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satya2029
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TheUltimateWinner
What values of x will satisfy the following inequality?
\(|x-3|> |x|-3\)

A. x∈ (-∞, 3)
B. x∈ (-3, 3)
C. x∈ (-∞, -3)
D. x∈ (∞, 3)
E. x∈ (-∞, -3) ∪ (3, ∞)

Any other solution to this problem?

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Can anyone solve this inequality algebraically

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This is the way I solved it algebraically:


[X - 3] > [X] - 3


The 2 Critical Boundaries over which the value inside each Modulus will change sign (“critical points”) will be:

X = 0 and + 3

So we need to open Each Modulus under 3 Assumed Scenarios:


Range 1: X >/= +3

X - 3 > X - 3

0 > 0 ——- false

No value that equals +3 or greater will satisfy the inequality


Range 2: 0 </= X < +3

-(X - 3) > X - 3

-X + 3 > X - 3

6 > 2X

X < 3

For all values from: 0 </= X < 3

The inequality will be satisfied. (Note: +3 does not work)


Range 3: X < 0

-(X - 3) > -(X) - 3

-X + 3 > -X - 3

+ 3 > -3

True——-test any value from 0 to negative infinite and you will see that the range will be satisfied.


Answer: X < 3

Any value in this range will satisfy the inequality


aka14
Can anyone solve this inequality algebraically

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Asked: What values of x will satisfy the following inequality?\(|x-3|> |x|-3\)
Case 1: x<0
3 - x > -x - 3
3 > -3 
ALWAYS TRUE

Case 2: 0<=x<=3
3-x > x-3
x < 3
ALWAYS TRUE

Case 3: x>3
x-3 > x -3
Not feasible since x -3 = x - 3

x∈ (-∞, 3)

IMO A
­
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Kinshook
Asked: What values of x will satisfy the following inequality?\(|x-3|> |x|-3\)
Case 1: x<0
3 - x > -x - 3
3 > -3 
ALWAYS TRUE

Case 2: 0<=x<=3
3-x > x-3
x < 3
ALWAYS TRUE

Case 3: x>3
x-3 > x -3
Not feasible since x -3 = x - 3

x∈ (-∞, 3)

IMO A
­
­Hi, how do we decide the critical points in this equation?
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Kinshook
Asked: What values of x will satisfy the following inequality?\(|x-3|> |x|-3\)
Case 1: x<0
3 - x > -x - 3
3 > -3 
ALWAYS TRUE

Case 2: 0<=x<=3
3-x > x-3
x < 3
ALWAYS TRUE

Case 3: x>3
x-3 > x -3
Not feasible since x -3 = x - 3

x∈ (-∞, 3)

IMO A
­
At x=3 both sides of equation becomes equal to zero and hence the the inequality fails

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