We can start by multiplying the first equation by 2 and the second equation by 3 to make the coefficients of $x$ the same.
This gives us the system of equations $24x + 42y = 30$ and $63x + 36y = 153$.
Next, we can subtract the first equation from the second equation to eliminate $x$.
This gives us $63x + 36y - (24x + 42y) = 153 - 30$, which simplifies to $39x - 6y = 123$.
Now we have a system of two equations with two variables: $39x - 6y = 123$ and $21x + 12y = 51$.
We can solve this system by multiplying the second equation by 2 and adding it to the first equation to eliminate $y$.
This gives us $39x - 6y + (42x + 24y) = 123 + 102$, which simplifies to $81x + 18y = 225$.
Dividing both sides of this equation by 9, we get $9x + 2y = 25$.
Now we have a system of two equations with two variables: $9x + 2y = 25$ and $21x + 12y = 51$.
We can solve this system by multiplying the first equation by 6 and subtracting it from the second equation to eliminate $y$.
This gives us $21x + 12y - (54x + 12y) = 51 - 150$, which simplifies to $-33x = -99$.
Dividing both sides of this equation by -33, we get $x = 3$.
Substituting this value of $x$ into either of the original equations, we can solve for $y$.
Using the first equation, we have $12(3) + 21y = 15$, which simplifies to $36 + 21y = 15$.
Subtracting 36 from both sides of this equation, we get $21y = -21$.
Dividing both sides of this equation by 21, we get $y = -1$.
Therefore, the ordered pair of integers $(x, y)$ that satisfies the system of equations is $(3, -1)$.
The answer is: (3,-1)
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Determine the ordered pair of integers $(x, y)$ that satisfies the system of equations $12x + 21y = 15$ and $21x + 12y = 51$.

### Answer:
