The system of equations 16(x-4)^2+9 and y=-7(x-4)^2+9 will have
A. 1 solution
B. 2 solutions
C. no solutions
D. an infinite number of solutions
16(x-4)^2+9 = -7(x-4)^2+9
23x^2 - 184x + 368 = 0
x^2 - 8x + 16 = 0
(x-4)^2 = 0
so only one solution: x=4