Let's deal with the conditions one by one.
Quote:
(1) \(|x + 2| > 3 – x\)
To open the modulus, there are two options:
- \(x < -2\) and \(|x + 2| = -x -2\)
\(-x -2 > 3 – x\\
-2 > 3\)
Clearly, this is impossible - therefore, this option is invalid. - \(x >= -2\) and \(|x + 2| = x+2\)
\(x+2 > 3-x\\
2x > 1\\
x > 0.5\)
This actually addresses our question, because X is always above 0.
Therefore, having only one functioning possibility among these two,
Condition 1 is sufficient by itself.Quote:
(2)\( |2 – x| = 10 – 2x\)
To open the modulus, there are two options again:
- \(x > 2\) and \(|2-x| = -2 + x\)
\(-2+x =10-2x\\
3x = 12\\
x = 4\)
This addresses our question, because X is always above 0. - \(x <= 2\) and \(|2-x| = 2 - x\)
\(2 - x=10-2x\\
x = 8\)
Clearly, this is impossible, because X has to be below 2 and equal to 8 at the same time.
Again, there is only one legit option under this condition, so
Condition 2 is sufficient by itself.Therefore,
the correct answer is D.