Consider the following system of equations.



w + x + y + z = 6
w + y + z = 4
w + y = 2
(∗)
(a) List the leading variables .
(b) List the free variables .
(c) The general solution of (∗) (expressed in terms of the free variables) is
( , , , ) .
(d) Suppose that a fourth equation −2w + y = 5 is included in the system (∗). What is
the solution of the resulting system? Answer: ( , , , ).
(e) Suppose that instead of the equation in part (d), the equation −2w − 2y = −3 is
included in the system (∗). Then what can you say about the solution(s) of the
resulting system? Answer: .

We are given the following system of equations:

\[
\begin{cases}
w + x + y + z = 6 \\
w + y + z = 4 \\
w + y = 2
\end{cases}
\]

Let's solve each part step by step.

---

### (a) List the **leading variables**:
Leading variables are variables that can be solved directly from the system of equations without any free parameters. By analyzing the system:

1. From \(w + y = 2\), we can directly solve for \(w\) or \(y\).
2. From \(w + y + z = 4\), once we have \(w\) or \(y\), we can solve for \(z\).
3. From \(w + x + y + z = 6\), we can solve for \(x\) once \(w, y, z\) are determined.

Thus, **leading variables**: \(w\), \(y\), \(z\), and \(x\).

---

### (b) List the **free variables**:
Free variables are variables that we cannot uniquely determine, and whose values can be arbitrary. In this case, there are no free variables, as all variables \(w, x, y, z\) can be uniquely determined by solving the system.

Thus, there are **no free variables**.

---

### (c) The general solution of \((*)\) (expressed in terms of the free variables):

Let's solve the system to find the general solution. Start with the third equation:

1. \( w + y = 2 \) ⟶ solve for \(w = 2 - y\).
2. Substitute \(w = 2 - y\) into the second equation:
   \[
   (2 - y) + y + z = 4 \implies 2 + z = 4 \implies z = 2.
   \]
3. Now substitute \(w = 2 - y\) and \(z = 2\) into the first equation:
   \[
   (2 - y) + x + y + 2 = 6 \implies 4 + x = 6 \implies x = 2.
   \]

So, the solution in terms of free variables is:  
\[
(w, x, y, z) = (2 - y, 2, y, 2).
\]

---

### (d) Adding the fourth equation \( -2w + y = 5 \):

Now, let's consider adding this fourth equation:
\[
-2w + y = 5.
\]
Substitute \(w = 2 - y\) into this equation:
\[
-2(2 - y) + y = 5 \implies -4 + 2y + y = 5 \implies 3y = 9 \implies y = 3.
\]

Once \(y = 3\), substitute into the previous expressions for \(w, x, z\):
- \( w = 2 - 3 = -1 \),
- \( z = 2 \) (from before),
- \( x = 2 \) (from before).

Thus, the solution is:
\[
(w, x, y, z) = (-1, 2, 3, 2).
\]

---

### (e) Adding the equation \( -2w - 2y = -3 \):

Now, consider the new fourth equation:
\[
-2w - 2y = -3.
\]
Substitute \(w = 2 - y\) into this equation:
\[
-2(2 - y) - 2y = -3 \implies -4 + 2y - 2y = -3 \implies -4 = -3,
\]
which is a **contradiction**.

Therefore, the system is **inconsistent**, and there is **no solution**.

---

### Summary of answers:
- (a) **Leading variables**: \(w, x, y, z\).
- (b) **Free variables**: None.
- (c) **General solution**: \( (w, x, y, z) = (2 - y, 2, y, 2) \).
- (d) **Solution of the system with \(-2w + y = 5\)**: \( (-1, 2, 3, 2) \).
- (e) **Solution of the system with \(-2w - 2y = -3\)**: The system is **inconsistent**, so there is **no solution**.

---

Let me know if you would like more details or have any further questions!

Here are 5 related questions:
1. How do you determine if a variable is leading or free in a system of equations?
2. Can a system with no free variables have infinitely many solutions?
3. How does a contradiction in a system of equations lead to no solution?
4. What conditions must a system of linear equations satisfy to be consistent?
5. How can you represent the general solution of a consistent system in terms of parameters?

**Tip**: To identify if a system is inconsistent, look for equations that reduce to a false statement (e.g., \( 0 = 1 \)).

Consider the following system of equations.

w + x + y + z = 6
w + y + z = 4
w + y = 2

(a) List the leading variables.
(b) List the free variables.
(c) The general solution of (*) (expressed in terms of the free variables) is ( , , , ).
(d) Suppose that a fourth equation −2w + y = 5 is included in the system (*). What is the solution of the resulting system? Answer: ( , , , ).
(e) Suppose that instead of the equation in part (d), the equation −2w − 2y = −3 is included in the system (*). Then what can you say about the solution(s) of the resulting system? Answer: .

Explore the solution to a system of linear equations involving the variables w, x, y, and z. Learn how to identify leading variables, free variables, and determine the consistency of the system when an additional equation is introduced. This problem is ideal for high school students in Grades 10-12.

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