We are given the system of linear equations:

$4x + 5y = 3$ $3x - 2y = 8$

To solve this system, we can use the substitution method or the elimination method. I'll use the elimination method here to solve for $x$ and $y$.

Step 1: Make the coefficients of $y$ the same in both equations.

We have the coefficients of $y$ as 5 and -2. To eliminate $y$, let's multiply the first equation by 2 and the second equation by 5. This will make the coefficients of $y$ equal to 10 and -10.

$(2) \cdot (4x + 5y) = 2 \cdot 3 \quad \Rightarrow \quad 8x + 10y = 6$ $(5) \cdot (3x - 2y) = 5 \cdot 8 \quad \Rightarrow \quad 15x - 10y = 40$

Step 2: Add the two equations to eliminate $y$.

$(8x + 10y) + (15x - 10y) = 6 + 40$ $8x + 15x = 46$ $23x = 46$

Step 3: Solve for $x$.

$x = \frac{46}{23} = 2$

Step 4: Substitute $x = 2$ into one of the original equations to find $y$.

Let's substitute $x = 2$ into the first equation:

$4x + 5y = 3$ $4(2) + 5y = 3$ $8 + 5y = 3$ $5y = 3 - 8$ $5y = -5$ $y = \frac{-5}{5} = -1$

Final Answer:

The solution to the system of equations is:

$x = 2, \quad y = -1$

Let me know if you'd like a further explanation or if you have any other questions!

---

Related Questions:

1. How do you solve a system of equations using substitution?
2. What are the graphical interpretations of a system of linear equations?
3. What happens if the system has no solution or infinitely many solutions?
4. How can we use matrices to solve a system of equations?
5. Can a system of linear equations have exactly one solution, no solution, or infinitely many solutions?

Tip: Always check your solution by substituting $x$ and $y$ back into both original equations to confirm the values satisfy both equations.