1. **State the problem:** Solve the system of linear equations: $$5x + 6y = 70$$ $$3x - 12y = -270$$ 2. **Formula and rules:** We can use the substitution or elimination method to solve this system. Here, we'll use elimination. 3. **Step 1: Make coefficients of $x$ or $y$ equal to eliminate one variable.** Multiply the first equation by 2 to align the $y$ coefficients: $$2(5x + 6y) = 2(70)$$ $$10x + 12y = 140$$ 4. **Step 2: Add the new equation to the second equation to eliminate $y$.** $$10x + 12y = 140$$ $$3x - 12y = -270$$ \cancel{+}$$ $$13x + \cancel{0} = -130$$ 5. **Step 3: Solve for $x$.** $$13x = -130$$ $$x = \frac{-130}{13}$$ $$x = -10$$ 6. **Step 4: Substitute $x = -10$ into one of the original equations to find $y$.** Using the first equation: $$5(-10) + 6y = 70$$ $$-50 + 6y = 70$$ $$6y = 70 + 50$$ $$6y = 120$$ $$y = \frac{120}{6}$$ $$y = 20$$ 7. **Final answer:** $$x = -10, \quad y = 20$$ This means the solution to the system is the point $(-10, 20)$ where both lines intersect.