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1. lx-3l >= y In this case, x could have any (-ve, 0 or +ve) value. Its not sufficient.

2. lx-3l =< -y

In this case, lx-3l is always equal to 0 or +ve. Since y is >=0, y cannot be +ve to have statement 2 to be true. So lx-3l must be 0 and hence x = 3. Its sufficient.

This statement will hold true for several values of x and y. for y = 0, x = to for y = 1, x = ( to 2) and (4 to ) for y = 2, x = ( to 1) and (5 to ) ... and so on

(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value. example, for y = 1, |x-3| can never be less than -y=-1. for y = 2, |x-3| can never be less than -y=-2. -y ≠ +ve value, because of precondition that y ≥ 0 The only value of y, for which this statement is true is y = 0. |x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.

This statement will hold true for several values of x and y. for y = 0, x = to for y = 1, x = ( to 2) and (4 to ) for y = 2, x = ( to 1) and (5 to ) ... and so on

(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value. example, for y = 1, |x-3| can never be less than -y=-1. for y = 2, |x-3| can never be less than -y=-2. -y ≠ +ve value, because of precondition that y ≥ 0 The only value of y, for which this statement is true is y = 0. |x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.

hope it helps.

Understood why B.

Can you explain how to solve general equation type l something l > something else? please explain this for statement 1

This statement will hold true for several values of x and y. for y = 0, x = to for y = 1, x = ( to 2) and (4 to ) for y = 2, x = ( to 1) and (5 to ) ... and so on

(2) |x-3| ≤ -y

A mod can never be less than or equal to a negative value. example, for y = 1, |x-3| can never be less than -y=-1. for y = 2, |x-3| can never be less than -y=-2. -y ≠ +ve value, because of precondition that y ≥ 0 The only value of y, for which this statement is true is y = 0. |x-3| ≤ -(0) → |x-3| ≤ 0 → x = 0, for no positive or negative value, this condition will hold true.

hope it helps.

Understood why B.

Can you explain how to solve general equation type l something l > something else? please explain this for statement 1

Most of the time, values have to be substituted in the variable to determine its RANGE. Like in Statement 1, we found that the range is ANY value. for example, |x+4| > 0, then range of x will be -4 < x <