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Re: M05-28 [#permalink]
Hello,

One quick question, I solved the algebraic part of the question and got the solution X = -1 or 3. Do I have enough information at this point to answer the question yes or no without needing to solve be plugging the answers to solve? For example, prompt #1, if X is greater than 0 do I need to apply 3? I can answer the question yes or no and not spend additional time with the equation X = 3 and continue to solve? Do I need to come up with the number 49 in each case?
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Re: M05-28 [#permalink]
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Jordon wrote:
Hello,

One quick question, I solved the algebraic part of the question and got the solution X = -1 or 3. Do I have enough information at this point to answer the question yes or no without needing to solve be plugging the answers to solve? For example, prompt #1, if X is greater than 0 do I need to apply 3? I can answer the question yes or no and not spend additional time with the equation X = 3 and continue to solve? Do I need to come up with the number 49 in each case?

Hello, Jordon. I have written at length about these matters above, in this post. In short, no, once you have worked your way through the initial system of equations and derived your two possible values for x, you do not have to work out each equation to its numerical end when encountering each statement. It is good enough to know that you could assess the question with certainty. (And yes, this is a tremendous time-saving technique, so as long as you do not get sloppy employing it, taking things for granted when you are not in a position to do so, I would highly recommend exploring such a method on these types of questions.)

Happy studies.

- Andrew
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Re: M05-28 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M05-28 [#permalink]
(2) x^2=4+5, so x can be +3 or -3 right? How is x not = -3?
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Re: M05-28 [#permalink]
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Zekkwal wrote:
(2) x^2=4+5, so x can be +3 or -3 right? How is x not = -3?


We have:

(i) \(x^2 = y + 5\);

(ii) \(z = 2x\);

(iii) \(y = z - 2\).

From (2) we have y = 4, hence z = 6 (from iii) and x = 3 (from ii). If you try substituting x = -3, you'd see that it won't work.

Hope it's clear.
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Re M05-28 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re M05-28 [#permalink]
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