rules of disclosure: 
Bringing similar. It is much easier here: you need to transfer the components through the sign of equality to ensure that on the one hand, only the components are with an unknown, and on the other - only numbers. The main rule is: each term, transferred through, changes its mark - if it was C, then it will become C, and vice versa. After successful transfer, you must count the final number of unknown, the final number of equality standing on the other side than the variables, and solve "Lukovichny equation".
