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We are given the system of equations:

\[
\begin{cases}
y = 4 - x \\
z + 1 = x \\
y + 1 = z
\end{cases}
\]

To solve for \(x\), we will use substitution.

### Step 1: Solve for \(z\) in terms of \(y\)
From the third equation \( y + 1 = z \), we can express \(z\) as:
\[
z = y + 1
\]

### Step 2: Substitute \(z = y + 1\) into the second equation
We know \( z + 1 = x \), so substituting \( z = y + 1 \):
\[
(y + 1) + 1 = x
\]
Simplifying:
\[
x = y + 2
\]

### Step 3: Substitute \(y = 4 - x\) into \(x = y + 2\)
Now substitute \( y = 4 - x \) from the first equation into \(x = y + 2\):
\[
x = (4 - x) + 2
\]
Simplify the right side:
\[
x = 6 - x
\]

### Step 4: Solve for \(x\)
Add \(x\) to both sides:
\[
2x = 6
\]
Now divide by 2:
\[
x = 3
\]

Thus, the value of \(x\) is \(3\).

Would you like more details or have any questions?

Here are 5 related questions:
1. How can we substitute values in systems of equations?
2. Can you verify if the solution \(x = 3\) satisfies all the original equations?
3. What happens when the system has no solutions or infinitely many solutions?
4. How would the solution change if one of the equations were altered slightly?
5. Can we solve the system graphically, and how would that process look?

**Tip:** Always check your solution by plugging the values back into the original equations to ensure consistency.

Solve the system of equations: y = 4 - x, z + 1 = x, y + 1 = z. What is the value of x?

Learn how to solve the system of equations y = 4 - x, z + 1 = x, y + 1 = z using substitution. This problem explores fundamental algebraic concepts and is suitable for students in Grades 8-10.

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