We have a system of two linear equations:

1. $2x - 3y = 12$
2. $-9x + 6y = 21$

We can solve this system using either substitution or elimination. Let's use the elimination method.

Step 1: Multiply the first equation by 2 and the second equation by 1

This will align the coefficients of $y$, so we can eliminate $y$.

$(2x - 3y = 12) \times 2 \implies 4x - 6y = 24$ $(-9x + 6y = 21) \times 1 \implies -9x + 6y = 21$

Step 2: Add the two equations

Now, add the modified equations together:

$(4x - 6y) + (-9x + 6y) = 24 + 21$

Simplifying:

$4x - 9x = 45$ $-5x = 45$ $x = -9$

Step 3: Substitute $x = -9$ into one of the original equations

We can substitute $x = -9$ into the first equation:

$2x - 3y = 12$

Substitute $x = -9$:

$2(-9) - 3y = 12$ $-18 - 3y = 12$ $-3y = 12 + 18$ $-3y = 30$ $y = -10$

Final Answer:

The solution to the system of equations is: $x = -9, \quad y = -10$

Would you like any further explanation on the steps? Or do you have any additional questions?

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Related Questions:

1. How can the substitution method be applied to solve this system?
2. What does the graphical representation of this system of equations look like?
3. How do you solve systems of equations when one equation is quadratic?
4. What happens when the system has no solution or infinitely many solutions?
5. How do you solve systems with three variables?

Tip:

When solving systems of equations, it's useful to always check the solution by substituting the values of $x$ and $y$ back into both original equations. This verifies that the solution is correct.