To solve the given system of equations:

1. $-2x + 5y = 10$
2. $6x - 4y = -8$

We will use the substitution or elimination method. Here, elimination seems more straightforward.

Step 1: Align the equations for elimination

Multiply the first equation by 3 to align $x$-terms: $3(-2x + 5y) = 3(10)$ $-6x + 15y = 30$

The system now is: $-6x + 15y = 30$ $6x - 4y = -8$

Step 2: Add the equations to eliminate $x$

$(-6x + 15y) + (6x - 4y) = 30 + (-8)$ $11y = 22$

Solve for $y$: $y = \frac{22}{11} = 2$

Step 3: Substitute $y = 2$ into one of the original equations

Substitute into $-2x + 5y = 10$: $-2x + 5(2) = 10$ $-2x + 10 = 10$ $-2x = 0 \implies x = 0$

Final Solution:

The solution to the system is: $(x, y) = (0, 2)$

Correct answer: (0, 2)

Would you like additional details or clarifications?

Related Questions:

1. How would the solution change if the equations were altered slightly?
2. What happens if we use substitution instead of elimination for this system?
3. Can you explain graphically what solving this system represents?
4. How do we determine if a system has no solution or infinitely many solutions?
5. What are some shortcuts to quickly verify a solution?

Tip:

Always double-check your substitution step to ensure no arithmetic errors!