To solve the system of equations using the elimination method, we aim to eliminate one variable by making its coefficients equal in both equations. The given system of equations is:

1. $-2x + 3y = -2$
2. $-4x + 7y = -10$

Step 1: Make the coefficients of $x$ equal

To eliminate $x$, we can multiply the first equation by 2, so that the coefficient of $x$ in the first equation matches the coefficient in the second equation:

$2(-2x + 3y) = 2(-2)$

$-4x + 6y = -4$

Now the system becomes:

$1'. \, -4x + 6y = -4$ $2. \, -4x + 7y = -10$

Step 2: Subtract one equation from the other

Subtract equation $1'$ from equation $2$:

$(-4x + 7y) - (-4x + 6y) = -10 - (-4)$

Simplify:

$y = -6$

Step 3: Substitute $y = -6$ into one of the original equations

Substitute $y = -6$ into the first equation, $-2x + 3y = -2$:

$-2x + 3(-6) = -2$

$-2x - 18 = -2$

$-2x = 16$

$x = -8$

Final Solution:

The solution to the system of equations is:

$x = -8, \, y = -6$

Verification:

Substitute $x = -8$ and $y = -6$ into both equations to ensure they are satisfied:

1. $-2(-8) + 3(-6) = 16 - 18 = -2$ ✅
2. $-4(-8) + 7(-6) = 32 - 42 = -10$ ✅

The solution is correct.

---

Do you have any questions, or would you like further details on any step?

Here are 5 related questions to deepen your understanding:

1. How would you solve this system using the substitution method instead?
2. What happens if both variables were eliminated during elimination?
3. How can you identify inconsistent systems using elimination?
4. Can elimination work efficiently with three equations and three variables?
5. How does elimination compare to graphing methods for solving systems?

Tip: Always verify your solution by substituting back into the original equations.