1. **State the problem:** Solve the simultaneous equations using the elimination method: $$2x + 3y = 13$$ $$3x + 2y = 12$$ 2. **Elimination method formula:** We aim to eliminate one variable by making the coefficients of either $x$ or $y$ the same (or opposites) in both equations, then subtract or add the equations. 3. **Eliminate $y$:** Multiply the first equation by 2 and the second by 3 to make the coefficients of $y$ equal: $$\text{Eq1}: 2(2x + 3y) = 2(13) \Rightarrow 4x + 6y = 26$$ $$\text{Eq2}: 3(3x + 2y) = 3(12) \Rightarrow 9x + 6y = 36$$ 4. **Subtract Eq1 from Eq2 to eliminate $y$:** $$ (9x + 6y) - (4x + 6y) = 36 - 26 $$ $$ 9x - \cancel{6y} - 4x - \cancel{6y} = 10 $$ $$ 5x = 10 $$ 5. **Solve for $x$:** $$ x = \frac{10}{5} = 2 $$ 6. **Substitute $x=2$ into the first original equation to find $y$:** $$ 2(2) + 3y = 13 $$ $$ 4 + 3y = 13 $$ $$ 3y = 13 - 4 = 9 $$ $$ y = \frac{9}{3} = 3 $$ 7. **Final solution:** $$ x = 2, \quad y = 3 $$ This means the solution to the system is the point $(2,3)$ where both equations intersect.