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aarish2102
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AndrewN
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kntombat
aarish2102, could you respond as soon as possible to what AndrewN posted on top.

It would be much appreciated.
Thank you, kntombat. I am next to positive that Statement (2) should read

\(x^2 - 8x + 15 = 0\)

Solving then yields one solution that overlaps with the one valid solution from Statement (1).

\((x - 3)(x - 5) = 0\)

x = 3 or 5

Since we know from Statement (1) that 5 is a valid solution, we only need to consider 3 in the original equation.

\(|(3) - 3| - |(3) - 4| = |(3) - 6|\)

\(|0| - |-1| = |-3|\)

\(0 - 1 = 3\) X

Hence, 3 cannot be a valid solution, and only 5 remains. The answer should, in fact, be (D), but only if the original Statement (2) shows x squared rather than cubed.

- Andrew
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AndrewN, I agree with you 100%.
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kntombat
aarish2102, could you respond as soon as possible to what AndrewN posted on top.

It would be much appreciated.
Well, it looks like Bunuel took care of the issue for us. Thank you, Bunuel. (I have no idea how you manage to peek into these corners of the site when the World Cup competition is going on, but your efforts are much appreciated.)

- Andrew
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