What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As 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 can not be negative. Not sufficient.

(2) \(|3 - y| = 11\):

\(y<3\) --> \(3-y=11\) --> \(y=-8\);
\(y\geq{3}\) --> \(-3+y=11\) --> \(y=14\).

Two values for \(y\). Not sufficient.

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

Answer: C.
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Hey, pls explain the reasoning of LHS

LHS is absolute value which is never negative, hence RHS als can not be negative.???
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1st equation:
x^2 - 4 can be positive, or negative ( say x = 1, then x^2 - 4 is -3, and say x = 3, then x^2-4 is 5. However since its absolute value is taken, it will always be a positive number, which when multiplied by 3 will give either zero (if x=2) or a positive number. This added to 2 brought pver fom RHS of the equation will have to be a positive number still. So while Statement 1 is insufficient in telling us the value of y, it does tell us that y is positive.
Equation 2:
by telling us that absolute value of 3-y is 11 implies that the vlue inside the absolute value lines is either 11 or -11, which means that y is either 14 or -8. since it gives two values of y it is insufficient by itself.
However since statement tells us for certain that y i s positive and statement two give sus two values of y one positive and one negative, both combined we can pick the positive value and be sure that that is the correct value of y
hence both together ae sufficient

1. Insufficient

because all we know here is y>2

2. Insufficient.

because y can be -8 or 14

together . Sufficient.

Answer is C.

[quote="Bunuel"]What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As 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 can not be negative. Not sufficient."

Bunuel,

In this case we said since the LHS has a mod so its value is positive always. But we also know mod function is negative for values of x < 0 , or we dont always assume for a given equation say |x-2| =2 , that LHS is positive always ? we say x-2 = x-2 , when x> 2 and 2-x when x < 2.
So this looks a contradiction, can you please explain ?

In this case for statement 1,

as it is in modulus so LHS will always be positive, its x whose value can be -ve or +ve, but LHS will always be +ve.
Therefore we conclude that y-2>=0 or y>=2

From statement 2 we get y=-8,14

Combining 1 and 2, we get y=14

Hope it is clear.[/quote]

Let us take an example. y =|X-2|
So y = x-2 , when x-2>0 or x-2=0
and y = 2-x, when x-2<0 ro x <2

If we take x = 3, then y = -1, how ? I am doing something silly here for sure!
If you see graph of mod function Y is always positive, so how is Y = -1 above at x =3 ?

Ok Thanks. I was missing this - of -x = +x, So basically in mod function x can range from - to +, but y will always be >0 or = 0. Correct me if I understood it wrongly.

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