1. **Problem:** Solve the linear system: $$\begin{cases} x - y + 4z = 0 \\ x - 2y - 4z = 1 \\ 2x - 5y - 4z = 2 \end{cases}$$ 2. **Formula and rules:** Use substitution or elimination to solve linear systems. 3. **Step 1:** Subtract the first equation from the second: $$ (x - 2y - 4z) - (x - y + 4z) = 1 - 0 $$ $$ x - 2y - 4z - x + y - 4z = 1 $$ $$ -y - 8z = 1 $$ 4. **Step 2:** Subtract twice the first equation from the third: $$ (2x - 5y - 4z) - 2(x - y + 4z) = 2 - 0 $$ $$ 2x - 5y - 4z - 2x + 2y - 8z = 2 $$ $$ -3y - 12z = 2 $$ 5. **Step 3:** From step 3 and 4, system reduces to: $$ \begin{cases} -y - 8z = 1 \\ -3y - 12z = 2 \end{cases} $$ 6. **Step 4:** Multiply first by 3: $$ -3y - 24z = 3 $$ 7. **Step 5:** Subtract second from this: $$ (-3y - 24z) - (-3y - 12z) = 3 - 2 $$ $$ -12z = 1 $$ $$ z = -\frac{1}{12} $$ 8. **Step 6:** Substitute $z$ into $-y - 8z = 1$: $$ -y - 8(-\frac{1}{12}) = 1 $$ $$ -y + \frac{2}{3} = 1 $$ $$ -y = 1 - \frac{2}{3} = \frac{1}{3} $$ $$ y = -\frac{1}{3} $$ 9. **Step 7:** Substitute $y$ and $z$ into first equation: $$ x - y + 4z = 0 $$ $$ x - (-\frac{1}{3}) + 4(-\frac{1}{12}) = 0 $$ $$ x + \frac{1}{3} - \frac{1}{3} = 0 $$ $$ x = 0 $$ **Final solution:** $$ (x,y,z) = \left(0, -\frac{1}{3}, -\frac{1}{12}\right) $$