x = -3 is not a solution.

y = 4, then z = 6 (from y = z - 2) and x = 3 (from z = 2x).
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If you read the solution carefully, you'll see that the system of equation we have (\(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\)) has the following roots: \(x=3\), \(y=4\), \(z=6\) OR \(x=-1\), \(y=-4\), \(z=-2\). No other values of x, y and z satisfy these three equations (so x can be neither 1 nor 2). If x = 3, then \(x^3+y^2+z\) IS divisible by 7 but if x = -1 then it's NOT.

(1) says that x > 0, so x = 3 and \(x^3+y^2+z\) IS divisible by 7.
(2) says that y = 4, so x = 3 and \(x^3+y^2+z\) IS divisible by 7.

Hope it's clear.
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Hey bonion

I am not sure if u still have the same doubt. But here is the simple logic.

If we take x = -3 then does it satisfy ALL the other equations?
That is IF \(x = -3\) ,
Plug it in here \(z = 2x\) ----- We get \(z = -6\)
Next plug it here \(y = z - 2\) ----- We get \(y = -8\)

Now most importantly check \(y = -8\) in the first equation
Plug it here \(x^2 = y + 5\)
\(x^2 = -8 + 5\)
\(x^2 = 3\)

Do u see now that we cannot take x = -3? Because it does not satisfy all the equations.

Hope it helps!

First of all X=5, Y=8 and Z=10 does not satisfy \(x^2 = y + 5\).

Also, (2) says that \(y = 4\)
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The whole point was to utilize the given

\(x^2 = y + 5\), \(y = z - 2\) and \(z = 2x\) when utilized, will give x^2 -2x-3=0

Which on solving will give 2 solutions for x i.e. -1 and 3

which when put back in y 2x -2 will give y = -4 or y = 4 and give z = -2 or z =6

from 1, x > 0

one case will work here
y = 4 & z = 6 , x = 3

\(x^3 + y^2 + z\) divisible by 7
Sufficient

from 2, y =4
one case will work here
y = 4 & z = 6 , x = 3

\(x^3 + y^2 + z\) divisible by 7
Sufficient

D
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I think this is a high-quality question and I agree with explanation.
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Another approach - It is a good idea in general to assess what you may be able to piece together from the given information before you jump in further by taking on either statement. Here, the three given equations each contain just two unknowns, so substitution is a straightforward process. As soon as I saw a quadratic (\(x^2\)), I thought that x would likely have two solutions. Thus, I looked to get the first equation in terms of x only. This led me from the second to the third equation and back again pretty quickly: if \(z = 2x\), then, backtracking a bit, \(y = (2x) - 2\), and, reaching back a little further, \(x^2 = (2x - 2) + 5\), or \(x^2 - 2x -3 = 0\). This is a factorable quadratic, with binomial roots of (x - 3) and (x + 1). The solutions will be 3 or -1. With all the hard work out of the way, I took on Statement (1) first, rather than what I consider to be the easier statement in (2).

Why would it be important that x would be greater than 0? Because that would lead only to the conclusion that x = 3. It is thus not even necessary to test what could happen if x were to equal -1. In fact, I did not even bother to see what would happen when I substituted 3 for x in the modified quadratic, since I knew that regardless of the outcome, it would be possible to tell whether the eventual answer would be divisible by 7. Statement (1) had to be SUFFICIENT, narrowing the answer choices down to (A) or (D).

Statement (2) requires little effort to justify, since all three given equations must be considered. The middle equation would tell us that z must equal 6, and then the third equation that x must equal 3. At that point, the original quadratic would no longer have two justifiable solutions for x, and we would also be able to answer the question. Thus, (D) had to be the correct response, and both the effort and time it took to arrive at such a conclusion were minimal.

In Yes/No questions, all you need to be able to determine is whether there is enough evidence to reach a certain conclusion. It may not be necessary to actually work the numbers out.

I sometimes call myself "the lazy mathematician" in my tutoring sessions. I love math, but I would rather work less to get a correct solution than put in more effort and get... the same thing. I have adopted this approach not only because of some innate tendency, but also no doubt due to knowing a brilliant mathematician in college, a guy who tied for 24th in the Putnam collegiate math competition as a sophomore, and who later finished first in the nation in another year. I discovered (via a friendship that extended into chess club) that his extraordinary abilities stemmed not from seeing something in the aether that no one else could see, but from finding ways to break down complex matters into simpler pieces that he could readily put together to lead to a solution.

Good luck with your studies.

- Andrew