thirst4edu wrote:

If x & y are integers and y=|x+3| + |4-x|, does y equals 7?

1) x < 4

2) x > -3

Had a hard time solving this, would like to know how to solve this using number picking approach as well as algebraic approach. Thanks.

OA is

\(y=|x+3|+|4-x|\) two check points: \(x=-3\) and \(x=4\) (check point: the value of \(x\) when expression in || equals to zero), hence three ranges to consider:

A. \(x<{-3}\) --> \(y=| x + 3| +|4-x| =

-x-3+4-x=-2x+1\), which means that when \(x\) is in the range {-infinity,-3} the value of \(y\) is defined by \(x\) (we would have multiple choices of \(y\) depending on \(x\) from the given range);

B. \(-3\leq{x}\leq{4}\) --> \(y=|x+3|+|4-x|=x+3+4-x=7\), which means that when \(x\) is in the range {-3,4} the value of \(y\) is \(7\) (value of y does not depend on value of \(x\), when \(x\) is from the given range);

C. \(x>{4}\) --> \(y=|x+3|+|4-x|=x+3

-4+x=2x-1\), which means that when \(x\) is in the range {4, +infinity} the value of \(y\) is defined by \(x\) (we would have multiple choices of \(y\) depending on \(x\) from the given range).

Hence we can definitely conclude that \(y=7\) if \(x\) is in the range {-3,4}

(1) \(x<4\) --> not sufficient (\(x<4\) but we don't know if it's \(\geq{-3}\));

(2) \(x>-3\) --> not sufficient (\(x>-3\) but we don't know if it's \(\leq{4}\));

(1)+(2) \(-3<x<4\) exactly the range we needed, so \(y=7\). Sufficient.

Answer: C.

OR: looking at \(y=|x+3|+|4-x|\) you can notice that \(y=7\) (\(y\) doesn't depend on the value of \(x\)) when \(x+3\) and \(4-x\) are both positive, in this case \(x-es\) cancel out each other and we would have \(y=|x+3|+|4-x|=x+3+4-x=7\). Both \(x+3\) and \(4-x\) are positive in the range \(-3<{x}<4\) (\(x+3>0\) --> \(x>-3\) and \(4-x>0\) --> \(x<4\)).

Hope it's clear.

Sorry I have a question. I understand that because of two check points, there are three possible ranges. But how |x+3| + |4-x| have different results in each scenario?