We are given a system of two linear equations:

1. $x + 2y = 6$
2. $2x - 3y = 12$

To solve this system, we can use either the substitution or elimination method. Let's proceed with the elimination method to eliminate one variable and solve for the other.

Step 1: Eliminate $x$

To eliminate $x$, we can multiply the first equation by 2, so that the coefficients of $x$ in both equations are the same:

$2(x + 2y) = 2(6) \implies 2x + 4y = 12$

Now the system becomes:

1. & \quad 2x + 4y = 12 \\ 2. & \quad 2x - 3y = 12 \end{aligned}$$ ### Step 2: Subtract the equations Now subtract the second equation from the first to eliminate $$x$$: $$(2x + 4y) - (2x - 3y) = 12 - 12$$ Simplifying: $$2x + 4y - 2x + 3y = 0 \implies 7y = 0$$ ### Step 3: Solve for $$y$$ $$y = 0$$ ### Step 4: Solve for $$x$$ Substitute $$y = 0$$ into the first equation: $$x + 2(0) = 6 \implies x = 6$$ ### Solution: The solution to the system of equations is $$x = 6$$ and $$y = 0$$. Would you like further details or have any questions? Here are some related questions you might explore: 1. How do you solve systems of equations using substitution? 2. What are the benefits of using the elimination method over substitution? 3. How can we graphically represent a system of linear equations? 4. What happens if a system of equations has no solution? 5. How can we identify a system with infinitely many solutions? **Tip:** Always check your solution by plugging the values back into both original equations!