1. **State the problem:** Solve the system of linear equations: $$\begin{cases}-4r - 6s - 5t = 1 \\ -r + s - 2t = 2 \\ r - 3s + t = -5 \end{cases}$$ 2. **Use substitution or elimination method:** We will use substitution here. 3. From the second equation, express $r$ in terms of $s$ and $t$: $$-r + s - 2t = 2 \implies r = s - 2t - 2$$ 4. Substitute $r = s - 2t - 2$ into the third equation: $$ (s - 2t - 2) - 3s + t = -5 $$ Simplify: $$ s - 2t - 2 - 3s + t = -5 $$ $$ -2s - t - 2 = -5 $$ $$ -2s - t = -3 $$ $$ t = -3 + 2s $$ 5. Substitute $r$ and $t$ into the first equation: $$ -4r - 6s - 5t = 1 $$ Replace $r$ and $t$: $$ -4(s - 2t - 2) - 6s - 5t = 1 $$ Substitute $t = -3 + 2s$: $$ -4(s - 2(-3 + 2s) - 2) - 6s - 5(-3 + 2s) = 1 $$ Simplify inside the parentheses: $$ s - 2(-3 + 2s) - 2 = s + 6 - 4s - 2 = -3s + 4 $$ So: $$ -4(-3s + 4) - 6s - 5(-3 + 2s) = 1 $$ Simplify: $$ 12s - 16 - 6s + 15 - 10s = 1 $$ Combine like terms: $$ (12s - 6s - 10s) + (-16 + 15) = 1 $$ $$ (-4s) - 1 = 1 $$ $$ -4s = 2 $$ $$ s = -\frac{1}{2} $$ 6. Find $t$: $$ t = -3 + 2s = -3 + 2(-\frac{1}{2}) = -3 - 1 = -4 $$ 7. Find $r$: $$ r = s - 2t - 2 = -\frac{1}{2} - 2(-4) - 2 = -\frac{1}{2} + 8 - 2 = \frac{11}{2} $$ **Final solution:** $$ r = \frac{11}{2}, \quad s = -\frac{1}{2}, \quad t = -4 $$