The system of equations in the image is:

$3x - 2y = 2$ $5x - 5y = 10$

We can solve this system using either substitution or elimination. I'll use the elimination method to find the solution.

Step 1: Multiply equations if needed to align coefficients

We want to eliminate one variable. Let's eliminate $x$ by multiplying the first equation by 5 and the second equation by 3, so that the coefficients of $x$ become the same.

$5(3x - 2y) = 5(2) \quad \Rightarrow \quad 15x - 10y = 10$ $3(5x - 5y) = 3(10) \quad \Rightarrow \quad 15x - 15y = 30$

Step 2: Subtract the equations

Now subtract the second equation from the first to eliminate $x$:

$(15x - 10y) - (15x - 15y) = 10 - 30$ $15x - 10y - 15x + 15y = -20$ $5y = -20$

Step 3: Solve for $y$

$y = \frac{-20}{5} = -4$

Step 4: Substitute $y = -4$ into one of the original equations

Let's substitute $y = -4$ into the first equation:

$3x - 2(-4) = 2$ $3x + 8 = 2$ $3x = 2 - 8$ $3x = -6$ $x = \frac{-6}{3} = -2$

Final Solution:

The solution to the system is: $x = -2, \quad y = -4$

Would you like further details or explanations on any step?

Here are 5 related questions to expand on this concept:

1. Can you solve the same system using the substitution method?
2. What is the graphical interpretation of this system of equations?
3. How would the solution change if the second equation was $5x - 5y = 12$?
4. What would happen if both equations were parallel (e.g., $3x - 2y = 2$ and $6x - 4y = 4$)?
5. How can you tell if a system of equations has no solution or infinite solutions?

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