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Question

There are two pairs (x,y) of real numbers that satisfy the equation x+y = 3xy = 4. Given that the solutions x are in the form x = a bcd where a, b, c, and d are positive integers and the expression is completely simplified, what is the value of a + b + c + d?

Accepted Answer

We begin by considering the equation x + y = 4. From this, we know that y = 4-x. We can then substitute this into the equation 3xy = 4 to get 3x(4-x) = 12x - 3x^2 = 4. This then becomes 3x^2 - 12x + 4 = 0. Using the quadratic formula, we find align* x &= -b b^2 - 4ac2a x &= 12 144-486 x &= 6 263. align*Thus, a + b + c + d = 6 + 2 + 6 + 3 = 17.