Given the system of equations \begin{cases} x(x+y) &=& 9 \\ y(x+y) &=& 16 \end{cases}
the value of xy can be written as \frac{a}{b} where a and b are positive coprime integers. Find a+b.
To find the value of xy, we can try to eliminate one variable from the equations.
Starting with the first equation, x(x+y) = 9, we can expand it to get x^2 + xy = 9.
Similarly, for the second equation, y(x+y) = 16, we can expand it to get xy + y^2 = 16.
Now, we have a system of two equations:
x^2 + xy = 9 -------- (1)
xy + y^2 = 16 -------- (2)
To eliminate xy, we can subtract equation (1) from equation (2):
(xy + y^2) - (x^2 + xy) = 16 - 9
This simplifies to y^2 - x^2 = 7.
Now, we can factorize the left side: (y + x)(y - x) = 7.
Since 7 is a prime number, it can only be factored as 1*7 or -1*(-7). Therefore, we have two cases to consider:
Case 1: y + x = 7 and y - x = 1
Adding these two equations gives us 2y = 8, which implies y = 4. Substituting this value into either of the equations, we get x = 3.
Case 2: y + x = -7 and y - x = -1
Adding these two equations gives us 2y = -8, which implies y = -4. Substituting this value into either of the equations, we get x = -3.
Since the question specifies positive coprime integers, we discard the solution for x = -3 and y = -4.
Therefore, the valid solution is x = 3 and y = 4.
To find the value of xy, we substitute these values into any of the original equations:
xy = 3(3+4) = 3(7) = 21.
Thus, xy = 21, and a+b = 21+1 = 22.