Question

Боря купил 4 книги.Без первой книги все остальные стоят 42 рубля,без второй книги 40 рублей,без 3 38 рублей,без 4 36 рублей!Сколько стоит каждая книга?

Answer 1

II+III+IV=42   (1), I+III+!V=40   (2), I+II+IV=38   (3), !+II+III+36  (4)

(3)-(4)   IV-III=2,  (2)-(3)  III-II=2,   (1)-(2)   II-!=2

II=I+2,  II=III - 2,   I+2=III-2,   I+4=III

I+II+I+4=36,  2I=II=32,  2I+I+2=32,   3I=30,  I=10 руб - 1 книга

10+4=14 руб 3 книга,  14-2=12 руб - 2 книга,  14+2=16 руб 4 книга

Answer 2

Calculation of the Cost of Each Book

To calculate the cost of each book, we can use the given information that without the first book, the remaining three books cost 42 rubles, without the second book they cost 40 rubles, without the third book they cost 38 rubles, and without the fourth book they cost 36 rubles.

Let's assign variables to represent the cost of each book. Let x be the cost of the first book, y be the cost of the second book, z be the cost of the third book, and w be the cost of the fourth book.

From the given information, we can set up the following equations:

1. Without the first book, the remaining three books cost 42 rubles: y + z + w = 42. 2. Without the second book, the remaining three books cost 40 rubles: x + z + w = 40. 3. Without the third book, the remaining three books cost 38 rubles: x + y + w = 38. 4. Without the fourth book, the remaining three books cost 36 rubles: x + y + z = 36.

We now have a system of four equations with four unknowns. We can solve this system to find the values of x, y, z, and w.

Let's solve the system of equations:

y + z + w = 42    (Equation 1)
x + z + w = 40    (Equation 2)
x + y + w = 38    (Equation 3)
x + y + z = 36    (Equation 4)

Subtracting Equation 2 from Equation 1, we get:

(y + z + w) - (x + z + w) = 42 - 40
y + z - x = 2

Subtracting Equation 3 from Equation 2, we get:

(x + z + w) - (x + y + w) = 40 - 38
z - y = 2

Subtracting Equation 4 from Equation 3, we get:

(x + y + w) - (x + y + z) = 38 - 36
w - z = 2

Now we have a system of three equations with three unknowns:

y + z - x = 2    (Equation 5)
z - y = 2        (Equation 6)
w - z = 2        (Equation 7)

Adding Equation 6 to Equation 7, we get:

(z - y) + (w - z) = 2 + 2
w - y = 4

Adding Equation 5 to Equation 6, we get:

(y + z - x) + (z - y) = 2 + 2
2z - x = 4

Now we have a system of two equations with two unknowns:

w - y = 4        (Equation 8)
2z - x = 4       (Equation 9)

From Equation 8, we can express w in terms of y:

w = y + 4

Substituting this into Equation 9, we get:

2z - x = 4

Now we have a system of two equations with two unknowns:

w = y + 4        (Equation 10)
2z - x = 4       (Equation 11)

To solve this system, we need one more equation. Unfortunately, the given information does not provide any additional equations. Therefore, we cannot determine the exact values of x, y, z, and w with the given information.

However, we can express the cost of each book in terms of one variable. Let's express the cost of each book in terms of y:

x = 2z - 4
y = y
z = z
w = y + 4

Therefore, the cost of each book can be represented as follows:

- The cost of the first book (x) is 2z - 4 rubles. - The cost of the second book (y) is y rubles. - The cost of the third book (z) is z rubles. - The cost of the fourth book (w) is y + 4 rubles.

Please note that without additional information or constraints, we cannot determine the exact values of x, y, z, and w.