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target is x>0

#1
|x + 2| > 3 – x
case 1 :
x+2>3-x
2x>1
x >1/2
for all -ve values #1 will not hold true as RHS will always > LHS
sufficient that x>0
#2
l2 – x| = 10 – 2x
case 1
2-x= 10-2x
x = 8
case 2
-2+x = 10-2x
3x = 12
x = 4
as at x=8 LHS is +ve but RHS is -ve
x>0
sufficient
OPTION D is correct

Bunuel
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Is x > 0 ?

(1) |x + 2| > 3 – x
(2) |2 – x| = 10 – 2x



 


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1) abs(x+2)>3-x
We can solve for the inequality
Case 1:
x+2>3-x
2x>1
x>1/2

Case 2:
-x-2>3-x
-2>3
This is always false, so we can ignore case 2. That leaves us with x>1/2. Clearly, X>0. Suff.

2)
abs(2-x)=10-2x
Case 1:
2-x=10-2x
x=8
But this is not a valid solution, since 10-2(8)<0, and abs cannot be less than 0. discard x=8
Case 2:
-2+x=10-2x
3x=12
x=4
This is a valid solution:
abs(2-4)=abs(-2)=2
10-2(4)=2.

therefore, x=2 which is clearly>0.

Both are suff, therefore D.
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Is x > 0 ?

(1) |x + 2| > 3 – x

It will solve to lead only to x > 0.5

Thus, X > O

YES is the solution.....So, Sufficient


(2) |2 – x| = 10 – 2x

It will solve to lead only to x = 4

Thus, X > 0

YES is the solution.....So, Sufficient

Thus, (D) will be correct
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Is x > 0?

Condition (1):

|x + 2| > 3 – x

Squaring both sides,

(x+2)^2 > (3-x)^2


4x+4 > 9-6x

10x > 5

x > 1/2.

Condition (1) alone is sufficient to prove x > 0.

So eliminate BCE and keep A and D.




Condition (2):


|2 – x| = 10 – 2x

(2-x)^2 = (10 -2x)^2

4+x^2-4x = 4x^2-40x+100

3x^2-36x+96 = 0

x^2-12x+32 = 0
(x-4)(x-8)= 0

Two possible value of x both greater than 0.

So we can say x > 0.

Condition (2) alone is sufficient.

Keep D and eliminate A.

So the best answer choice is D.
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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.
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