Bunuel wrote:
Tough and Tricky questions: Algebra.
If 3x – 2y – z = 32 + z and \(\sqrt{3x} - \sqrt{2y + 2z} = 4\), what is the value of x + y + z?
(A) 3
(B) 9
(C) 10
(D) 12
(E) 14
Kudos for a correct solution.From equation 1, we know that 3x-32 = 2y+2z.
Now, we also know that \(\sqrt{3x} - \sqrt{2y + 2z} = 4\)
or
\(\sqrt{3x} - 4 = \sqrt{2y + 2z}\) -->\(\sqrt{3x} - 4 = \sqrt{3x-32}\)
Square on both sides, to get \(3x+16 -8\sqrt{3x} = 3x-32\), cancel 3x on both sides, and we get
\(8\sqrt{3x} = 48 --> x = 12\)
Thus, 3x-32 = 2y+2z, hence, y+z = 2 and x+y+z = 14.
E.
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