To solve this system of equations using the elimination method, we need to eliminate one of the variables by adding or subtracting the equations. Let's work through this step by step.

The system is: $6x - 10y = -8$ $6x - 5y = 2$

Step 1: Set up for elimination

To eliminate $x$, we subtract the second equation from the first. We multiply the second equation by $-1$ to make the coefficients of $x$ cancel each other out:

$6x - 10y = -8 \quad \text{(unchanged)}$ $-6x + 5y = -2 \quad \text{(multiplied by -1)}$

Step 2: Add the equations together

Now, we add the two equations:

$(6x - 10y) + (-6x + 5y) = -8 + (-2)$

This simplifies to:

$-5y = -10$

Step 3: Solve for $y$

Now, solve for $y$ by dividing both sides of the equation by $-5$:

$y = \frac{-10}{-5} = 2$

Step 4: Substitute $y = 2$ into one of the original equations

Now that we have $y = 2$, we substitute this value into the second original equation $6x - 5y = 2$:

$6x - 5(2) = 2$ $6x - 10 = 2$

Step 5: Solve for $x$

Add 10 to both sides:

$6x = 12$

Now divide by 6:

$x = \frac{12}{6} = 2$

Final Solution

The solution to the system is $x = 2$ and $y = 2$.

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Filling in the missing numbers:

1. $\left(6x - 10y = -8\right) \rightarrow -6x + 10y = -2$
2. $5y = -10$
3. $y = 2$
4. $6x - 5(2) = 2$
5. $6x = 12$
6. $x = 2$

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Would you like more details on any of these steps?

Here are some related questions to explore:

1. How would you solve the system if you wanted to eliminate $y$ first instead of $x$?
2. Can you solve the system using substitution instead of elimination?
3. What happens if the coefficients of $x$ and $y$ do not match easily? How do you modify the method?
4. What is the geometric interpretation of solving a system of linear equations?
5. How does the solution method change for a system with no solutions or infinitely many solutions?

Tip: Always double-check the final solution by substituting both $x$ and $y$ into the original equations.