1. **State the problem:** Solve the first system of linear equations using the reduction method: $$\begin{cases} 13x - 6y = -2 \\ 8x + 3y = -22 \end{cases}$$ 2. **Goal:** Eliminate one variable by making the coefficients of $y$ opposites. 3. Multiply the first equation by 3 and the second by 6 to align $y$ coefficients: $$\begin{cases} 3(13x - 6y) = 3(-2) \\ 6(8x + 3y) = 6(-22) \end{cases}$$ which gives $$\begin{cases} 39x - 18y = -6 \\ 48x + 18y = -132 \end{cases}$$ 4. Add the two equations to eliminate $y$: $$ (39x - 18y) + (48x + 18y) = -6 + (-132) $$ $$ 39x + 48x + \cancel{-18y} + \cancel{18y} = -138 $$ $$ 87x = -138 $$ 5. Solve for $x$: $$ x = \frac{-138}{87} $$ Simplify the fraction by dividing numerator and denominator by 3: $$ x = \frac{\cancel{-138}^\div3}{\cancel{87}^\div3} = \frac{-46}{29} $$ 6. Substitute $x = \frac{-46}{29}$ into the second original equation to find $y$: $$ 8x + 3y = -22 $$ $$ 8\left(\frac{-46}{29}\right) + 3y = -22 $$ $$ \frac{-368}{29} + 3y = -22 $$ 7. Move $\frac{-368}{29}$ to the right side: $$ 3y = -22 + \frac{368}{29} $$ Convert $-22$ to a fraction with denominator 29: $$ -22 = \frac{-638}{29} $$ So, $$ 3y = \frac{-638}{29} + \frac{368}{29} = \frac{-270}{29} $$ 8. Solve for $y$: $$ y = \frac{\frac{-270}{29}}{3} = \frac{-270}{29 \times 3} = \frac{-270}{87} $$ Simplify by dividing numerator and denominator by 3: $$ y = \frac{\cancel{-270}^\div3}{\cancel{87}^\div3} = \frac{-90}{29} $$ **Final solution:** $$ x = \frac{-46}{29}, \quad y = \frac{-90}{29} $$