1. **State the problem:** Solve the system of simultaneous linear equations using the substitution method: $$2x + 3y = -12$$ $$5x + 2y = 14$$ 2. **Explain the substitution method:** We solve one equation for one variable and substitute that expression into the other equation. 3. **Solve the first equation for $x$:** $$2x + 3y = -12 \implies 2x = -12 - 3y \implies x = \frac{-12 - 3y}{2}$$ 4. **Substitute $x$ into the second equation:** $$5x + 2y = 14 \implies 5\left(\frac{-12 - 3y}{2}\right) + 2y = 14$$ 5. **Simplify and solve for $y$:** $$\frac{5(-12 - 3y)}{2} + 2y = 14$$ $$\frac{-60 - 15y}{2} + 2y = 14$$ Multiply both sides by 2 to clear the denominator: $$-60 - 15y + 4y = 28$$ $$-60 - 11y = 28$$ Add 60 to both sides: $$-11y = 88$$ Divide both sides by -11: $$y = -8$$ 6. **Substitute $y = -8$ back into the expression for $x$:** $$x = \frac{-12 - 3(-8)}{2} = \frac{-12 + 24}{2} = \frac{12}{2} = 6$$ 7. **Final solution:** $$x = 6, \quad y = -8$$ This means the two lines intersect at the point $(6, -8)$.