1. **State the problem:** Solve the system of linear equations: $$-\frac{1}{3}x + \frac{1}{2}y = 14$$ $$2x + \frac{1}{5}y = -\frac{4}{5}$$ 2. **Rewrite the equations for clarity:** Equation 1: $$-\frac{1}{3}x + \frac{1}{2}y = 14$$ Equation 2: $$2x + \frac{1}{5}y = -\frac{4}{5}$$ 3. **Goal:** Find values of $x$ and $y$ that satisfy both equations simultaneously. 4. **Eliminate fractions by multiplying each equation by the least common denominator (LCD):** - For Equation 1, LCD is 6: $$6 \times \left(-\frac{1}{3}x + \frac{1}{2}y\right) = 6 \times 14$$ $$-2x + 3y = 84$$ - For Equation 2, LCD is 5: $$5 \times \left(2x + \frac{1}{5}y\right) = 5 \times -\frac{4}{5}$$ $$10x + y = -4$$ 5. **Rewrite the system without fractions:** $$-2x + 3y = 84$$ $$10x + y = -4$$ 6. **Use substitution or elimination. Here, use substitution:** From second equation: $$y = -4 - 10x$$ 7. **Substitute $y$ into the first equation:** $$-2x + 3(-4 - 10x) = 84$$ Simplify: $$-2x - 12 - 30x = 84$$ $$-32x - 12 = 84$$ 8. **Solve for $x$:** $$-32x = 84 + 12$$ $$-32x = 96$$ $$x = \frac{96}{-32} = -3$$ 9. **Find $y$ using $x = -3$:** $$y = -4 - 10(-3) = -4 + 30 = 26$$ 10. **Final solution:** $$x = -3, \quad y = 26$$ This means the point $(-3, 26)$ satisfies both equations.