Problems solved using the equation. Solving problems in mathematics

In the course of school mathematics there are always problems. Some are tamed in several acts, others require some puzzle.

The problems solved with the help of the equation are only at first glance difficult. If you practice, then this process will reach the automatism.

Geometric figures

In order to understand the question, you need to understand the essence. Carefully read the condition, it's better to reread it several times. Problems for equations are only at first glance difficult. Let's consider an example for the beginning the most simple.

Given a rectangle, you need to find its area. Given: the width is 48% smaller than the length, the perimeter of the rectangle is 7.6 centimeters.

Solving problems in mathematics requires careful reading, logic. Let's handle it together. What do you need to take into account first of all? Denote the length of x. Therefore, in our equation, the width is 0.52x. We are given a perimeter - 7.6 centimeters. Let's find a half -perimeter, for this 7.6 centimeters we divide by 2, it is equal to 3.8 centimeters. We have obtained an equation with the help of which we find the length and width:

0.52x + x = 3.8.

When we get x (length), it will not be difficult to find 0.52x (width). If we know these two quantities, then we find the answer to the main question.

The problems solved with the help of the equation are not as complex as they seem, we could understand this from the first example. We found the length x = 2.5 centimeters, the width (we shall designate y) 0.52x = 1.3 centimeters. We pass to the square. It is found from the simple formula S = x * y (for rectangles). In our problem, S = 3.25. This will be the answer.

Let us consider some examples of solving problems with finding an area. And this time we take a rectangle. Solving problems in mathematics on finding the perimeter, the area of different figures is quite often. We read the condition of the problem: a rectangle is given, its length is 3.6 centimeters greater than the width, which is 1/7 of the perimeter of the figure. Find the area of this rectangle.

It will be convenient to denote the width of the variable x, and the length of the ( x + 3.6) centimeter. Let's find the perimeter:

P = 2x + 3.6 .

We can not solve the equation, since we have two variables in it. Therefore we look again at the condition. It says that the width is 1/7 of the perimeter. We obtain the equation:

1/7 (2x + 3.6) = x .

For the convenience of the solution, multiply each part of the equation by 7, so we get rid of the fraction:

2x + 3.6 = 7x.

After the solution, we get x (width) = 0.72 centimeters. Knowing the width, we find the length:

0.72 + 3.6 = 4.32 cm.

Now we know the length and width, answer the main question about what is equal to the area of the rectangle.

S = x * y , S = 3,1104 cm.

Tubes with milk

Solving problems with the help of equations causes a lot of difficulties for schoolchildren, despite the fact that this topic begins in the fourth grade. There are many examples, we looked at finding the area of the figures, now a little distracted from the geometry. Let's look at simple tasks with tabulation, they help visually: so the data that helps in the solution is better seen.

Invite the children to read the condition of the problem and create a table to help you compose the equation. Here's the condition: there are two cans, in the first three times more milk than in the second. If the first pour five liters into the second, the milk will be equally divided. Question: how much milk was in each can?

To help with the solution, you need to create a table. What should it look like?

Decision

	It was	Became
1 can	3x	3x5
2 cans	X	X + 5

How does this help in the formulation of the equation? We know that as a result of the milk has become equal, so the equation will look like this:

3x-5 = x + 5;

2x = 10;

X = 5.

We found the initial quantity of milk in the second can, which means that in the first one there were: 5 * 3 = 15 liters of milk.

Now a little explanation on the compilation of the table.

Why did we designate the first can for 3x: in the condition it is stipulated that the second milk cannon is three times less. Then we read that 5 liters were drained from the first canister, hence it became 3x-5 , and in the second they poured: x + 5 . Why did we equate these conditions? In the condition of the task it is said that the milk has become equal.

So we get the answer: the first cannon is 15 liters, the second - 5 liters of milk.

Determination of depth

By the condition of the problem: the depth of the first well is 3.4 meters greater than the second. The first well was increased by 21.6 meters, and the second - three times, after these actions the wells have the same depth. It is necessary to calculate what depth each well had originally.

Methods for solving problems are numerous, one can do actions, create equations or their system, but the second option is most convenient. To go to the solution, we create a table, as in the previous example.

Decision

	It was	Became
1 well	X + 3.4	X + 3.4 + 21.6
2 wells	X	3x

We now turn to the formulation of the equation. Since the wells are of the same depth, it has the following form:

X + 3.4 + 21.6 = 3x;

X = 3x = -25;

-2x = -25;

X = -25 / -2;

X = 12.5

We found the original depth of the second well, now we can find the first:

12.5 + 3.4 = 15.9 m.

After the done actions, we write down the answer: 15.9 m, 12.5 m.

Two brothers

Note that this task differs from all previous ones, because by the condition initially there was the same number of objects. Proceeding from this, the auxiliary table is drawn up in the reverse order, that is, from "it became" to "was."

Condition: two brothers were given an equal number of nuts, but the elder gave his brother 10, after that the nuts of the younger became five times as large. How many nuts are there for every boy?

Decision

	It was	Became
Older	Х + 10	X
Jr	5x - 10	5x

We form the equation:

X + 10 = 5x - 10;

-4x = -20;

Х = 5 - it became nuts in the elder brother;

5 * 5 = 25 - the younger brother.

Now you can write down the answer: 5 nuts; 25 nuts.

Purchases

The school needs to buy books and notebooks, the first more expensive than the second at 4.8 rubles. You need to calculate how much one copybook and one book cost, if you bought the same amount of money with five books and twenty-one notebooks.

Before proceeding to a solution, it is worth answering the following questions:

• What is the problem in the problem?
• How much did they pay?
• What did you buy?
• What values can be leveled?
• What do you need to know?
• What is the value of x ?

If you answered all the questions, then we turn to the solution. In this example, the value of x can be taken as the price of one notebook, and the cost of the book. Let's consider two possible variants:

1. X is the cost of one notebook, then x + 4.8 is the price of the book. Proceeding from this, we obtain the equation: 21x = S (x + 4.8).
2. X is the cost of the book, then x is 4.8 is the price of the notebook. The equation has the form: 21 (x - 4.8) = 5x.

You can choose a more convenient option for yourself, then solve two equations and compare the answers, they should coincide as a result.

The first way

The solution of the first equation:

21x = 5 (x + 4.8);

4,2х = х + 4,8;

4,2х - х = 4,8;

3,2х = 4,8;

Х = 1,5 (rubles) - cost of one notebook;

4.8 + 1.5 = 6.3 (rubles) - the cost of one book.

Another way to solve this equation (opening parentheses):

21x = 5 (x + 4.8);

21x = 5x + 24;

16x = 24;

Х = 1,5 (rubles) - cost of one notebook ;

1,5 + 4,8 = 6,3 (rubles) - the cost of one book.

The second way

5x = 21 (x = 4.8);

5x = 21x - 100.8;

16x = 100.8;

Х = 6,3 (rubles) - cost of 1 book;

6.3 - 4.8 = 1.5 (rubles) - the cost of one notebook.

As can be seen from the examples, the answers are identical, hence, the problem is solved correctly. Watch for the correctness of the solution, in our example, the answers should not be negative.

There are other problems that can be solved with the help of an equation, for example, on motion. Let's consider them in more detail in the following examples.

Two cars

In this section we will discuss the tasks of motion. To be able to solve them, you need to know the following rule:

S = V * T,

S is the distance, V is the speed, and T is the time.

Let's try to consider an example.

Two cars left simultaneously from point A to point B. The first traveled all the distance at the same speed, the second first half of the road was traveling at a speed of 24 km / h, and the second - 16 km / h. It is necessary to determine the speed of the first motorist, if in point B they came simultaneously.

What we need to compose the equation: main variable V ₁ (speed of the first car), secondary: S - path, T ₁ - time in the path of the first car. Equation: S = V ₁ * T _{1 .}

Next: the second car the first half of the way (S / 2) drove at a speed of V ₂ = 24 km / h. We get the expression: S / 2 = 24 * T _{2 .}

The next part of the way he drove at a speed V ₃ = 16 km / h. We obtain S / 2 = 16 * T ₃ .

Further from the condition it is clear that the cars arrived at the same time, hence T ₁ = T ₂ + T ₃ . Now we have to express the variables T ₁ , T ₂ , T ₃ from our previous conditions. We obtain the equation: S / V ₁ = (S / 48) + (S / 32).

S is taken as unity and we solve the equation:

1 / V ₁ = 1/48 + 1/32;

1 / V ₁ = (2/96) + (3/96);

1 / V ₁ = 5/96;

V ₁ = 96/5;

V ₁ = 19.2 km / h.

This is the answer. Problems solved using the equation are complex only at first glance. In addition to the above, you can meet work tasks, what it is, consider in the next section.

Job Challenge

To solve this type of task, you need to know the formula:

A = VT ,

Where A is the work, V is the productivity.

For a more detailed description, you need to give an example. The topic "Solving problems by an equation" (Grade 6) may not contain such problems, since this is a more complex level, but nevertheless we give an example for acquaintance.

Carefully read the condition: two workers work together and plan to perform for twelve days. It is necessary to determine how long it will take the first employee to perform the same norm on his own. It is known that he performs the amount of work for two days as a second employee in three days.

Solving problems for the formulation of equations requires a careful reading of the condition. The first thing we understood from the task, that the work is not defined, means, we take it as a unit, that is, A = 1 . If the problem refers to a certain number of parts or liters, then the work should be taken from these data.

We denote the productivity of the first and second workers through V ₁ and V _2, respectively, at this stage, the following equation is possible:

1 = 12 (V ₁ + V ₂ ) .

What does this equation tell us? That all work is done by two people in twelve hours.

Further we can state: 2V ₁ = 3V ₂ . Because the first for two days does as much as the second in three. We have obtained a system of equations:

1 = 12 (V1 + V2);

2V ₁ = 3V _2.

On the basis of the solution of the system, we obtained an equation with one variable:

1 - 8V ₁ = 12V _1;

V ₁ = 1/20 = 0.05.

This is the labor productivity of the first worker. Now we can find the time for which the first person will cope with all the work:

A = V ₁ * T _1;

1 = 0.05 * T ₁ ;

T ₁ = 20.

Since the day was taken as the unit of time, the answer is: 20 days.

Reformulation of the problem

If you have mastered the skill to solve traffic problems, and you have some difficulties with tasks for work, then it is possible to get traffic from work. How? If we take the last example, the condition is as follows: Oleg and Dima move towards each other, they meet in 12 hours. For how many will overcome the path independently Oleg, if it is known that in two hours he travels the way equal to Dima's way in three hours.