1. **State the problem:** We are given a system of three linear equations with three variables $r$, $s$, and $t$: $$\begin{cases} 7r - 2s + t = 2 \\ 2r + 2s - t = -2 \\ 2r - s - 2t = 1 \end{cases}$$ We need to find the values of $r$, $s$, and $t$ that satisfy all three equations simultaneously. 2. **Write the system clearly:** $$\begin{cases} 7r - 2s + t = 2 \\ 2r + 2s - t = -2 \\ 2r - s - 2t = 1 \end{cases}$$ 3. **Add the first two equations to eliminate $t$:** $$ (7r - 2s + t) + (2r + 2s - t) = 2 + (-2) $$ $$ 7r - 2s + t + 2r + 2s - t = 0 $$ $$ (7r + 2r) + (-2s + 2s) + (t - t) = 0 $$ $$ 9r + 0 + 0 = 0 $$ $$ 9r = 0 $$ $$ r = 0 $$ 4. **Substitute $r=0$ into the first and second equations:** - First equation: $$ 7(0) - 2s + t = 2 $$ $$ -2s + t = 2 $$ - Second equation: $$ 2(0) + 2s - t = -2 $$ $$ 2s - t = -2 $$ 5. **Add these two new equations to eliminate $t$ again:** $$ (-2s + t) + (2s - t) = 2 + (-2) $$ $$ (-2s + 2s) + (t - t) = 0 $$ $$ 0 + 0 = 0 $$ This confirms the system is consistent so far. 6. **Solve for $t$ from the first new equation:** $$ -2s + t = 2 $$ $$ t = 2 + 2s $$ 7. **Substitute $t = 2 + 2s$ into the third original equation:** $$ 2r - s - 2t = 1 $$ $$ 2(0) - s - 2(2 + 2s) = 1 $$ $$ -s - 4 - 4s = 1 $$ $$ -5s - 4 = 1 $$ $$ -5s = 5 $$ $$ s = -\cancel{\frac{5}{5}}{1} $$ 8. **Find $t$ using $s = -1$:** $$ t = 2 + 2(-1) = 2 - 2 = 0 $$ 9. **Final solution:** $$ r = 0, \quad s = -1, \quad t = 0 $$ This solution satisfies all three equations.