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# If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z?

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Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
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Hello fskilnik

So its possible just to find the value of z by taking any of the roots?

Like:

$$(z-3)^2 = 0$$

$$z = 3$$

And then just substitute?

Kind regards!
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Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
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jfranciscocuencag wrote:
Hello fskilnik

So its possible just to find the value of z by taking any of the roots?

Like:

$$(z-3)^2 = 0$$

$$z = 3$$

And then just substitute?

Kind regards!

Hi jfranciscocuencag !!

Thank you for your interest in my problem/solution.

Please be sure you understand the importance of EVEN POWERS, in the sense that (real number)^(positive even power) is always nonnegative...

That´s the reason why (for instance) $$a^2+(2b)^4+(c+3)^6 = 0$$ implies $$a^2=0$$ and $$(2b)^4=0$$ and $$(c+3)^6=0$$ ...

That mentioned/understood, note that there is no need to "take (square, fourth or sixth) roots" in each addend above, because a square (or a fourth or a sixth power) of a real number is zero only if the number itself is zero... (If the number is negative or positive, the square, fourth and sixth powers are positive.) In other words, we get immediately $$a=0$$ and $$2b=0$$ and $$c+3=0$$ in the scenario created... nice, isn´t it?!

I hope everything is clear now!

Regards and success in your studies,
Fabio.

P.S.: do not forget to press "Mention this user" palette before you call someone here (in this case, me, fskilnik). Reason: I came to your question by "luck", only...
(If you press the "Mention this user" button, then the person you call receive a message... the same way I called you here... got it?!) Thank you!
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Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
To tackle this problem, we break it down by setting each squared term in the equation to zero since the sum of squares on the left must equal the negative square on the right, which only makes sense if they're all zero. This gives us:

1. 2x + y - z = 0
2. x - y = 0
3. z - 3 = 0
4. 3y - z = 0

From these, it quickly unfolds:
x = y and z = 3
Plugging z = 3 into 3y - z = 0 simplifies to y = 1, and hence, x = 1

Finally, substituting x, y, and z into 3x + 2y + z calculates to 3(1) + 2(1) + 3 = 8

So, the answer is 8, choice (A). This method shows how breaking problems into smaller parts can clarify and simplify the solution process.
Re: If (2x+y-z)^2+(x-y)^2+(z-3)^2=-(3y-z)^2, what is the value of 3x+2y+z? [#permalink]
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