Solve the system



x + y + 2z + w = 10
x − y − 3z + 2w = 3
4x − 2y − 7z + 7w = −2

We are given a system of three linear equations with four unknowns:

\[
\begin{cases}
x + y + 2z + w = 10 \\
x - y - 3z + 2w = 3 \\
4x - 2y - 7z + 7w = -2
\end{cases}
\]

Since there are three equations and four unknowns, we expect to find a family of solutions with one free variable.

### Step 1: Write the system in matrix form
The system can be written as a matrix equation \( A \vec{x} = \vec{b} \), where \( A \) is the coefficient matrix, \( \vec{x} \) is the vector of unknowns \([x, y, z, w]\), and \( \vec{b} \) is the constants on the right-hand side:

\[
\begin{bmatrix}
1 & 1 & 2 & 1 \\
1 & -1 & -3 & 2 \\
4 & -2 & -7 & 7
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ w
\end{bmatrix}
=
\begin{bmatrix}
10 \\ 3 \\ -2
\end{bmatrix}
\]

### Step 2: Apply Gaussian elimination
We will now use Gaussian elimination to reduce this system to row echelon form.

Start by subtracting the first row from the second and third rows to eliminate \(x\) in the second and third rows.

- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - 4R_1 \)

After this step, the system becomes:

\[
\begin{bmatrix}
1 & 1 & 2 & 1 \\
0 & -2 & -5 & 1 \\
0 & -6 & -15 & 3
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ w
\end{bmatrix}
=
\begin{bmatrix}
10 \\ -7 \\ -42
\end{bmatrix}
\]

### Step 3: Eliminate the \(y\)-term from the third row
To eliminate \(y\) in the third row, perform the operation:

\[
R_3 \leftarrow R_3 - 3R_2
\]

This yields the system:

\[
\begin{bmatrix}
1 & 1 & 2 & 1 \\
0 & -2 & -5 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z \\ w
\end{bmatrix}
=
\begin{bmatrix}
10 \\ -7 \\ 0
\end{bmatrix}
\]

### Step 4: Solve the system
We now have a simplified system of equations:

\[
\begin{cases}
x + y + 2z + w = 10 \\
-2y - 5z + w = -7
\end{cases}
\]

This is a system of two equations with four unknowns, so we expect two free variables.

#### Solve for \(w\) in terms of \(y\) and \(z\):
From the second equation:

\[
-2y - 5z + w = -7 \implies w = 2y + 5z - 7
\]

#### Substitute into the first equation:
Substitute \(w = 2y + 5z - 7\) into the first equation:

\[
x + y + 2z + (2y + 5z - 7) = 10
\]

Simplify:

\[
x + 3y + 7z = 17
\]

### Step 5: Express the general solution
From the equation \(x + 3y + 7z = 17\), we solve for \(x\):

\[
x = 17 - 3y - 7z
\]

Thus, the general solution is:

\[
\begin{cases}
x = 17 - 3y - 7z \\
y = y \\
z = z \\
w = 2y + 5z - 7
\end{cases}
\]

where \(y\) and \(z\) are free parameters.

### Final Solution:
The solution can be written as:

\[
(x, y, z, w) = (17 - 3y - 7z, y, z, 2y + 5z - 7)
\]

where \(y\) and \(z\) are free variables.

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Would you like to explore this solution further? Here are some questions related to this topic:

1. How can we check the consistency of this solution?
2. What happens when specific values are assigned to \(y\) and \(z\)?
3. How can the solution be represented in vector form?
4. How does the number of free variables relate to the number of equations and unknowns?
5. What are the implications of having free variables in a system of equations?

**Tip:** In systems of equations with more unknowns than equations, expect free variables, leading to infinitely many solutions.

Solve the system of equations: x + y + 2z + w = 10, x − y − 3z + 2w = 3, 4x − 2y − 7z + 7w = −2.

Explore how to solve the system of linear equations involving four unknowns using Gaussian elimination. The solution provides a step-by-step explanation of parametric forms, matrix representation, and free variables, suitable for college-level or advanced high school students.

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