9514 1404 393
Answer:
Explanation:
Let x, y, z represent the numbers of A330, B767, and B787 aircraft to order, respectively. The problem statement suggests 3 equations:
330x +270y +240z = 6750 . . . . . total seats
250x +200y +250z = 5550 . . . . . total cost (millions)
2x -y -z = 0 . . . . . . . . . . . . . . . . . . . ratio of orders
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These can be solved by a variety of methods, including on-line equation solvers. If you want to do it by hand, substitution can help.
First divide by common factors:
11x +9y +8z = 225
5x +4y +5z = 111
Write an expression for z and substitute:
z = 2x -y
11x +9y +8(2x -y) = 225 ⇒ 27x +y = 225
5x +4y +5(2x -y) = 111 ⇒ 15x -y = 111
Adding the second of these to the first gives ...
(27x +y) +(15x -y) = (225) +(111)
42x = 336 . . . simplify
x = 8 . . . . . . . . divide by 42
15(8) -111 = y = 9
z = 2(8) -9 = 7
The solution to the equations is (x, y, z) = (8, 9, 7).
The order should be for 8 A330s, 9 B767s, and 7 B787s.