We are given the system of linear equations:

$x_1 + 2x_2 = -1 \quad (1)$ $4x_1 + 5x_2 = -10 \quad (2)$

To solve this, we'll use the substitution or elimination method. Let's proceed with the elimination method.

Step 1: Eliminate $x_1$

Multiply equation (1) by 4 to align the coefficients of $x_1$:

$4(x_1 + 2x_2) = 4(-1)$ $4x_1 + 8x_2 = -4 \quad (3)$

Now subtract equation (2) from equation (3):

$(4x_1 + 8x_2) - (4x_1 + 5x_2) = -4 - (-10)$ $(4x_1 - 4x_1) + (8x_2 - 5x_2) = -4 + 10$ $3x_2 = 6$

Step 2: Solve for $x_2$

$x_2 = \frac{6}{3} = 2$

Step 3: Substitute $x_2 = 2$ into equation (1)

Substitute $x_2 = 2$ into equation (1):

$x_1 + 2(2) = -1$ $x_1 + 4 = -1$ $x_1 = -1 - 4 = -5$

Final Solution:

The solution to the system of equations is:

$(x_1, x_2) = (-5, 2)$

Do you have any questions or need further details?

Here are some related questions to explore:

1. How would you solve the system using substitution instead of elimination?
2. Can the determinant method (Cramer's rule) be used to solve this system?
3. How do you graphically represent this system of equations?
4. What happens if the system has no solutions or infinite solutions?
5. How can matrix methods (e.g., Gaussian elimination) solve larger systems of equations?

Tip: When using elimination, always try to align coefficients by multiplication to simplify cancellation of variables.