If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value : Problem Solving (PS)

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If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value

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Joined: 02 Sep 2009

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If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value [#permalink]

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06 Nov 2014, 08:56

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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

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Spoiler: OA

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Re: If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value [#permalink]

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06 Nov 2014, 11:59

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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Re: If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value [#permalink]

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10 Nov 2014, 01:32

\(\sqrt{3x} - \sqrt{2y + 2z} = 4\)

Say \(\sqrt{2y + 2z} = 2\)

then y = 1, z = 1

Say \(\sqrt{3x} = 6\)

then x = 12

These values also satisfy equation 3x – 2y – z = 32 + z

x+y+z = 12+1+1 = 14

Answer = E
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Re: If 3x – 2y – z = 32 + z and √(3x) - √(2y + 2z) = 4, what is the value [#permalink]

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22 Feb 2016, 05:30

3x – 2y – z = 32 + z means 3x – 2y – 2z = 32
Or 3x – (2y +2z) = 32

This is of the form \(a^2 - b^2 = 32\), i.e., (a+b)*(a-b) = 32

The second equation is of the form \(a - b = 4\). This means \(a + b = 32/4 = 8\)

Now a+b = 8, a-b=4. So 2a = 12, or a=6. (a is \sqrt{(3x)}). So x is 12.

Now, \sqrt{(3x)} - \sqrt{(2y + 2z)} = 4, or 6-\sqrt{(2y + 2z)} = 4
\sqrt{(2y + 2z)} = 2, or: (2y + 2z) = 4
(2y + 2z) + 2x = 4 + 2*12 = 28, or x+y+z = 14.