1. **State the problem:** Solve the system of linear equations: $$-3y + 4x = 13$$ $$-5y - 6x = -67$$ Find the values of $x$ and $y$. 2. **Rewrite the system for clarity:** Equation 1: $$4x - 3y = 13$$ Equation 2: $$-6x - 5y = -67$$ 3. **Use the elimination method to solve for one variable.** Multiply Equation 1 by 2 to align coefficients of $x$: $$2(4x - 3y) = 2(13)$$ $$8x - 6y = 26$$ Equation 2 is: $$-6x - 5y = -67$$ 4. **Add the two equations to eliminate $x$: ** $$8x - 6y + (-6x - 5y) = 26 + (-67)$$ $$ (8x - 6x) + (-6y - 5y) = -41$$ $$2x - 11y = -41$$ 5. **We made a mistake in elimination; we wanted to eliminate $x$, so instead multiply Equation 2 by 4 and Equation 1 by 6:** Multiply Equation 1 by 6: $$6(4x - 3y) = 6(13)$$ $$24x - 18y = 78$$ Multiply Equation 2 by 4: $$4(-6x - 5y) = 4(-67)$$ $$-24x - 20y = -268$$ 6. **Add the two equations to eliminate $x$: ** $$24x - 18y + (-24x - 20y) = 78 + (-268)$$ $$0x - 38y = -190$$ $$-38y = -190$$ 7. **Solve for $y$: ** $$y = \frac{-190}{-38} = \frac{190}{38}$$ Simplify the fraction: $$y = \frac{\cancel{190}^{5} \times 38^{\cancel{5}}}{\cancel{38}^{1} \times 38^{\cancel{1}}} = 5$$ 8. **Substitute $y=5$ into Equation 1 to solve for $x$: ** $$4x - 3(5) = 13$$ $$4x - 15 = 13$$ $$4x = 13 + 15$$ $$4x = 28$$ 9. **Solve for $x$: ** $$x = \frac{28}{4} = 7$$ **Final answer:** $$x = 7, \quad y = 5$$