The given system of equations is:

1. $2x + 3y = 5$
2. $5x - 2y = -16$

Let's solve this system of linear equations using either substitution or elimination. I'll use the elimination method.

Step 1: Eliminate one variable

To eliminate a variable, we need to multiply the equations so that the coefficients of one variable (either $x$ or $y$) become the same. Let's eliminate $y$.

We can multiply the first equation by 2 and the second equation by 3 to match the coefficients of $y$:

$(2x + 3y = 5) \times 2 \Rightarrow 4x + 6y = 10$ $(5x - 2y = -16) \times 3 \Rightarrow 15x - 6y = -48$

Now, we have the system: 4x + 6y = 10 \tag{3} 15x - 6y = -48 \tag{4}

Step 2: Add the equations

Now, we add equations (3) and (4) to eliminate $y$:

$(4x + 6y) + (15x - 6y) = 10 + (-48)$ $4x + 15x = -38$ $19x = -38$

Solve for $x$:

$x = \frac{-38}{19} = -2$

Step 3: Substitute $x = -2$ into one of the original equations

Substitute $x = -2$ into equation (1) to find $y$:

$2(-2) + 3y = 5$ $-4 + 3y = 5$ $3y = 5 + 4 = 9$ $y = \frac{9}{3} = 3$

Final Solution:

The solution to the system of equations is: $x = -2, \quad y = 3$

Would you like a detailed breakdown of this process or have any questions?

Here are 5 related questions:

1. How can the substitution method be used to solve this system?
2. What happens if you try to solve the system graphically?
3. How do you check if the solution is correct?
4. Can you solve the system using matrices?
5. What other real-world applications rely on solving linear systems?

Tip: Always check your solution by substituting both values back into the original equations to verify correctness.