To solve this system of linear equations:

$5x + 10y = 10$ $-x - 8y = 4$

we can use either substitution or elimination. Here, elimination might be efficient. Let's proceed with that:

1. Multiply the second equation by 5 to align terms with $x$: $-5x - 40y = 20$

2. Now, add this result to the first equation: $(5x + 10y) + (-5x - 40y) = 10 + 20$ $-30y = 30$

3. Solve for $y$: $y = -1$

4. Substitute $y = -1$ back into the first equation to find $x$: $5x + 10(-1) = 10$ $5x - 10 = 10$ $5x = 20$ $x = 4$

The solution to the system is: $(x, y) = (4, -1)$

Would you like further details or have any questions?

Here are five related questions to explore further:

1. How would the solution change if we used substitution instead?
2. Can we verify this solution by plugging the values back into the original equations?
3. How does elimination compare to substitution for solving linear systems?
4. What would happen if both equations were multiples of each other?
5. How can we use matrix methods to solve this system?

Tip: When using elimination, look for coefficients that allow easy cancellation of variables.