Self-taught equation with three steps. How to solve algebraic equations in two steps

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.

Solving 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 0х = - 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:
- 22х = - 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 simpler equations, you have to start not from the first, but from the second ( Example. 2), third ( Example. 13) 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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LESSON SCRIPT

using a computer.

Educational institution - Municipal educational institution "Severskaya Gymnasium" ZATO Seversk.

Item - mathematics.

Class - third.

Subject: Solving equations in several steps.

Lesson type- discovery of new knowledge.

Lesson form – combined lesson with elements of problem-search learning.

Forms of organizing educational activities: collective activity to solve a problem, individual tasks of choice, work in pairs, independent work.

Lesson objectives:

Educational and methodological support – textbook for third grade in 3 parts “Mathematics”, part 2, L.G. Peterson.

Lesson duration- 45 minutes.

13 slides (Power Point, Word).

Necessary equipment and materials for the lesson:

Computer, media projector, screen.

Blackboard, textbook, workbooks, media product.

Methods:

Problem

Comparative

Observation

Using schematization ( drawing up an algorithm)

Forms of work:

Collective activities

Work on options, mutual verification

Performing an optional task

Independent work

Equation, components of actions, order of actions, algorithm.

Bibliography:

Textbook for third grade “Mathematics” L.G. Peterson in 3 parts, part two, M.: Yuventa Publishing House, 2008.

L.G. Peterson “The activity approach and its implementation in mathematics lessons in elementary school,” article in the magazine “Elementary School: Plus or Minus,” No. 5 1999.

Internet resources: http:// www. cwer. ru/ files ( Pictures)

During the classes:

Lesson objectives: systematize knowledge about equations of various types;

To develop the skill of finding an unknown component, to train students in commenting on equations through action components;

Introduce the algorithm for solving compound equations;

Develop computational skills, practice solving problems of the types studied;

Develop correct mathematical speech and logical thinking;

Teach self-assessment of your activities, compare the results of your activities with a model.

Organizational moment (Slide No. 1).

Oral exercises (Slide No. 2).

Consider the expressions. Determine the order of actions, highlight the last action.

k m + n: 3 (5 + b) : 16

a 4 – 8 (15: x) (8 – y)

Read the expressions based on the last action.

Introduction of new material.

(Slide No. 3)

Read the entries. Remember what each entry is called?

26 + 37 (D: expression)

236 – 21 = 215 (D: true equality)

48: x (D: variable expression)

At what values A inequality will be true?

What mathematical concept have we not named? (D: equation)

I suggest you solve several equations, but first we will repeat the rules for finding an unknown component:

Cards:

(Students repeat the rules for finding an unknown component using the cards).

Now write down the number in your notebooks and solve the following equations:

(Slide No. 4)

a – 86 = 9 56: c = 2 4 (4 b – 16) : 2 = 10

Who did the job?

How many equations did you solve? (D: two equations).

Let's check the solved equations. (Slide No. 4a).

What is the root of the first equation? (D: a = 95).

What is the root of the second equation? (D: c = 7).

What problem arose in solving the third equation?

(D: There is nothing to simplify on the right side).

Maybe someone can formulate the topic of the lesson?

(D: Solving equations in several steps).

Yes, that's right, today we will learn how to solve equations in several steps. (Slide No. 5)

Let's take a closer look at our equation again. Think about what you and I know well? What can we already do?

Children's answers (Slide No. 6):

We know how to determine the order of actions.

We can solve simple equations and find unknown components.

We know how to perform operations (direct and inverse).

Let's do what we know how to do, it should help us. And I will record our actions. (The teacher directs the activities of the students with an introductory dialogue; they pronounce the actions and solve the equation in their notebooks). Slide number 7

(4 ·b – 16) : 2 = 10 1. Determine the order of actions.

2. Select the last action.

3. Determine the unknown component.

4 · b – 16 = 10 · 2 4. Apply the rule.

4 ·b – 16 = 20 5. Simplify the right side.

6. We arrange the order of actions.

7. Select the last action.

8. Determine the unknown component.

4 · b = 20 + 16 9. Apply the rule.

4 · b = 36 10. Simplify the right side.

11. Determine the unknown component.

b = 36: 4 12. Apply the rule.

b = 9 13. Find the root.

Look carefully, what program of action have we come up with?

What interesting things did you notice?

Is it possible to shorten our program somehow?

Let's create an algorithm of actions:

(Slide No. 8)

Physical education minute (Slide No. 9).

Gymnastics for the eyes.

Primary consolidation (pronunciation).

(Slide number 10).

Now, using the algorithm, let's try to explain the following equation:

(2 + x: 7) · 8 = 72

2 + x: 7 = 72: 8

2 + X : 7 = 9 Students comment step by step

x: 7 = 9 – 2 solution to the equation.

Raise your hand, who clearly understands how to solve the equation in several steps? Tell us about your actions.

Who else is experiencing difficulties and needs help?

Self-control.

Check your solution, exchange notebooks, help your neighbor check.

Whoever thinks that the solution is correct, that he coped with the work, put “+” in the margin.

Check students' work. Who got the same root of the equation?

The result of the work.

Guys, what is the topic of today's lesson?

What problem did you encounter at the beginning of the lesson?

How did you cope with difficulties?

Repeat the algorithm of actions.

Do you think, while doing work now, is it only equations that we learn to solve? (D: we learn to plan our activities, practice counting, calculations, learn to complete tasks).

Can our knowledge and skills be useful in life? Where? When?

What keywords would you highlight in the lesson?

(D: Equation, procedure, unknown component, rule for finding the unknown component, expressions) – Slide number 11.

8. Self-assessment of your activities.

If it was easy in the lesson, you figured it all out – the color green. If there were difficulties, doubts - yellow. If you didn’t understand the topic, it was difficult - the color red. – Slide “12.

9. Homework (Slide No. 13)

Compose your example equation in several steps;

p. 36, No. 7 (according to options).

Slide number 14 – end of the lesson.

Content:

You can solve simple algebraic equations in just two steps. To do this, it is enough to isolate a variable using addition, subtraction, multiplication or division. Want to know different ways to solve algebraic equations? Read on.

Steps

1 Solving equations with one unknown

1. 1 Write down the equations. To solve an algebraic equation, the first thing you need to do is write it down, so everything will immediately become clearer. Let's say we are dealing with the following equation: -4x + 7 = 15.
2. 2 We decide what action we will use to isolate the variable. The next step is to figure out how to store "-4x" on one side and constants (integers) on the other. To do this, we use the “law of symmetry” and find the number opposite to +7, this is -7. Now we subtract 7 from both sides of the equation so that the “+7” in the part where the variable is located turns into 0. We simply write “-7” under 7 on one side and under 15 on the other so that the equation essentially does not change.
   • Remember the Golden Rule of Algebra. Whatever we do to one side of the equation, we also do to the other. That's why we subtracted 7 from 15 too.
3. 3 We add or subtract a constant on both sides of the equation. This way we isolate the variable. Subtracting 7 from +7 we get 0 on the left. Subtracting 7 from +15 we get 8 on the right.
   • -4x + 7 = 15 =
   • -4x = 8
4. 4 By dividing or multiplying we get rid of the coefficient of the variable. In this example the coefficient is -4. To get rid of it you need to divide both sides of the equation by -4.
   • Again, all actions are carried out on both sides, which is why you see ÷ -4 twice.
5. 5 Find the variable. To do this, divide the left side (-4x) by -4, you get x. Divide the right side of (8) by -4 to get -2. Thus x = -2. The equation is solved in two steps: -- subtraction and division --.

2 Solving equations with variables on both sides

1. 1 Write down the equation. We will solve the equation: -2x - 3 = 4x - 15. First, make sure the variables are the same: in this case x.
2. 2 Translate the constants to the right side of the equation. To do this you need to use addition or subtraction. The constant is -3, so we take the opposite of +3 and add it to both sides.
   • By adding +3 to the left side (-2x -3) we get -2x.
   • Adding +3 to the right side (4h -15) we get 4x -12.
   • So (-2x - 3) +3 = (4x - 15) +3 = -2x = 4x - 12
   • Modified equation: -2x = 4x -12
3. 3 We move the variables to the left with a change of sign. We get -6x = -12
   • -2x - 4x = (4x - 12) - 4x = -6x = -12
4. 4 Finding the variable. To do this, divide both sides by -6 and get x = 2.
   • -6x ÷ -6 = -12 ÷ -6
   • x = 2

3 Other ways to solve equations in two steps

1. 1 The equation can be solved and leaving the variable on the right, it doesn't matter. Let's take the equation 11 = 3 - 7x. First, let's get rid of the 3 on the right, to do this we subtract 3 from both sides. Then divide both sides by -7 and get x:
   • 11 = 3 - 7x =
   • 11 - 3 = 3 - 3 - 7x =
   • 8 = - 7x =
   • 8/-7 = -7/7x
   • -8/7 = x or -1.14 = x
2. 2 We solve the equation by the second action by multiplying, not dividing. The principle is the same. Let's take the equation x/5 + 7 = -3. First, subtract 7 from both sides and then multiply both sides by 5 to get x:
   • x/5 + 7 = -3 =
   • (x/5 + 7) - 7 = -3 - 7 =
   • x/5 = -10
   • x/5 * 5 = -10 * 5
   • x = -50