Question

Рыбаки поймали 29 щук.когда один рыбак сварил уху из 6 щук,другой из 2,а третий из 3,то у каждого осталось равное количество рыб.Сколько рыб поймал каждый рыбак

Answer 1

Х- оставшееся количество щук
(х+6) - первый рыбак
(х+2) - второй рыбак
(х+3) - третий рыбак

(х+6)+(х+2)+(х+3)=29
х+6+х+2+х+3=29
3х+11=29
3х=29-11
3х=18
х=18/3
х=6

(х+6) = 6+6=12 первый рыбак
(х+2) =6+2=8 второй рыбак
(х+3)=6+3=9 третий рыбак 
всего 29

или намного проще, без уравнения
29 -(6+2+3) =29-11=18
18/3=6

Answer 2

Problem Analysis

Three fishermen caught a total of 29 pikes. One of the fishermen cooked a fish soup using 6 pikes, another used 2 pikes, and the third used 3 pikes. After cooking, each fisherman had an equal number of fish remaining. We need to determine how many fish each fisherman caught.

Solution

Let's assume that each fisherman caught x pikes. After cooking, each fisherman had an equal number of fish remaining, so we can set up the following equation:

x - 6 = x - 2 = x - 3

Simplifying the equation, we get:

x - 6 = x - 2 x - 6 = x - 3

Solving the first equation, we find:

x - x = 6 - 2 0 = 4

This equation has no solution, which means our assumption that each fisherman caught the same number of pikes is incorrect. Let's try a different approach.

Let's assume that the first fisherman caught x pikes, the second fisherman caught y pikes, and the third fisherman caught z pikes. We can set up the following equations based on the given information:

x + y + z = 29 x - 6 = y - 2 = z - 3

From the second equation, we can rewrite it as:

x = y + 6 y = z + 2

Substituting these values into the first equation, we get:

(y + 6) + y + (z + 2) = 29 2y + z + 8 = 29 2y + z = 21

Now we have a system of two equations:

2y + z = 21 x + y + z = 29

We can solve this system of equations to find the values of x, y, and z.

Solution Steps

1. Solve the system of equations: - 2y + z = 21 - x + y + z = 29

2. Substitute the values of y and z into the equation x = y + 6 to find the value of x.

3. Calculate the values of y and z.

Solution

Let's solve the system of equations:

2y + z = 21 (Equation 1) x + y + z = 29 (Equation 2)

From Equation 2, we can rewrite it as:

x = 29 - y - z (Equation 3)

Substituting Equation 3 into Equation 1, we get:

2y + z = 21 (Equation 1) 29 - y - z + y + z = 21 29 = 21

This equation is not possible, which means there is no solution that satisfies all the given conditions. It seems there is an error in the problem statement or the information provided.

Please double-check the information and let me know if there are any additional details or corrections.