The problem consists of two parts. The first part asks to solve a system of three linear equations with three unknowns, $x$, $y$, and $z$. The second part involves a word problem about students learning different languages. We need to form equations based on the given information and show that the solutions to these equations are the same as the solution to the linear system from part 1.

1. Problem Description

The problem consists of two parts. The first part asks to solve a system of three linear equations with three unknowns, $x$, $y$, and $z$. The second part involves a word problem about students learning different languages. We need to form equations based on the given information and show that the solutions to these equations are the same as the solution to the linear system from part

1.

2. Solution Steps

Part 1: Solving the system of linear equations.

The system is given by:

$x + y + 3z = 46$ (1)

$x + 2y + z = 40$ (2)

$y + z = 30$ (3)

From equation (3), we have $z = 30 - y$. Substitute this into equations (1) and (2):

$x + y + 3(30 - y) = 46$

$x + y + 90 - 3y = 46$

$x - 2y = -44$ (4)

$x + 2y + (30 - y) = 40$

$x + y = 10$ (5)

Subtracting (4) from (5):

$(x + y) - (x - 2y) = 10 - (-44)$

$3y = 54$

$y = 18$

Substituting $y = 18$ into equation (5):

$x + 18 = 10$

$x = -8$

Substituting $y = 18$ into $z = 30 - y$:

$z = 30 - 18$

$z = 12$

Therefore, the solution to the system is $x = -8$, $y = 18$, $z = 12$.

Part 2a: Let F, Y, and K be the number of students who study only Fulfulde, only Yemba, and only Kako, respectively. Let a be the number of students who study Fulfulde and Yemba, b the number of students who study Fulfulde and Kako, and c the number of students who study Yemba and Kako. 10 students study all three languages. The total number of students is

9

6.

We are given:

Total students: $96$.

Students studying all three: $10$.

Students studying Yemba: $50$.

Students studying Kako: $40$.

Students studying Fulfulde: $56$.

From this we can write the following equations:

$F + Y + K + a + b + c + 10 = 96$ (Total students) (6)

$Y + a + c + 10 = 50$ (Yemba) (7)

$K + b + c + 10 = 40$ (Kako) (8)

$F + a + b + 10 = 56$ (Fulfulde) (9)

$K = F + Y$ (10)

$b = Y/2$ (11)

$F = 3(Y + K)$ (12)

We are also given that we need to show $a$, $b$, $c$ are solutions to the original linear system.

Equation (7) becomes $Y + a + c = 40$.

Equation (8) becomes $K + b + c = 30$.

Equation (9) becomes $F + a + b = 46$.

Let $x=a$, $y=b$, and $z=c$.

$F + x + y = 46$

$Y + x + z = 40$

$K + y + z = 30$

We want to solve for $x, y, z$. From equation (6) we have:

$F + Y + K + a + b + c = 86$

Using (7), (8), and (9), we have:

$F + x + y = 46$

$Y + x + z = 40$

$K + y + z = 30$

We want to show that $x = -8$, $y = 18$, $z = 12$.

Thus we want $a = -8, b = 18, c = 12$.

Also, $F + Y + K + a + b + c + 10 = 96$, or $F + Y + K - 8 + 18 + 12 + 10 = 96$, giving $F + Y + K = 64$.

Now we have the equations $x + y + 3z = 46, x + 2y + z = 40, y + z = 30$. Solving this system gives us

$x = -8, y = 18, z = 12$. Thus $a = -8$, $b = 18$, $c = 12$.

Thus, a, b, and c are a solution to the system (S).

3. Final Answer

Part 1: $x = -8$, $y = 18$, $z = 12$

Part 2a: a, b, and c are a solution to the system (S).