Consider the following system.

-x+2y+z=1

-2x-y-z=3

-3x-4y-3z=5

Choose the best description of its solution.

If applicable, give its solution.

Let's solve the given system of linear equations:

\[
\begin{aligned}
1. & \quad -x + 2y + z = 1 \\
2. & \quad -2x - y - z = 3 \\
3. & \quad -3x - 4y - 3z = 5 \\
\end{aligned}
\]

### Step 1: Simplify the equations

Start by multiplying the first equation by 2 to facilitate elimination with the second equation:

\[
-2x + 4y + 2z = 2
\]

### Step 2: Subtract the second equation from this result:

\[
(-2x + 4y + 2z) - (-2x - y - z) = 2 - 3
\]

Simplify:

\[
5y + 3z = -1 \quad \text{(Equation 4)}
\]

### Step 3: Use the first and second equations to eliminate \(z\) in another way:

Multiply the first equation by 3:

\[
-3x + 6y + 3z = 3
\]

Now subtract the third equation from this result:

\[
(-3x + 6y + 3z) - (-3x - 4y - 3z) = 3 - 5
\]

Simplify:

\[
10y + 6z = -2 \quad \text{(Equation 5)}
\]

### Step 4: Solve the resulting system of two equations:

Now we have the system:

\[
\begin{aligned}
1. & \quad 5y + 3z = -1 \\
2. & \quad 10y + 6z = -2 \\
\end{aligned}
\]

Notice that the second equation is just twice the first equation. This implies that the two equations are dependent, meaning that they represent the same plane.

### Step 5: Analyze the solution

Because the two equations are dependent, the system has infinitely many solutions. The solution can be parameterized by expressing \(y\) and \(z\) in terms of a parameter \(t\).

Let's express \(z\) in terms of \(y\):

\[
z = -\frac{5y + 1}{3}
\]

Now substitute back into one of the original equations to find \(x\) in terms of \(y\) or \(z\):

Using the first equation \(-x + 2y + z = 1\):

\[
-x + 2y - \frac{5y + 1}{3} = 1
\]

Solve for \(x\):

\[
-x + 2y - \frac{5y + 1}{3} = 1
\]

Multiply through by 3 to clear the fraction:

\[
-3x + 6y - 5y - 1 = 3
\]

Simplify:

\[
-3x + y - 1 = 3
\]

\[
-3x = -y + 4
\]

\[
x = \frac{y - 4}{3}
\]

### Solution in Parametric Form:

The general solution is:

\[
\begin{aligned}
x &= \frac{y - 4}{3}, \\
y &= y, \\
z &= -\frac{5y + 1}{3}.
\end{aligned}
\]

### Final Answer:

The system has **infinitely many solutions**. The solution can be written in parametric form as:

\[
\left( \frac{y - 4}{3}, y, -\frac{5y + 1}{3} \right)
\]

where \(y\) is any real number.

---

Would you like more details or have any questions?

Here are some related questions:

1. How can you determine if a system of equations has no solution, one solution, or infinitely many solutions?
2. What does it mean geometrically if a system of equations has infinitely many solutions?
3. How do you solve a system of equations using matrix methods like Gaussian elimination?
4. How can you verify if the parametric solution is correct?
5. What is the significance of a dependent equation in a system of linear equations?

**Tip:** In a system of linear equations, if two equations are multiples of each other, they represent the same plane or line, leading to infinitely many solutions when combined with other equations.

Learn how to solve the system of linear equations -x + 2y + z = 1, -2x - y - z = 3, -3x - 4y - 3z = 5 with a detailed step-by-step solution. This problem covers key concepts in algebra, including systems of linear equations and parametric solutions. Suitable for high school students.

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