1. **State the problem:** Solve the system of linear equations: $$\begin{cases} 2x_1 - 5x_2 - 6x_3 = 5 \\ -3x_1 + 2x_2 + 5x_3 = 1 \\ 2x_1 + 4x_2 - 3x_3 = -16 \end{cases}$$ 2. **Method:** We will use the method of elimination or substitution to find $x_1$, $x_2$, and $x_3$. 3. **Step 1: Eliminate one variable.** Multiply the first and third equations to align coefficients for elimination. Multiply equation 1 by 1 and equation 3 by -1: $$2x_1 - 5x_2 - 6x_3 = 5$$ $$-2x_1 - 4x_2 + 3x_3 = 16$$ Add these two equations: $$ (2x_1 - 2x_1) + (-5x_2 - 4x_2) + (-6x_3 + 3x_3) = 5 + 16 $$ $$ 0x_1 - 9x_2 - 3x_3 = 21 $$ Simplify: $$ -9x_2 - 3x_3 = 21 $$ Divide both sides by -3: $$ 3x_2 + x_3 = -7 $$ 4. **Step 2: Express $x_3$ in terms of $x_2$:** $$ x_3 = -7 - 3x_2 $$ 5. **Step 3: Substitute $x_3$ into the second equation:** $$ -3x_1 + 2x_2 + 5(-7 - 3x_2) = 1 $$ $$ -3x_1 + 2x_2 - 35 - 15x_2 = 1 $$ $$ -3x_1 - 13x_2 = 36 $$ 6. **Step 4: Express $x_1$ in terms of $x_2$:** $$ -3x_1 = 36 + 13x_2 $$ $$ x_1 = -\frac{36 + 13x_2}{3} $$ 7. **Step 5: Substitute $x_1$ and $x_3$ into the first equation:** $$ 2\left(-\frac{36 + 13x_2}{3}\right) - 5x_2 - 6(-7 - 3x_2) = 5 $$ $$ -\frac{72 + 26x_2}{3} - 5x_2 + 42 + 18x_2 = 5 $$ Multiply through by 3 to clear denominator: $$ -72 - 26x_2 - 15x_2 + 126 + 54x_2 = 15 $$ Combine like terms: $$ (-72 + 126) + (-26x_2 - 15x_2 + 54x_2) = 15 $$ $$ 54 + 13x_2 = 15 $$ 8. **Step 6: Solve for $x_2$:** $$ 13x_2 = 15 - 54 $$ $$ 13x_2 = -39 $$ $$ x_2 = -3 $$ 9. **Step 7: Find $x_3$ using $x_2 = -3$:** $$ x_3 = -7 - 3(-3) = -7 + 9 = 2 $$ 10. **Step 8: Find $x_1$ using $x_2 = -3$:** $$ x_1 = -\frac{36 + 13(-3)}{3} = -\frac{36 - 39}{3} = -\frac{-3}{3} = 1 $$ **Final solution:** $$ (x_1, x_2, x_3) = (1, -3, 2) $$