Solve the system of equations. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, enter INFINITELY MANY.) x + 1 3 y − 2z = −1 1 3 x + y + 1 2 z = 0 1 2 x + 1 3 y − 1 2 z = −1

We are given the following system of equations:

\[
1. \quad x + \frac{1}{3}y - 2z = -1
\]
\[
2. \quad \frac{1}{3}x + y + \frac{1}{2}z = 0
\]
\[
3. \quad \frac{1}{2}x + \frac{1}{3}y - \frac{1}{2}z = -1
\]

We will solve this system using substitution or elimination. First, let's convert these equations into a more standard form.

### Step 1: Eliminate fractions by multiplying each equation by an appropriate factor.

- Multiply equation 1 by 3:
  \[
  3x + y - 6z = -3 \quad \text{(Equation 1')}
  \]
- Multiply equation 2 by 6:
  \[
  2x + 6y + 3z = 0 \quad \text{(Equation 2')}
  \]
- Multiply equation 3 by 6:
  \[
  3x + 2y - 3z = -6 \quad \text{(Equation 3')}
  \]

The new system is:

\[
1'. \quad 3x + y - 6z = -3
\]
\[
2'. \quad 2x + 6y + 3z = 0
\]
\[
3'. \quad 3x + 2y - 3z = -6
\]

### Step 2: Solve the system using elimination.

Let's eliminate \(x\) by subtracting equations.

#### Subtract Equation 2' from Equation 1':
\[
(3x + y - 6z) - (2x + 6y + 3z) = -3 - 0
\]
\[
x - 5y - 9z = -3 \quad \text{(Equation 4)}
\]

#### Subtract Equation 3' from Equation 1':
\[
(3x + y - 6z) - (3x + 2y - 3z) = -3 - (-6)
\]
\[
-y - 3z = 3 \quad \text{(Equation 5)}
\]
\[
y + 3z = -3 \quad \text{(Rearranged Equation 5)}
\]

### Step 3: Substitute and solve.

We now have two new equations:
\[
4. \quad x - 5y - 9z = -3
\]
\[
5. \quad y + 3z = -3
\]

Substitute \(y = -3 - 3z\) (from Equation 5) into Equation 4:
\[
x - 5(-3 - 3z) - 9z = -3
\]
\[
x + 15 + 15z - 9z = -3
\]
\[
x + 6z = -18
\]
\[
x = -18 - 6z \quad \text{(Equation 6)}
\]

### Step 4: Find the solution.

Now, we have \(x = -18 - 6z\) and \(y = -3 - 3z\). These are parametric solutions in terms of \(z\), meaning there are infinitely many solutions.

### Conclusion:
Since we have a parametric solution for \(x\), \(y\), and \(z\) in terms of \(z\), the system has **infinitely many solutions**.

---

Would you like further details or have any questions?

Here are 5 related problems you might find interesting:
1. Solve a system of equations with two variables.
2. Solve a system of three linear equations using matrices.
3. Solve a system of nonlinear equations.
4. Solve a system of equations using Gaussian elimination.
5. Determine whether a system of equations is inconsistent.

**Tip:** When solving systems of equations, always check for special cases like "no solution" or "infinitely many solutions" based on the consistency of the equations.

Solve the system of equations. (If there is no solution, enter NO SOLUTION. If there are infinitely many solutions, enter INFINITELY MANY.) x + 1/3 y − 2z = −1, 1/3 x + y + 1/2 z = 0, 1/2 x + 1/3 y − 1/2 z = −1

Explore how to solve the system of linear equations with three variables, leading to infinitely many solutions. Learn key concepts in linear algebra such as parametric solutions, the substitution method, and criteria for infinite solutions. Suitable for Grades 10-12.

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