https://r2.erweima.ai/mathbot/image/1726476881528c0fzy9fn.jpg

The image contains two systems of linear equations from problems 47 and 48. Let's solve each system step by step.

### Problem 47:
\[
\begin{aligned}
x + y + z &= 6 \quad (1) \\
2x - y + z &= 3 \quad (2) \\
3x - z &= 0 \quad (3)
\end{aligned}
\]

**Step 1:** Solve equation (3) for \(z\):
\[
3x - z = 0 \quad \Rightarrow \quad z = 3x
\]
Substitute \(z = 3x\) into equations (1) and (2).

**Step 2:** Substituting into equation (1):
\[
x + y + 3x = 6 \quad \Rightarrow \quad 4x + y = 6 \quad (4)
\]
Now, substitute \(z = 3x\) into equation (2):
\[
2x - y + 3x = 3 \quad \Rightarrow \quad 5x - y = 3 \quad (5)
\]

**Step 3:** Solve the system of equations (4) and (5):
\[
4x + y = 6 \quad (4)
\]
\[
5x - y = 3 \quad (5)
\]
Add both equations:
\[
(4x + y) + (5x - y) = 6 + 3 \quad \Rightarrow \quad 9x = 9 \quad \Rightarrow \quad x = 1
\]

**Step 4:** Substitute \(x = 1\) into equation (4):
\[
4(1) + y = 6 \quad \Rightarrow \quad 4 + y = 6 \quad \Rightarrow \quad y = 2
\]

**Step 5:** Substitute \(x = 1\) into the expression for \(z\):
\[
z = 3(1) = 3
\]

So, the solution to Problem 47 is:
\[
x = 1, \quad y = 2, \quad z = 3
\]

---

### Problem 48:
\[
\begin{aligned}
x + y + z &= 2 \quad (1) \\
-x + 3y + 2z &= 8 \quad (2) \\
4x + y &= 4 \quad (3)
\end{aligned}
\]

**Step 1:** Solve equation (3) for \(y\):
\[
4x + y = 4 \quad \Rightarrow \quad y = 4 - 4x
\]
Substitute \(y = 4 - 4x\) into equations (1) and (2).

**Step 2:** Substituting into equation (1):
\[
x + (4 - 4x) + z = 2 \quad \Rightarrow \quad x + 4 - 4x + z = 2 \quad \Rightarrow \quad -3x + z = -2 \quad (4)
\]

Substituting into equation (2):
\[
-x + 3(4 - 4x) + 2z = 8 \quad \Rightarrow \quad -x + 12 - 12x + 2z = 8 \quad \Rightarrow \quad -13x + 2z = -4 \quad (5)
\]

**Step 3:** Solve the system of equations (4) and (5):
\[
-3x + z = -2 \quad (4)
\]
\[
-13x + 2z = -4 \quad (5)
\]

Solve equation (4) for \(z\):
\[
z = 3x - 2
\]
Substitute this into equation (5):
\[
-13x + 2(3x - 2) = -4 \quad \Rightarrow \quad -13x + 6x - 4 = -4 \quad \Rightarrow \quad -7x = 0 \quad \Rightarrow \quad x = 0
\]

**Step 4:** Substitute \(x = 0\) into the equation for \(z\):
\[
z = 3(0) - 2 = -2
\]

**Step 5:** Substitute \(x = 0\) into the expression for \(y\):
\[
y = 4 - 4(0) = 4
\]

So, the solution to Problem 48 is:
\[
x = 0, \quad y = 4, \quad z = -2
\]

---

Would you like more details or have any questions?

Here are 5 related questions for further practice:
1. How would you solve a system of three equations using the substitution method?
2. Can a system of three linear equations have no solution? Explain why.
3. What is the graphical interpretation of a solution to a system of three equations in three variables?
4. How do we use matrices to solve systems of linear equations?
5. Can a system of linear equations have an infinite number of solutions? Under what conditions?

**Tip:** When solving systems of equations, always start by simplifying one of the equations to express one variable in terms of others—this makes substitution easier!

Solve the following systems of equations:

Problem 47:
1. x + y + z = 6
2. 2x - y + z = 3
3. 3x - z = 0

Problem 48:
1. x + y + z = 2
2. -x + 3y + 2z = 8
3. 4x + y = 4

Learn how to solve systems of linear equations with three variables using the substitution and elimination methods. This example covers two problems: Problem 47 and Problem 48. Suitable for high school students (Grades 10-12).

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