Solving linear equations with examples. Online equations Accuracy assessment based on adjustment materials

One of the most important skills when admission to 5th grade is the ability to solve simple equations. Since 5th grade is not yet so far from primary school, then there are not so many types of equations that a student can solve. We will introduce you to all the basic types of equations that you need to be able to solve if you want enter a physics and mathematics school.

Type 1: "bulbous"
These are equations that you are almost likely to encounter when admission to any school or a 5th grade club as a separate task. They are easy to distinguish from others: in them the variable is present only once. For example, or.
They are solved very simply: you just need to “get” to the unknown, gradually “removing” everything unnecessary that surrounds it - as if peeling an onion - hence the name. To solve it, just remember a few rules from the second class. Let's list them all:

Addition

1. term1 + term2 = sum
2. term1 = sum - term2
3. term2 = sum - term1

Subtraction

1. minuend - subtrahend = difference
2. minuend = subtrahend + difference
3. subtrahend = minuend - difference

Multiplication

1. factor1 * factor2 = product
2. factor1 = product: factor2
3. factor2 = product: factor1

Division

1. dividend: divisor = quotient
2. dividend = divisor * quotient
3. divisor = dividend: quotient

Let's look at an example of how to apply these rules.

Note that we are dividing on and we receive . In this situation, we know the divisor and the quotient. To find the dividend, you need to multiply the divisor by the quotient:

We have become a little closer to ourselves. Now we see that is added and it turns out . This means that to find one of the terms, you need to subtract the known term from the sum:

And another “layer” has been removed from the unknown! Now we see the situation with known value product () and one known factor ().

Now the situation is “minuend - subtrahend = difference”

And the last step - famous work() and one of the multipliers ()

Type 2: equations with brackets
Equations of this type most often found in tasks - 90% of all tasks for admission to 5th grade. Unlike "onion equations" the variable here can appear several times, so it is impossible to solve it using the methods from the previous paragraph. Typical equations: or
The main difficulty is opening the brackets correctly. After you have managed to do this correctly, you should reduce similar terms (numbers to numbers, variables to variables), and after that we get the simplest "onion equation" which we can solve. But first things first.

Expanding parentheses. We will give several rules that should be used in this case. But, as practice shows, the student begins to open the brackets correctly only after 70-80 completed problems. The basic rule is this: any factor outside the brackets must be multiplied by each term inside the brackets. And the minus sign in front of the bracket changes the sign of all the expressions inside. So, the basic rules of disclosure:

Bringing similar. Here everything is much easier: you need, by transferring the terms through the equal sign, to ensure that on one side there are only terms with the unknown, and on the other - only numbers. The basic rule is this: each term transferred through changes its sign - if it was with, it will become with, and vice versa. After a successful transfer, it is necessary to count the total number of unknowns, the total number on the other side of the equality than the variables, and solve a simple "onion equation".

An equation is an equality in which there is an unknown term - x. Its meaning must be found.

The unknown quantity is called the root of the equation. Solving an equation means finding its root, and to do this you need to know the properties of the equations. The equations for grade 5 are not difficult, but if you learn to solve them correctly, you will not have problems with them in the future.

The main property of the equations

When both sides of an equation change by the same amount, it continues to be the same equation with the same root. Let's solve some examples to better understand this rule.

How to Solve Equations: Addition or Subtraction

Suppose we have an equation of the form:

• a + x = b - here a and b are numbers, and x is the unknown term of the equation.

If we add (or subtract from them) the value c to both sides of the equation, it will not change:

• a + x + c = b + c
• a + x - c = b - c.

Example 1

Let's use this property to solve the equation:

• 37+x=51

Subtract the number 37 from both sides:

• 37+x-37=51-37

we get:

• x=51-37.

The root of the equation is x=14.

If we look closely at the last equation, we can see that it is the same as the first one. We simply moved term 37 from one side of the equation to the other, replacing plus with 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=37+22

Let's carry out the same action, move the number 37 from the left side of the equation to the right:

• x=37-37+22

Since 37-37=0, we simply reduce this and get:

• x =22.

Identical terms of an equation with the same sign, located in different parts equations can be reduced (crossed out).

Multiplying and dividing equations

Both sides of the equality can also be multiplied or divided by the same number:

If the equality a = b is divided or multiplied by c, it does not change:

• a/c = b/c,
• ac = bс.

Example 3

• 5x = 20

Let's divide both sides of the equation by 5:

• 5x/5 = 20/5.

Since 5/5 = 1, we reduce these multiplier and divisor on the left side of the equation and get:

• x = 20/5, x=4

Example 4

• 5x = 5a

If both sides of the equation are divided by 5, we get:

• 5x/5 = 5a/5.

The 5 in the numerator and denominator of the left and right sides are canceled, resulting in x = a. This means that identical factors on the left and right sides of the equations cancel.

Let's solve another example:

• 13 + 2x = 21

We move term 13 from the left side of the equation to the right with the opposite sign:

• 2x = 21 - 13
• 2x = 8.

Dividing both sides of the equation by 2, we get:

• x = 4.

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

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

For example, all equations:

2x + 3= 7 – 0.5x; 0.3x = 0; x/2 + 3 = 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 = 13 instead of the unknown x we ​​substitute the number 2, we obtain the correct equality 3 2 +7 = 13. This means that the value x = 2 is the solution or root of the equation.

And the value x = 3 does not turn the equation 3x + 7 = 13 into a true equality, since 3 2 +7 ≠ 13. This means that the value x = 3 is not a solution or a root of the equation.

Solution of any linear equations reduces to solving equations of the form

ax + b = 0.

Let's move 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 x = ‒ b/a .

Example 1. Solve the equation 3x + 2 =11.

Let's move 2 from the left side of the equation to the right, changing the sign in front of 2 to the opposite, we get
3x = 11 – 2.

Let's do the subtraction, then
3x = 9.

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

This means that the value x = 3 is the solution or root of the equation.

Answer: x = 3.

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

Example 2. Solve the equation 5(x – 3) + 2 = 3 (x – 4) + 2x ‒ 1.

Let's expand the brackets:
5x – 15 + 2 = 3x – 12 + 2x ‒ 1.

5x – 3x ‒ 2x = – 12 ‒ 1 + 15 ‒ 2.

Here are some similar terms:
0x = 0.

Answer: x - any number.

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

Example 3. Solve the equation x + 8 = x + 5.

Let’s group terms containing unknowns on the left side, and free terms on the right side:
x – x = 5 – 8.

Here are some similar terms:
0х = ‒ 3.

Answer: no solutions.

On Figure 1 shows a diagram for solving a linear equation

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

Example 4. Suppose we need to solve the equation

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

2) After reduction we get
4 (x – 4) + 3 2 (x + 1) ‒ 12 = 6 5 (x – 3) + 24x – 2 (11x + 43)

3) To separate terms containing unknown and free terms, open the brackets:
4x – 16 + 6x + 6 – 12 = 30x – 90 + 24x – 22x – 86.

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

5) Let us present similar terms:
- 22x = - 154.

6) Divide by – 22, We get
x = 7.

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

Generally such equations can be solved using the following scheme:

a) bring the equation to its integer form;

b) open the brackets;

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

d) bring similar members;

e) solve an equation of the form aх = b, which was obtained after bringing similar terms.

However, this scheme is not necessary for every equation. When solving many more simple equations you have to start not from the first, but from the second ( Example. 2), third ( Example. 1, 3) and even from the fifth stage, as in example 5.

Example 5. Solve the equation 2x = 1/4.

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

Let's look at solving some linear equations found in the main state exam.

Example 6. Solve the equation 2 (x + 3) = 5 – 6x.

2x + 6 = 5 – 6x

2x + 6x = 5 – 6

Answer: - 0.125

Example 7. Solve the equation – 6 (5 – 3x) = 8x – 7.

– 30 + 18x = 8x – 7

18x – 8x = – 7 +30

Answer: 2.3

Example 8. Solve the equation

3(3x – 4) = 4 7x + 24

9x – 12 = 28x + 24

9x – 28x = 24 + 12

Example 9. Find f(6) if f (x + 2) = 3 7's

Solution

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

We solve the linear equation x + 2 = 6,
we get x = 6 – 2, x = 4.

If x = 4 then
f(6) = 3 7-4 = 3 3 = 27

Answer: 27.

If you still have questions or want to understand solving equations more thoroughly, sign up for my lessons in the SCHEDULE. I will be glad to help you!

TutorOnline also recommends watching a new video lesson from our tutor Olga Alexandrovna, which will help you understand both linear equations and others.

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Multiplying the system of normal equations NttXt1 + Bt1 = 0 by the inverse matrix N-1

receive:

(34)

(35)

Solving normal equations using the inversion method.

By definition of an inverse matrix, N-1N = E. This equality is used to justify the method for determining the elements of an inverse matrix. Let t = 2.

It follows from this:

- 1st system of weighted normal equations.

- 2nd system of weighted normal equations.

In the general case, as a result of such actions, t systems of weighted normal equations will be obtained, with t equations in each system. These systems have the same matrix of coefficients as the main one, with unknowns δхj and differ from it only in the columns of free terms. In the j-th equation of the j-th system, the free term is -1, the rest are equal to zero. Systems of weighted normal equations are solved in parallel with the main system, in a general scheme, using additional columns for the free terms of these systems (Table 9). For control, the calculated values ​​of the elements of the inverse matrix Qij are substituted into the summary equations compiled for weighting systems. For example, for t = 2 these equations will look like:

( + [rab])Q11 + ( + )Q12 - 1 = 0;

( + )Q21 + ( + )Q22 - 1 = 0.

For preliminary control, the equalities Qij = Qji (i ≠ j) are used.

The elements of the inverse matrix Qij are called weighting coefficients.

Table 9

Determining the elements of the inverse matrix in the Gaussian scheme

3.6. Accuracy assessment based on adjustment materials

The root mean square error of the parameter function is determined by the formula:

Where

(36)

Mean square error of unit weight;

(37)

Inverse weight of a function of parameters or in matrix form:

(38)

The inverse weight of the parameter, equal to the diagonal element of the inverse matrix.

3.7. Block diagram of the parametric adjustment method

1. Analyze the set of measurements yi, determine t - the number of required measurements. Set up a system of measurement scales pi (i = 1, 2, ..., n).

2. Select independent parameters x1, x2, ..., xt, the number of which is equal to t.

3. Compose parametric communication equations. The equalized values ​​of all measured quantities are expressed as functions of the selected parameters.

4. Find approximate values ​​of the parameters x0j.

5. Parametric equations connections lead to linear form, calculate the coefficients and free terms of the parametric correction equations.

6. Construct a function of parameters to evaluate its accuracy. The weighting function is linearized.

7. Compose normal equations, calculate coefficients and free terms of normal equations.

8. Solve normal equations, calculate corrections to approximate values ​​of parameters and control them.

9. Corrections vi to the measurement results are calculated, and νi and are monitored.

10. Calculate parameters, adjusted measurement results and perform adjustment control.

11. Calculate the inverse weights of parameters and functions of parameters.

12. Evaluate the accuracy of the measurement results and calculate the mean square error of a unit of weight.

13. Calculate the mean square errors of the equalized quantities.