If the area of the region bounded by 16x^2 - 9y^2 = 144 and 8x - 3y = 24 is A, then 3(A + 6 _e(3)) is equal to _______.

Step-by-step solution

The given equations are: Hyperbola: 16x^2 - 9y^2 = 144 x^2 9 - y^2 16 = 1 Line: 8x - 3y = 24 y = 8 3 (x - 3) To find the points of intersection, substitute y from the line equation into the hyperbola equation: 16x^2 - 9 ( 8 3 (x - 3) )^2 = 144 16x^2 - 64(x^2 - 6x + 9) = 144 x^2 - 4(x^2 - 6x + 9) = 9 -3x^2 + 24x - 45 = 0 x^2 - 8x + 15 = 0 (x - 3)(x - 5) = 0 x = 3, x = 5 For x [3, 5] , the upper curve is the hyperbola y = 4 3 x^2 - 9 and the lower curve is the line y = 8 3 (x - 3) . The area A of the bounded region i