Solve the simultaneous equation using elimination method: 2x+y−z=1, x−y+2z=4, 3x+2y−z=2

Add equation (1) and equation (2):

$$(2x+y-z) + (x-y+2z) = 1+4$$ $$3x + z = 5 \quad \text{(Equation 4)}$$

Multiply equation (1) by 2:

$$2(2x+y-z) = 2(1)$$ $$4x+2y-2z=2 \quad \text{(Equation 1')}$$

Subtract equation (1') from equation (3):

$$(3x+2y-z) - (4x+2y-2z) = 2-2$$ $$3x+2y-z-4x-2y+2z = 0$$ $$-x+z=0 \quad \text{(Equation 5)}$$

We have: 4) $3x+z=5$ 5) $-x+z=0$ From equation (5), we can see that $z=x$. Substitute $z=x$ into equation (4):

$$3x+x=5$$ $$4x=5$$ $$x=\frac{5}{4}$$

Since $z=x$:

$$z=\frac{5}{4}$$

Let's use equation (1):

$$2x+y-z=1$$ $$2\left(\frac{5}{4}\right)+y-\frac{5}{4}=1$$ $$\frac{10}{4}+y-\frac{5}{4}=1$$ $$\frac{5}{2}+y-\frac{5}{4}=1$$

To combine the fractions, find a common denominator:

$$\frac{10}{4}+y-\frac{5}{4}=1$$ $$\frac{10-5}{4}+y=1$$ $$\frac{5}{4}+y=1$$ $$y=1-\frac{5}{4}$$ $$y=\frac{4}{4}-\frac{5}{4}$$ $$y=-\frac{1}{4}$$

The solution to the system of equations is:

$$x=\frac{5}{4}, \quad y=-\frac{1}{4}, \quad z=\frac{5}{4}$$

The final answer is $\boxed{x=\frac{5}{4}, y=-\frac{1}{4}, z=\frac{5}{4}}$.

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