The problem consists of two sub-problems. The first sub-problem asks to solve the following system of linear equations: $x + y = 12$ $2x + 3y = 32$ The second sub-problem is a word problem. A box contains 12 copybooks. Some are priced at 500 L.L. each and the others at 750 L.L. each. The total price is 8000 L.L. We need to find the number of copybooks of each type.

1. Problem Description

The problem consists of two sub-problems.

The first sub-problem asks to solve the following system of linear equations:

$x + y = 12$

$2x + 3y = 32$

The second sub-problem is a word problem. A box contains 12 copybooks. Some are priced at 500 L.L. each and the others at 750 L.L. each. The total price is 8000 L.L. We need to find the number of copybooks of each type.

2. Solution Steps

Sub-problem 1:

We are given the system:

$x + y = 12$ (1)

$2x + 3y = 32$ (2)

From equation (1), we have $x = 12 - y$.

Substitute this into equation (2):

$2(12 - y) + 3y = 32$

$24 - 2y + 3y = 32$

$y = 32 - 24$

$y = 8$

Substitute $y = 8$ back into equation (1):

$x + 8 = 12$

$x = 12 - 8$

$x = 4$

Thus, $x = 4$ and $y = 8$.

Sub-problem 2:

Let $n_1$ be the number of copybooks priced at 500 L.L. each, and $n_2$ be the number of copybooks priced at 750 L.L. each.

We are given that the total number of copybooks is 12, so

$n_1 + n_2 = 12$ (3)

The total price is 8000 L.L., so

$500n_1 + 750n_2 = 8000$ (4)

From equation (3), we have $n_1 = 12 - n_2$.

Substitute this into equation (4):

$500(12 - n_2) + 750n_2 = 8000$

$6000 - 500n_2 + 750n_2 = 8000$

$250n_2 = 8000 - 6000$

$250n_2 = 2000$

$n_2 = \frac{2000}{250}$

$n_2 = 8$

Substitute $n_2 = 8$ back into equation (3):

$n_1 + 8 = 12$

$n_1 = 12 - 8$

$n_1 = 4$

Thus, there are 4 copybooks priced at 500 L.L. each, and 8 copybooks priced at 750 L.L. each.

3. Final Answer

For the first sub-problem, $x = 4$ and $y = 8$.

For the second sub-problem, there are 4 copybooks of price 500 L.L. and 8 copybooks of price 750 L.L.