�� ���� ��� �������� �������� ������ ���� ��������� x, y � z�
� ���������������� ��� �������� ��
���  8 x + 24 y + 4 z = 12,
6 x + 18 y + 3 z = 9.
���  24 x + 4 y + 12 z = 8,
18 x + 3 y + 9 z = 6.

���



2 x + 6 y + z = 3,
4 y + 5 z = 8,
z = 7.

���



2 x + 6 y + z = 3,
10 x + 30 y + 5 z = 15,
4 y + 5 z = 8.

Let's analyze and solve the given systems of equations. I'll start by solving them one by one.

### System 1:
\[
\begin{aligned}
8x + 24y + 4z &= 12 \quad (1) \\
6x + 18y + 3z &= 9 \quad (2)
\end{aligned}
\]

We can simplify these equations by dividing each one by its greatest common divisor:

- Divide equation (1) by 4:
\[
2x + 6y + z = 3 \quad (3)
\]
- Divide equation (2) by 3:
\[
2x + 6y + z = 3 \quad (4)
\]

Since equations (3) and (4) are identical, this means that the two equations are essentially the same and do not provide enough information to solve for \(x\), \(y\), and \(z\) uniquely. So, the solution set for this system is not unique — it describes a family of solutions.

### System 2:
\[
\begin{aligned}
24x + 4y + 12z &= 8 \quad (5) \\
18x + 3y + 9z &= 6 \quad (6)
\end{aligned}
\]

We simplify these equations by dividing each one by its greatest common divisor:

- Divide equation (5) by 4:
\[
6x + y + 3z = 2 \quad (7)
\]
- Divide equation (6) by 3:
\[
6x + y + 3z = 2 \quad (8)
\]

Again, both equations (7) and (8) are identical, so this system also describes a family of solutions rather than a unique solution.

### System 3:
\[
\begin{aligned}
2x + 6y + z &= 3 \quad (9) \\
4y + 5z &= 8 \quad (10) \\
z &= 7 \quad (11)
\end{aligned}
\]

From equation (11), we know that:
\[
z = 7
\]

Substitute \(z = 7\) into equation (10):
\[
4y + 5(7) = 8 \\
4y + 35 = 8 \\
4y = 8 - 35 \\
4y = -27 \\
y = -\frac{27}{4}
\]

Now, substitute \(z = 7\) and \(y = -\frac{27}{4}\) into equation (9):
\[
2x + 6\left(-\frac{27}{4}\right) + 7 = 3 \\
2x - \frac{162}{4} + 7 = 3 \\
2x - \frac{162}{4} + \frac{28}{4} = 3 \\
2x - \frac{134}{4} = 3 \\
2x = 3 + \frac{134}{4} \\
2x = \frac{12}{4} + \frac{134}{4} \\
2x = \frac{146}{4} \\
x = \frac{73}{4}
\]

Thus, the solution for System 3 is:
\[
x = \frac{73}{4}, \quad y = -\frac{27}{4}, \quad z = 7
\]

### System 4:
\[
\begin{aligned}
2x + 6y + z &= 3 \quad (12) \\
10x + 30y + 5z &= 15 \quad (13) \\
4y + 5z &= 8 \quad (14)
\end{aligned}
\]

Equation (13) is a multiple of equation (12). Dividing equation (13) by 5:
\[
2x + 6y + z = 3 \quad (15)
\]

Since equation (12) and equation (15) are the same, we are left with two independent equations:
\[
\begin{aligned}
2x + 6y + z &= 3 \quad (16) \\
4y + 5z &= 8 \quad (17)
\end{aligned}
\]

Solving these two equations is the same as solving System 3. Therefore, the solution for System 4 is also:
\[
x = \frac{73}{4}, \quad y = -\frac{27}{4}, \quad z = 7
\]

### Conclusion:
- **System 1** and **System 2** have infinitely many solutions (they represent families of solutions).
- **System 3** and **System 4** have a unique solution: \(x = \frac{73}{4}\), \(y = -\frac{27}{4}\), \(z = 7\).

Would you like more details, or do you have any questions?

Here are some related questions to explore further:
1. How do you find a general solution for a system with infinitely many solutions?
2. Can matrix methods be used to solve systems of equations like these?
3. How does substitution work when solving systems of equations with multiple unknowns?
4. What are some conditions under which a system of linear equations has no solution?
5. How would these solutions change if the coefficients were altered slightly?

**Tip:** Simplifying equations by dividing by common factors early on helps make solving much easier and reduces the chances of calculation errors.

���� ��� �������� �������� ������ ���� ��������� x, y � z � ���������������� ��� �������� �� ��� 8 x + 24 y + 4 z = 12, 6 x + 18 y + 3 z = 9. ��� 24 x + 4 y + 12 z = 8, 18 x + 3 y + 9 z = 6. ���    2 x + 6 y + z = 3, 4 y + 5 z = 8, z = 7. ���    2 x + 6 y + z = 3, 10 x + 30 y + 5 z = 15, 4 y + 5 z = 8.

Explore how to solve systems of linear equations with variables x, y, and z. This detailed guide breaks down the process step-by-step, showing the simplification of equations and identifying when solutions are unique or represent families of solutions. Ideal for students in Grades 10-12.

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