The problem has two parts: Part 1: Solve the system of linear equations: $x + y = 12$ $2x + 3y = 32$ Part 2: A box contains 12 copybooks. Some cost 500 L.L. each and the others cost 750 L.L. each. The total price is 8000 L.L. We want to find the number of copybooks of each type.

1. Problem Description

The problem has two parts:

Part 1: Solve the system of linear equations:

$x + y = 12$

$2x + 3y = 32$

Part 2: A box contains 12 copybooks. Some cost 500 L.L. each and the others cost 750 L.L. each. The total price is 8000 L.L. We want to find the number of copybooks of each type.

2. Solution Steps

Part 1: Solving the system of equations.

We can use substitution or elimination. Let's use substitution.

From the first equation, we have $x = 12 - y$.

Substitute this into the second equation:

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

$24 - 2y + 3y = 32$

$y = 32 - 24$

$y = 8$

Now, substitute $y = 8$ back into $x = 12 - y$:

$x = 12 - 8$

$x = 4$

So the solution is $x = 4$ and $y = 8$.

Part 2: Determining the number of each type of copybook.

Let $a$ be the number of copybooks that cost 500 L.L. each, and let $b$ be the number of copybooks that cost 750 L.L. each.

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

$a + b = 12$

The total cost is 8000 L.L., so

$500a + 750b = 8000$

Divide the second equation by 250:

$2a + 3b = 32$

Now we have a system of two linear equations:

$a + b = 12$

$2a + 3b = 32$

From the first equation, $a = 12 - b$. Substitute this into the second equation:

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

$24 - 2b + 3b = 32$

$b = 32 - 24$

$b = 8$

Now, substitute $b = 8$ back into $a = 12 - b$:

$a = 12 - 8$

$a = 4$

Therefore, there are 4 copybooks that cost 500 L.L. each and 8 copybooks that cost 750 L.L. each.

3. Final Answer

Part 1: $x = 4$, $y = 8$

Part 2: There are 4 copybooks of price 500 L.L. and 8 copybooks of price 750 L.L.