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

Question

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

mau5 · Accepted Answer

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.

PareshGmat · Answer

\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

rava · Answer

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.

stonecold · Answer

The values 1,1 would not always work ..

SPatel1992 · Answer

Hey! Great solutions. I got this after 2 tries. Anyone know a shortcut to this sum?

the27s · Answer

From the 2nd equation, it is clear that 'x' should be even.

Further, 3x should be perfect square & √(y+z) should give √2 form when reduced.

The Lowest values that satisfy these are , x=12, y=1, z=1.

I don't think one needs to look at any other option beyond E once x=12 is deduced as a possibility.

Posted from my mobile device

Kinshook · Answer

Asked: If 3x – 2y – z = 32 + z and  \sqrt{3x} - \sqrt{2y + 2z} = 4 , what is the value of x + y + z?

3x - 2y - 2z = 32
 \sqrt{3x} - 4 = \sqrt{2y + 2z} 
Squaring...
 3x + 16 - 8\sqrt{3x} = 2y + 2z 
 3x - 2y - 2z = 8\sqrt{3x} - 16 = 32 
 \sqrt{3x} = 6 
3x = 36
x = 12

36 - 2y - 2z = 32
y + z = 2

x + y + z = 12 + 2 = 14

IMO E

Paras10 · Answer

Here is a mathematical solution :
given 3x - 2y -z = 32 + z and √(3x) - √(2y+2z) = 4 ........ (1)

3x - 2y -z = 32 + z
3x - (2y - 2z ) = 32
√(3x)^2 - √(2y+2z)^2 = 32 -> m^2 - n^2 = (m-n )*(m+n)

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

(√(3x) + √(2y+2z)) = 8 .............(2)

now adding and subtracting eqn 1 and eqn 2
we get x = 12 and y+z = 2

Answer = 14 (E)

adewale223 · Answer

3x - 2y -z = 32

/3x - /2y+2z = 4..........i

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

4(/3x + /2y + 2z) = 32

/3x + /2y + 2z = 8............ii

Add equation i and ii to eliminate / 2y + 2z

2/3x = 12

/3x = 6

Square both sides

3x = 36

x = 12

Substitute for x = 12 in equation i

6 - /2y + 2z = 4

- /2y + 2z = -2

/2y + 2z = 2

Square both sides

2y + 2z = 4

y + z = 2

Therefore x + y + z 
12 + 2 = 14 option E

ZIX · Answer

Here's a simpler approach:

3x = 32 + 2z + 2y  (given) -- (1)
 \sqrt{3x}-\sqrt{2z+2y}=4  (given) -- (2)

Substitute value of 3x
 \sqrt{32+2z+2y}-\sqrt{2z+2y}=4

Let 2z + 2y = a
 \sqrt{32+a}-\sqrt{a}=4

Bring sqrt(a) to RHS and square both sides to remove square root:
 (\sqrt{32+a})^2=(4+\sqrt{a})^2 
 32+a=16+a+8\sqrt{a} 
 2 = \sqrt{a} 
 4 = a

OR
 4 = 2z + 2y  -- (3)
 2 = z + y

Substitute (3) in equation (1) and solve:
 3x = 32 + 4 
 x = 12

Therefore,  x+y+z = 12 + 2 = 14

sippy01032003 · Answer

Hi, I could see that you did (a+b)^2 for the left hand side where you had Under root 3x - 4 but did you not do the same for underroot 3x-32 on the right hand side? or you did? I am very confused so if you could explain what was the thought process behind it, I would be grateful.

Bunuel · Answer

Squaring  \sqrt{3x} - 4 = \sqrt{3x-32}  yields:

(\sqrt{3x} - 4)^2 = (\sqrt{3x-32})^2

(\sqrt{3x})^2 -2*\sqrt{3x}*4 + 4^2 = 3x-32

3x -8\sqrt{3x} +16 = 3x-32

8\sqrt{3x} = 48

\sqrt{3x} = 6

3x = 36

x = 12

Dbrunik · Answer

solution

suburbMBA · Answer

I understand the solution but I'm not sure that I'd be able to come up this within the time allotted for it, during the exam. Is there anyway to create a system that ensures I get more of these right?

Any help would be appreciated.

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

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

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