X + 2Y + Z = 3
2X - 3Y - Z = 4
3X - Y - 2Z = 1

Q: X + 2Y + Z = 3 2X - 3Y - Z = 4 3X - Y - 2Z = 1

To solve the system of equations: 1. $ X + 2Y + Z = 3 $ 2. $ 2X - 3Y - Z = 4 $ 3. $ 3X - Y - 2Z = 1 $ We can use the method of elimination or substitution. Here, I'll use elimination: Step 1: Eliminate $ Z $ from equations (1) and (2). Add equation (1) and equation (2): \[ (X + 2Y + Z) + (2X - 3Y - Z) = 3 + 4 \] \[ 3X - Y = 7 \quad \text{(Equation 4)} \] Step 2: Eliminate $ Z $ from equations (1) and (3). Multiply equation (1) by 2: \[ 2(X + 2Y + Z) = 2 \times 3 \] \[ 2X + 4Y + 2Z = 6 \] Subtract equation (3) from this result: \[ (2X + 4Y + 2Z) - (3X - Y - 2Z) = 6 - 1 \] \[ -X + 5Y + 4Z = 5 \quad \text{(Equation 5)} \] Step 3: Solve the system of equations (4) and (5). From equation (4): \[ 3X - Y = 7 \quad \Rightarrow \quad Y = 3X - 7 \] Substitute $ Y = 3X - 7 $ into equation (5): \[ -X + 5(3X - 7) + 4Z = 5 \] \[ -X + 15X - 35 + 4Z = 5 \] \[ 14X + 4Z = 40 \] \[ 7X + 2Z = 20 \quad \text{(Equation 6)} \] Step 4: Solve equations (4) and (6). From equation (6): \[ 2Z = 20 - 7X \] \[ Z = 10 - \frac{7}{2}X \] Substitute $ Z = 10 - \frac{7}{2}X $ back into equation (4): \[ 3X - Y = 7 \] \[ Y = 3X - 7 \] Finally, substitute $ X = 2 $ in: \[ Y = 3(2) - 7 = 6 - 7 = -1 \] \[ Z = 10 - \frac{7}{2}(2) = 10 - 7 = 3 \] Thus, the solution is: \[ X = 2, \quad Y = -1, \quad Z = 3 \]

Question

Accepted Answer

To solve the system of equations:

1. $ X + 2Y + Z = 3 $
2. $ 2X - 3Y - Z = 4 $
3. $ 3X - Y - 2Z = 1 $

We can use the method of elimination or substitution. Here, I'll use elimination:

Step 1: Eliminate $ Z $ from equations (1) and (2).

Add equation (1) and equation (2):
$$
(X + 2Y + Z) + (2X - 3Y - Z) = 3 + 4
$$
$$
3X - Y = 7 \quad \text{(Equation 4)}
$$

Step 2: Eliminate $ Z $ from equations (1) and (3).

Multiply equation (1) by 2:
$$
2(X + 2Y + Z) = 2 \times 3
$$
$$
2X + 4Y + 2Z = 6
$$

Subtract equation (3) from this result:
$$
(2X + 4Y + 2Z) - (3X - Y - 2Z) = 6 - 1
$$
$$
-X + 5Y + 4Z = 5 \quad \text{(Equation 5)}
$$

Step 3: Solve the system of equations (4) and (5).

From equation (4):
$$
3X - Y = 7 \quad \Rightarrow \quad Y = 3X - 7
$$

Substitute $ Y = 3X - 7 $ into equation (5):
$$
-X + 5(3X - 7) + 4Z = 5
$$
$$
-X + 15X - 35 + 4Z = 5
$$
$$
14X + 4Z = 40
$$
$$
7X + 2Z = 20 \quad \text{(Equation 6)}
$$

Step 4: Solve equations (4) and (6).

From equation (6):
$$
2Z = 20 - 7X
$$
$$
Z = 10 - \frac{7}{2}X
$$

Substitute $ Z = 10 - \frac{7}{2}X $ back into equation (4):
$$
3X - Y = 7
$$
$$
Y = 3X - 7
$$

Finally, substitute $ X = 2 $ in:
$$
Y = 3(2) - 7 = 6 - 7 = -1
$$
$$
Z = 10 - \frac{7}{2}(2) = 10 - 7 = 3
$$

Thus, the solution is:
$$
X = 2, \quad Y = -1, \quad Z = 3
$$