https://r2.erweima.ai/mathbot/image/1726419169919zqc361c1.jpg

The image contains a system of equations with three variables, \(x\), \(y\), and \(z\), which are as follows:

1. \(1 - x + y + z = 1\)
2. \(8 \times 50 = 10x + 6y + 8z\)
3. \(x = 3y\)

Let me solve this system of equations step by step:

### Step 1: Simplify the equations

1. The first equation simplifies to:
   \[
   1 - x + y + z = 1 \quad \Rightarrow \quad -x + y + z = 0 \quad \Rightarrow \quad x = y + z
   \]

2. The second equation simplifies as follows:
   \[
   8 \times 50 = 400 \quad \Rightarrow \quad 400 = 10x + 6y + 8z
   \]

3. The third equation is \(x = 3y\).

### Step 2: Substitution

Substitute \(x = 3y\) into both the first and second equations:

1. From the first equation \(x = y + z\):
   \[
   3y = y + z \quad \Rightarrow \quad z = 2y
   \]

2. Substitute \(x = 3y\) and \(z = 2y\) into the second equation \(400 = 10x + 6y + 8z\):
   \[
   400 = 10(3y) + 6y + 8(2y) \quad \Rightarrow \quad 400 = 30y + 6y + 16y \quad \Rightarrow \quad 400 = 52y
   \]
   Solving for \(y\):
   \[
   y = \frac{400}{52} = \frac{100}{13}
   \]

### Step 3: Solve for \(x\) and \(z\)

Now that we know \(y = \frac{100}{13}\), we can find \(x\) and \(z\):

1. \(x = 3y = 3 \times \frac{100}{13} = \frac{300}{13}\)
2. \(z = 2y = 2 \times \frac{100}{13} = \frac{200}{13}\)

### Final Solution:
- \(x = \frac{300}{13}\)
- \(y = \frac{100}{13}\)
- \(z = \frac{200}{13}\)

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you solve a system of linear equations with substitution?
2. What are the alternative methods for solving such systems, like using matrices or elimination?
3. How do you interpret the result of a system with fractional solutions?
4. What is the significance of a consistent and independent system of equations?
5. Can you apply these methods to a system with more than three variables?

**Tip:** Always simplify the equations as much as possible before substituting or using other methods for solving systems.

Solve the system of equations: 1 - x + y + z = 1, 8 × 50 = 10x + 6y + 8z, x = 3y.

Solve a system of three equations using substitution, with variables x, y, and z. Learn how to simplify and substitute equations like x = 3y, 1 - x + y + z = 1, and 400 = 10x + 6y + 8z, suitable for students in Grades 8-10.

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