1. **State the problem:** Solve the system of linear equations: $$7x + 5y = -13$$ $$6x + 7y = -3$$ 2. **Method:** We will use the method of elimination to find $x$ and $y$. 3. **Eliminate one variable:** Multiply the first equation by 6 and the second by 7 to align coefficients of $x$: $$6(7x + 5y) = 6(-13) \Rightarrow 42x + 30y = -78$$ $$7(6x + 7y) = 7(-3) \Rightarrow 42x + 49y = -21$$ 4. **Subtract the first new equation from the second:** $$\cancel{42x} + 49y - (\cancel{42x} + 30y) = -21 - (-78)$$ $$49y - 30y = -21 + 78$$ $$19y = 57$$ 5. **Solve for $y$:** $$y = \frac{57}{19} = 3$$ 6. **Substitute $y=3$ into the first original equation:** $$7x + 5(3) = -13$$ $$7x + 15 = -13$$ 7. **Solve for $x$:** $$7x = -13 - 15$$ $$7x = -28$$ $$x = \frac{-28}{7} = -4$$ **Final answer:** $$x = -4, \quad y = 3$$