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(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.

Answer: C.

Hope it's clear.

Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0?

|some value|= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0?

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

I gives you \(y\geq{2}\)... II gives you two cases.. a) \(y<3\) --> \(3-y=11\) --> \(y=-8\); this means if y<3, it can have ONLY -8 as the value, so when we combine this with \(y\geq{2}\),which does not fit between 2 and 3.. hence WRONG

b)\(y\geq{3}\) --> \(-3+y=11\) --> \(y=14\) this means if \(y\geq{3}\), it can have ONLY 14 as the value, so when we combine this with \(y\geq{2}\), 14 fits in both the ranges.. hence CORRECT
_________________

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.

(1)+(2) Since from (1) \(y\geq{2}\), then from (2) \(y=14\). Sufficient.

Answer: C.

Hope it's clear.

Why is Y>=0? Can't we write 'Y>0'? If LHS is positive and RHS must also be positive then, RHS has to be greater than 0. As 0 is neither positive nor negative, how can we write Y>=0?

|some value|= some value. Since LHS is an absolute value, RHS has to be positive as well. Why is RHS>=0 and not RHS>0?

Hi, an ABSOLUTE value is NOT necessarily positive, it is NON-NEGATIVE, so it can include both 0 and +ive value.. if in statement I x is 2 or -2, LHS will be 0 and hence y=2..
_________________

While at first glance, seeing two variables you might think you need two equations, statement 1 alone is actually sufficient. By factoring out common terms (2y on top, 3 on bottom), you're left with: 2y(2x+1)3(2x+1)=y−3 So the (2x+1) terms cancel, leaving just: 2y3=y−3, a linear equation that allows you to solve for the value of y. Statement 1 is sufficient

(1) \(3|x^2-4|=y-2\). Now, since we are asked to find the value of y, from this statement we can conclude only that \(y\geq{2}\), as LHS is absolute value which is never negative, hence RHS als cannot be negative. Not sufficient.