Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0

Is \(|x| < 1\)?

Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?

(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of \(x\) when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).

So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.

Alternately we could just expand absolute values of RHS and LHS with same sign (for example both LHS and RHS positive: \(x + 1 = 2(x - 1)\)) and then with different sign (for example LHS positive and RHS negative \(x + 1 = 2(-x + 1)\)), solve for \(x\) both equations, and finally check whether the solutions satisfy \(|x + 1| = 2|x - 1|\).

OR: we can square given equation to get rid of the modulus: \((x + 1)^2 = 4(x - 1)^2\) --> \(3x^2-10x+3=0\) --> \(x=3\) or \(x=\frac{1}{3}\).

(2) \(|x - 3|>{0}\). Absolute value is always non-negative, more than or equal to zero: \(|some \ expression|\geq{0}\). We are told that absolute value of \(x-3\) is MORE than zero, so just it says that \(|x-3|\neq{0}\), which simply means that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.

(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.

Answer: C.

Hope it helps.