Let's solve this system of equations using the elimination method step by step.

System of equations:

1. $-12x - 3y = 3$
2. $4x + 2y = -6$

Step 1: Eliminate one variable.

To eliminate one variable, we can manipulate the equations to have coefficients of one variable cancel out. Let's eliminate $y$.

To do this, we make the coefficients of $y$ equal by multiplying the first equation by 2 and the second equation by 3:

• $2(-12x - 3y) = 2(3)$ → $-24x - 6y = 6$
• $3(4x + 2y) = 3(-6)$ → $12x + 6y = -18$

Now the system becomes:

1. $-24x - 6y = 6$
2. $12x + 6y = -18$

Step 2: Add the equations to eliminate $y$.

Adding the two equations: $(-24x - 6y) + (12x + 6y) = 6 + (-18)$ Simplify: $-12x = -12$

Solve for $x$: $x = 1$

Step 3: Substitute $x = 1$ into one of the original equations.

Using the second equation $4x + 2y = -6$: $4(1) + 2y = -6$ $4 + 2y = -6$ $2y = -10$ $y = -5$

Final Answer:

The solution is $x = 1$ and $y = -5$. The system has one solution.

Would you like more details or have any further questions?

Here are 5 related questions to deepen understanding:

1. How does the elimination method differ from substitution?
2. What does it mean when the system has no solution geometrically?
3. How do you verify a solution for a system of equations?
4. What are the conditions for a system to have an infinite number of solutions?
5. Could elimination fail to solve a system? If yes, when and why?

Tip: Always verify the solution by substituting it back into both original equations!