Manhattan Challenge Problem of the week! - 26/03/07

If 3x - 2y - z = 32 + z and (3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4 , what is the value of x + y + z ?

(A) 3
(B) 9
(C) 10
(D) 12
(E) 14

Answer: E

Each equation represents one of the elements in the common quadratic form:
Rewrite the given equation as follows:

3x - 2y - z = 32 + z
3x - (2y + 2z) = 32

Then, notice its relationship to the second given equation:

(3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4

The second equation is in the form a - b = 4,

while the first equation is in the form a ^ 2 - b ^ 2 = 32.

Since we know that and that a ^ 2 - b ^ 2 = 32 and that a - b = 4
we can solve for a + b , which must equal 8.

This gives us a third equation:

(3x) ^ 1/2 + (2y + 2z) ^ 1/2 = 8

Adding the second and third equations allows us to solve for x as follows:

(3x) ^ 1/2 - (2y + 2z) ^ 1/2 = 4
(3x) ^ 1/2 + (2y + 2z) ^1/2 = 8
_______________________

2{3 ^(1/2)} = 12
{3 ^(1/2)} = 6
3x = 36
x = 12

Plugging this value for x into the first equation allows us to solve for y + z as follows:

3x - 2y - 2z = 32
3(12) - 2y - 2z = 32
- 2y - 2z = -4
y + z = 2

The question asks for the value of x + y + z.

If x = 12 and y + z = 2, then x + y + z = 12 + 2 = 14.

The correct answer is E.