6x-9y-20z=-4
4x-15y+10z=-1
2x-3y-5z=-1

Answer

To solve the system of linear equations:

1. **Equations Given:**
\[ \begin{align*} (1) & \quad 6x - 9y - 20z = -4 \\ (2) & \quad 4x - 15y + 10z = -1 \\ (3) & \quad 2x - 3y - 5z = -1 \end{align*} \]

2. **Step 1: Simplify the equations.**
We can multiply the third equation by 2 to make it easier to work with:
\[ 2(2x - 3y - 5z) = 2(-1) \implies 4x - 6y - 10z = -2 \quad (4) \]

3. **Step 2: Eliminate \(x\) from equations (1) and (2).**
We can manipulate equations (1) and (2) to eliminate \(x\). First, we can multiply equation (2) by 1.5 to align the coefficients of \(x\):
\[ 1.5(4x - 15y + 10z) = 1.5(-1) \implies 6x - 22.5y + 15z = -1.5 \quad (5) \]

4. **Step 3: Subtract equation (1) from equation (5).**
\[ (6x - 22.5y + 15z) - (6x - 9y - 20z) = -1.5 - (-4) \]
Simplifying gives:
\[ -22.5y + 15z + 9y + 20z = 2.5 \\ -13.5y + 35z = 2.5 \quad (6) \]

5. **Step 4: Eliminate \(x\) from equations (4) and (1).**
Now, we can also eliminate \(x\) using equations (4) and (1). We can multiply equation (4) by 1.5:
\[ 1.5(4x - 6y - 10z) = 1.5(-2) \implies 6x - 9y - 15z = -3 \quad (7) \]

6. **Step 5: Subtract equation (1) from equation (7).**
\[ (6x - 9y - 15z) - (6x - 9y - 20z) = -3 - (-4) \]
Simplifying gives:
\[ -15z + 20z = 1 \\ 5z = 1 \\ z = \frac{1}{5} \]

7. **Step 6: Substitute \(z\) back to find \(y\).**
Substitute \(z = \frac{1}{5}\) into equation (6):
\[ -13.5y + 35\left(\frac{1}{5}\right) = 2.5 \\ -13.5y + 7 = 2.5 \\ -13.5y = 2.5 - 7 \\ -13.5y = -4.5 \\ y = \frac{-4.5}{-13.5} = \frac{1}{3} \]

8. **Step 7: Substitute \(y\) and \(z\) back to find \(x\).**
Substitute \(y = \frac{1}{3}\) and \(z = \frac{1}{5}\) into equation (3):
\[ 2x - 3\left(\frac{1}{3}\right) - 5\left(\frac{1}{5}\right) = -1 \\ 2x - 1 - 1 = -1 \\ 2x - 2 = -1 \\ 2x = 1 \\ x = \frac{1}{2} \]

9. **Final Solution:**
The values are:
\[ x = \frac{1}{2}, \quad y = \frac{1}{3}, \quad z = \frac{1}{5} \]