To solve the given system of equations, we can use the method of substitution. Let's start by rearranging one of the equations to solve for one variable in terms of the other variable.
From the equation X - 3Y = -2, we can solve for X:
X = 3Y - 2.
Now we can substitute this expression for X in the first equation:
3(3Y - 2)^2 + 4Y^2 + 59(3Y - 2) - 4Y - 24 = 0.
Simplifying the equation, we get:
27Y^2 - 36Y + 12 + 4Y^2 + 177Y - 118 - 4Y - 24 = 0.
Combining like terms, we have:
31Y^2 + 137Y - 130 = 0.
Now we can solve this quadratic equation for Y. Factoring or using the quadratic formula, we find that Y = -5/31 or Y = 10/31.
Substituting these values of Y back into the equation X = 3Y - 2, we can find the corresponding values of X:
For Y = -5/31, X = 3(-5/31) - 2 = -15/31 - 62/31 = -77/31.
For Y = 10/31, X = 3(10/31) - 2 = 30/31 - 62/31 = -32/31.
Therefore, the solutions to the given system of equations are:
X = -77/31 and Y = -5/31, or
X = -32/31 and Y = 10/31.
Hence, the answer is: The solutions to the given system of equations are X = -77/31, Y = -5/31, and X = -32/31, Y = 10/31.