Content:
Solve simple ones algebraic equations You can do it in just two steps. To do this, it is enough to isolate a variable using addition, subtraction, multiplication or division. Want to know various ways solutions to algebraic equations? Read on.
Steps
1 Solving equations with one unknown
- 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
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.
- We remember Golden Rule 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
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
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 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 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
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.
- 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
We move the variables to the left with a change of sign. We get -6x = -12
- -2x - 4x = (4x - 12) - 4x = -6x = -12
- 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
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
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
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 organization 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
Equation, components of actions, order of actions, algorithm.
Bibliography:
Textbook for third grade “Mathematics” by 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 “ Primary School: plus or minus", No. 5 1999
Internet resources: http:// www. cwer. ru/ files ( Pictures)
During the classes:
Lesson objectives: systematize knowledge about equations different 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, logical thinking;
Teach self-assessment of your activities, compare the results of your activities with a model.
Organizing time(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?
Whatever keywords highlighted in class?
(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.
Class: 4
Target: Consider practical ways to solve equations that require more than one arithmetic operation.
Lesson equipment: computer presentation of mental arithmetic, cards with equations, cards of three levels for independent work on tasks, feedback cube
During the classes
1. Organizational moment
Checking readiness for the lesson. The number is written in the notebooks, cool work.
2. Oral counting(computer presentation, slide No. 1)
Game "Snail Competition"
Your favorite dog Alik at the snail competition. Two snails must climb to the top of the mountain. Which one of them will come out first? Our snail is number 1 on the left. The snail takes a step only if we find the meaning of the expression correctly.
You are ready?
The signal to start has already sounded. We repeat the procedure and name the correct meanings of the expressions.
(122 + 18) : 70 = 2
(64: 8 + 20) : 7 = 4
20 · (26 + 14) : 100 = 8
1 (30 + 2) – 4 4 = 16
5 4 + 12 = 32
(400 – 300) – 36 = 64
We have a series of numbers.
2, 4, 8, 16, 32, 64
What pattern did you notice in the compilation of this series? (each subsequent number is doubled)
Continue this series of numbers and name at least the next three numbers. (128, 256, 512…)
Well done! We decided everything correctly, so our snail is on top of the mountain.
Each number has an encrypted letter. Let's turn them over and read the topic of today's lesson.
2 4 8 16 32 64 128 256 512
THE EQUATION
What is the equation called?
What is the root of an equation?
What does it mean to solve an equation?
We already know how to solve simple equations, and today we will get acquainted with solving complex equations where we need to perform several arithmetic operations.
3. Solution simple equations. Preparation for the introduction of new material.
On a magnetic board in random order there are cards with equations.
What groups can all these equations be divided into? (equations are distributed in 3 columns)
1) 7000 – x = 2489
7000 – x = 3489
7000 – x = 1689
Why did we put these equations in the first group? (simple equations With identically reduced) Can we solve them?
Find among them the equation with the largest root and solve it (one student at the board)
2) 71: x = 20 + 7
x: 3 = 16 + 11 ( these are equations on the right side of which the expression)
Can we solve the equations of the second column?
Solve any of the equations, but replace the sum on the right side with the difference. The root of the equation should remain the same. (two students at the blackboard)
3) (490 – x) – 250 = 70
Look at the remaining equation. Is it easy for us to solve it? Why?
4. Working on new material. (frontal conversation with the class, during which the solution to the equation is considered)
(490 – x) – 250 = 70
490 – x = 70 + 250
490 – x = 320
x = 490 – 320
x = 170
(490 – 170) – 250 = 70
70 = 70
Answer: 70
5. Consolidation.
1) Solving the equation (one of the strong students at the blackboard)
5 a + 500 = 4500: 5
5 a + 500 = 900
5 a = 900 – 500
5 a = 400
a = 400: 5
a = 80
5 80 + 500 = 900
900 = 900
Answer: 80
Solve the equations.
A+ 156 = 17 ∙ 20 (1604 – y) – 108 = 800
252: 36 ∙ x = 560 103300: (x + 297) = 25 ∙2
We solved two new complex equations. Look at the equations in front of you. Are they all complex? Which equation is the odd one out? Why? The rest are on the left side an expression in several actions. Find among them a sequence of actions that has already been encountered today.
(1604 – y) – 108 = 800
1604 – y = 800 + 108
1604 – y = 908
y = 1604 – 908
y = 696
(1604 – 696) – 108 = 800
800 = 800
Answer: 696
Solve the equation in pairs. One student turns the board around for later checking.
6. Solving the problem
Independent work using cards of 3 levels. Having completed the task of the first stage, the student proceeds to complete the task of the second stage, then the third. (Various methods of differentiated work)
Frontal check
1) 25700 – x = 12350
x = 25700 – 12350
x = 13350
25700 – 13350 = 12350
12350 = 12350
Answer: 13350 seedlings.
2) 25700 – x = 12000 + 350
3) 25700 – (x + 8580) = 12350
x + 8580 = 25700 – 12350
x + 8580 = 13350
x = 13350 – 8580
x = 4770
25700 – (4770 + 8580) =12350
12350 = 12350
Answer: 4770 limes.
4) What other equation could be made?
(25700 – x) – 8580 = 12350
We solved three problems by composing three equations. Which equation is considered complex? Why?
7. Homework.
Consider how the equations were solved in the textbook on page 106 and solve the equation in the printed notebook No. 44 (a).
Solve problem No. 47. Additional task: what other questions can be asked about this problem?
8. Lesson summary.
What equations did you learn to solve in class?
Was it difficult?
Who had it easy?
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 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 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 figure out how to linear equations, and so with others.
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Recently, the mother of a schoolchild with whom I study calls and asks me to explain mathematics to the child, because he does not understand, but she does not shout at him and the conversation with her son does not work out.
I don’t have a mathematical mind, this is not typical for creative people, but I said that I would see what they were going through and try. And this is what happened.
I took a sheet of A4 paper, plain white, felt-tip pens, a pencil in my hands and began to highlight what was worth understanding, remembering, paying attention to. And so that you can see where this figure goes and how it changes.
Explanation of examples from the left side to the right side.
Example #1
An example of an equation for 4th grade with a plus sign.
The very first step is to look at what we can do in this equation? Here we can do the multiplication. We multiply 80*7 and get 560. Rewrite it again.
X + 320 = 560 (highlighted the numbers with a green marker).
X = 560 – 320. We put a minus because when we transfer a number, the sign in front of it changes to the opposite. Let's do the subtraction.
X = 240 Be sure to check. Checking will show whether we solved the equation correctly. Instead of x we insert the number that we received.
Examination:
240 + 320 = 80*7 We add the numbers and multiply them on the other side.
That's right! So we solved the equation correctly!
Example No. 2
Example of an equation for 4th grade with a minus sign.
X – 180 = 240/3
The first step is to look at what we can do in this equation? In this example we can divide. We divide 240 divided by 3 to get 80. Rewrite the equation again.
X – 180 = 80 (highlighted the numbers with a green marker).
Now we see that we have x (unknown) and numbers, but not next to each other, but separated by an equal sign. X in one direction, numbers in the other.
X = 80 + 180 We put the plus sign because when transferring a number, the sign that was before the number changes to the opposite. We count.
X = 260 Execute test work. Checking will show whether we solved the equation correctly. Instead of x we insert the number that we received.
Examination:
260 – 180 = 240/3
That's right!
Example No. 3
400 – x = 275 + 25 Add the numbers.
400 – x = 300 Numbers are separated by an equal sign, x is negative. To make it positive, we need to move it through the equal sign, we collect the numbers on one side, x on the other.
400 - 300 = x The number 300 was positive, but when moved to the other side, it changed sign and became a minus. We count.
Since it is not customary to write this way, and the first one in the equation should be x, we simply swap them.
Examination:
400 – 100 = 275 + 25 Let’s count.
That's right!
Example No. 4
An example of an equation for 4th grade with a minus sign, where x is in the middle, in other words, an example of an equation where x is negative in the middle.
72 – x = 18 * 3 We perform multiplication. Let's rewrite the example.
72 – x = 54 We line up the numbers in one direction, x in the other. The number 54 changes sign to the opposite because it jumps over the equal sign.
72 – 54 = x Let’s count.
18 = x Swap places for convenience.
Examination:
72 – 18 = 18 * 3
That's right!
Example No. 5
Example of an x equation with subtraction and addition for 4th grade.
X – 290 = 470 + 230 Add.
X – 290 = 700 We put the numbers on one side.
X = 700 + 290 Let's count.
Examination:
990 – 290 = 470 + 230 We perform addition.
That's right!
Example No. 6
An example of an x equation for multiplication and division for 4th grade.
15 * x = 630/70 We perform division. Let's rewrite the equation.
15 * x = 90 This is the same as 15x = 90 We leave x on one side, numbers on the other. This equation takes the following form.
X = 90/15, when the number 15 is transferred, the multiplication sign changes to division. We count.
Examination:
15*6 = 630 / 7 We perform multiplication and subtraction.
That's right!
Now let's talk about the basic rules:
- Multiply, add, divide or subtract;
Doing what we can do will make the equation a little shorter.
- X in one direction, numbers in the other.
An unknown variable in one direction (it’s not always x, it could be another letter), numbers in the other.
- When you transfer x or a number through an equal sign, their sign changes to the opposite.
If the number was positive, then when transferring it we put a minus sign in front of the number. And vice versa, if the number or x had a minus sign, then when transferring through equals we put a plus sign.
- If at the end the equation begins with a number, then simply swap it.
- We always check!
By doing homework, class work, tests, you can always take a sheet and write on it first and check it.
Additionally, we find similar examples on the Internet, additional books, and manuals. It’s easier not to change the numbers, but to take ready-made examples.
The more the child decides for himself and studies on his own, the faster he will learn the material.
If a child does not understand examples with an equation, it is worth explaining the example and telling him to do the rest according to the model.
Given detailed description, how to explain equations with x to a schoolchild for:
- parents;
- schoolchildren;
- tutors;
- grandparents;
- teachers;
Children need to do everything in color, with different crayons on the board, but alas, not everyone does this.
From my practice
The boy wrote the way he wanted, despite existing rules mathematics. When checking the equations were different numbers and one number (on the left side) was not equal to the other (what with right side), he spent time searching for the error.
When asked why he does this? The answer was that he was trying to guess and thinking, what if he does it right.
In this case, you need to solve similar examples every day (every other day). Bringing actions to automaticity, and of course, all children are different, may not be achieved from the first lesson.
If parents do not have time, and this is often the case because parents earn money, then it is better to find a tutor in your city who can explain the material covered to the child.
Now is the age of the Unified State Examination, tests, tests, there are additional collections and manuals. When doing homework for a child, parents should remember that they will not be included in the school exam. It is better to explain it clearly to the child once, so that the child can solve the examples independently.
