The system of linear equations shown in the image is:

$\ell_1: 5x + 2y = -11$ $\ell_2: 3x + 2y = -9$

To solve this system of equations, let's use the elimination method by subtracting the second equation from the first to eliminate $y$.

1. Subtract equation $\ell_2$ from $\ell_1$: $(5x + 2y) - (3x + 2y) = -11 - (-9)$

This simplifies to: $2x = -2$

Now, solve for $x$: $x = \frac{-2}{2} = -1$

2. Substitute $x = -1$ back into one of the original equations to find $y$. Using $\ell_2: 3x + 2y = -9$: $3(-1) + 2y = -9$ $-3 + 2y = -9$ $2y = -9 + 3 = -6$ $y = \frac{-6}{2} = -3$

Thus, the solution to the system is: $x = -1, \quad y = -3$

Would you like more details or have any questions?

Here are 5 related questions to expand on this concept:

1. What happens if the coefficients of $y$ were different? How would we solve the system?
2. Can we solve this system using the substitution method?
3. What is the graphical interpretation of the solution to this system of equations?
4. How would the solution change if we modified the constant terms in each equation?
5. Can you express this system of equations in matrix form and solve using matrix operations?

Tip: Always check your solution by plugging the values of $x$ and $y$ back into both original equations.