You have a system of linear equations:

1. $2x + 3y = 12$
2. $x - y = 1$

To solve this system, we can use either substitution or elimination. Let's solve it using substitution.

Step 1: Solve one equation for one variable

From equation (2): $x - y = 1$, solve for $x$:

$x = y + 1$

Step 2: Substitute into the other equation

Substitute $x = y + 1$ into equation (1):

$2x + 3y = 12$

Substitute $x = y + 1$:

$2(y + 1) + 3y = 12$

Step 3: Solve for $y$

Simplify the equation:

$2y + 2 + 3y = 12$

Combine like terms:

$5y + 2 = 12$

Now, subtract 2 from both sides:

$5y = 10$

Finally, divide by 5:

$y = 2$

Step 4: Solve for $x$

Now that we know $y = 2$, substitute it into the equation $x = y + 1$:

$x = 2 + 1 = 3$

Solution:

The solution is $x = 3$ and $y = 2$.

Final Answer:

$x = 3, \quad y = 2$

Would you like more details on the solution process? Or have any questions about the steps?

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Here are 5 related questions for further practice:

1. Solve the system of equations $4x - y = 7$ and $2x + 3y = 10$.
2. Use the substitution method to solve the system $3x + 2y = 16$ and $5x - 4y = 2$.
3. Solve the system of equations using the elimination method: $2x + 5y = 15$ and $3x - 5y = 5$.
4. Find the values of $x$ and $y$ for the system: $x + 2y = 9$ and $4x - y = 5$.
5. How do you determine if a system of linear equations has no solution, one solution, or infinitely many solutions?

Tip: Always try to simplify the equations as much as possible before substitution or elimination—it can make the process much quicker!