Question

Solve the equations:
1=-3x-2y
0=2x-y-3z
0=x+y+z

Answer 1

To solve the system of equations:

1. \( 1 = -3x - 2y \) (Equation 1)
2. \( 0 = 2x - y - 3z \) (Equation 2)
3. \( 0 = x + y + z \) (Equation 3)

Let's rewrite the equations for clarity:

\[
-3x - 2y = 1 \quad \text{(From Equation 1)}
\]
\[
2x - y - 3z = 0 \quad \text{(From Equation 2)}
\]
\[
x + y + z = 0 \quad \text{(From Equation 3)}
\]

### Step 1: Solve for \( z \) in terms of \( x \) and \( y \)

From Equation 3, we can express \( z \) as:

\[
z = -x - y
\]

### Step 2: Substitute \( z \) into Equation 2

Substituting \( z = -x - y \) into Equation 2 gives:

\[
2x - y - 3(-x - y) = 0
\]
\[
2x - y + 3x + 3y = 0
\]
\[
(2x + 3x) + (-y + 3y) = 0
\]
\[
5x + 2y = 0
\]

We can rearrange this as:

\[
2y = -5x \quad \Rightarrow \quad y = -\frac{5}{2}x \quad \text{(Equation 4)}
\]

### Step 3: Substitute \( y \) back into Equation 1

Now, substitute \( y = -\frac{5}{2}x \) into Equation 1:

\[
-3x - 2\left(-\frac{5}{2}x\right) = 1
\]
\[
-3x + 5x = 1
\]
\[
2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}
\]

### Step 4: Solve for \( y \) and \( z \)

Substituting \( x = \frac{1}{2} \) back into Equation 4 to find \( y \):

\[
y = -\frac{5}{2}\left(\frac{1}{2}\right) = -\frac{5}{4}
\]

Now, substitute \( x = \frac{1}{2} \) and \( y = -\frac{5}{4} \) into the expression for \( z \):

\[
z = -\left(\frac{1}{2}\right) - \left(-\frac{5}{4}\right)
\]
\[
z = -\frac{1}{2} + \frac{5}{4}
\]
\[
z = -\frac{2}{4} + \frac{5}{4} = \frac{3}{4}
\]

### Final Solution

The solution to the system of equations is:

\[
x = \frac{1}{2}, \quad y = -\frac{5}{4}, \quad z = \frac{3}{4}
\]

Thus, the answer is \( (x, y, z) = \left( \frac{1}{2}, -\frac{5}{4}, \frac{3}{4} \right) \).