To solve the system of equations:

1. $x + y + z = 3$
2. $x - y + z = 2$
3. $x + y - 2z = 0$

Let's go through the steps.

Step 1: Eliminate $y$

Add equations (1) and (2):

$(x + y + z) + (x - y + z) = 3 + 2$

This simplifies to:

$2x + 2z = 5 \Rightarrow x + z = \frac{5}{2} \quad \text{(Equation A)}$

Now, add equations (1) and (3):

$(x + y + z) + (x + y - 2z) = 3 + 0$

This simplifies to:

$2x + 0y - z = 3 \Rightarrow 2x - z = 3 \quad \text{(Equation B)}$

Step 2: Solve for $x$ and $z$

From Equation A: $x + z = \frac{5}{2}$

From Equation B: $2x - z = 3$

Now, add these two new equations:

$(x + z) + (2x - z) = \frac{5}{2} + 3$

This simplifies to:

$3x = \frac{11}{2} \Rightarrow x = \frac{11}{6}$

Step 3: Substitute $x$ back to find $z$

Using $x + z = \frac{5}{2}$:

$\frac{11}{6} + z = \frac{5}{2}$

Solving for $z$:

$z = \frac{5}{2} - \frac{11}{6} = \frac{15 - 11}{6} = \frac{4}{6} = \frac{2}{3}$

Step 4: Substitute $x$ and $z$ back to find $y$

Using the first equation $x + y + z = 3$:

$\frac{11}{6} + y + \frac{2}{3} = 3$

Converting $\frac{2}{3}$ to $\frac{4}{6}$:

$\frac{11}{6} + y + \frac{4}{6} = 3$

$\frac{15}{6} + y = 3$

$y = 3 - \frac{15}{6} = \frac{18 - 15}{6} = \frac{3}{6} = \frac{1}{2}$

Solution

The solution to the system is:

$x = \frac{11}{6}, \quad y = \frac{1}{2}, \quad z = \frac{2}{3}$

Would you like more details on any step, or have any questions?

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1. How do you eliminate variables in a system of equations?
2. Can matrix methods solve this system?
3. What happens if we change one of the equations?
4. How can we verify the solution?
5. What other methods are available to solve this?

Tip: Always double-check each substitution step to ensure accuracy in solving systems.