Solve the system: x + y + z = 3, 2x – y + 3z = 3, x - 2y + 3z = 0

Question 27Select one:

a.
(1, 1, 1)

b.
(1, 2, 0)

c.
(2, 1, 0)

d.
No solution

e.
None of these

To solve the system of equations:

\[
\begin{aligned}
1. & \quad x + y + z = 3 \\
2. & \quad 2x - y + 3z = 3 \\
3. & \quad x - 2y + 3z = 0
\end{aligned}
\]

we can use the method of substitution, elimination, or matrix operations. I'll use elimination here.

**Step 1: Subtract equation 1 from equation 2 and equation 3 to eliminate \( x \):**

\[
(2x - y + 3z) - (x + y + z) = 3 - 3
\]
\[
x - 2y + 2z = 0 \quad \text{(Equation 4)}
\]

\[
(x - 2y + 3z) - (x + y + z) = 0 - 3
\]
\[
-3y + 2z = -3 \quad \text{(Equation 5)}
\]

**Step 2: Solve equation 5 for \( y \):**

\[
-3y + 2z = -3
\]
\[
3y = 2z + 3
\]
\[
y = \frac{2z + 3}{3}
\]

**Step 3: Substitute \( y \) from equation 5 into equation 4:**

\[
x - 2\left(\frac{2z + 3}{3}\right) + 2z = 0
\]

Multiply the entire equation by 3 to eliminate the fraction:

\[
3x - 4z - 6 + 6z = 0
\]
\[
3x + 2z = 6
\]
\[
x = \frac{6 - 2z}{3}
\]

**Step 4: Substitute \( x \) and \( y \) into equation 1 to find \( z \):**

\[
\frac{6 - 2z}{3} + \frac{2z + 3}{3} + z = 3
\]

Multiply the entire equation by 3 to clear the fractions:

\[
(6 - 2z) + (2z + 3) + 3z = 9
\]
\[
6 + 3z + 3 = 9
\]
\[
3z = 0
\]
\[
z = 0
\]

**Step 5: Find \( x \) and \( y \) using \( z = 0 \):**

Substitute \( z = 0 \) into the equations for \( x \) and \( y \):

\[
x = \frac{6 - 2(0)}{3} = 2
\]
\[
y = \frac{2(0) + 3}{3} = 1
\]

So, \( x = 2 \), \( y = 1 \), \( z = 0 \).

**Answer:**

The solution is \( (x, y, z) = (2, 1, 0) \).

The correct answer is:

**c. (2, 1, 0)**

Would you like any more details or have any questions? Here are some related questions:

1. How would you solve this system using matrix methods?
2. What happens if we change one of the coefficients in the system?
3. Can this method be applied to larger systems of equations?
4. How does this solution compare to solving with substitution?
5. What are the benefits of using elimination over other methods?

**Tip:** Always check your solution by substituting the values back into the original equations to ensure they satisfy all the equations.

Learn how to solve the system of linear equations x + y + z = 3, 2x - y + 3z = 3, x - 2y + 3z = 0 step-by-step. Understand the elimination method used to find the solution (2, 1, 0). Suitable for high school students studying systems of equations.

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