The image contains a system of linear equations. Let's solve the system:

1. $2x + 3y = 4$
2. $3x - 8y = -9$

Step 1: Solve for one variable using substitution or elimination.

We will use the elimination method to eliminate one variable. To eliminate $x$, we will multiply the first equation by 3 and the second equation by 2, so that the coefficients of $x$ become the same:

$(3) \cdot (2x + 3y) = 3 \cdot 4$ $6x + 9y = 12$

$(2) \cdot (3x - 8y) = 2 \cdot (-9)$ $6x - 16y = -18$

Step 2: Subtract the second equation from the first.

$(6x + 9y) - (6x - 16y) = 12 - (-18)$ $6x - 6x + 9y + 16y = 12 + 18$ $25y = 30$

Step 3: Solve for $y$.

$y = \frac{30}{25} = \frac{6}{5} = 1.2$

Step 4: Substitute $y = 1.2$ into one of the original equations to find $x$.

Using the first equation $2x + 3y = 4$: $2x + 3(1.2) = 4$ $2x + 3.6 = 4$ $2x = 4 - 3.6$ $2x = 0.4$ $x = \frac{0.4}{2} = 0.2$

Final Solution:

$x = 0.2, \quad y = 1.2$

Let me know if you'd like further details or have any questions!

Here are some related questions:

1. What is the graphical interpretation of solving a system of linear equations?
2. How can you check the solution to a system of equations?
3. Can all systems of linear equations be solved by elimination or substitution?
4. What are the possible outcomes when solving systems of linear equations (one solution, no solution, infinite solutions)?
5. How would solving this system change if one of the equations were non-linear?

Tip: The elimination method is often easier when both equations have similar coefficients for one variable, but you can also rearrange and solve using substitution if preferred.