Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is |x+y|<|x|+|y|?
1) x<0
2) y>0
CONCEPTUAL (THEORETICAL) TOOL:
$$\left| {x + y} \right| \le \left| x \right| + \left| y \right|$$
is always valid and it is famous in Mathematics. It is the so-called
first triangle inequality.
It is useful in the GMAT to know it and, also, to know that EQUALITY is valid if, and only if, we don´t have "opposing forces" (xy<0 does not occur):
$$\left| {x + y} \right| = \left| x \right| + \left| y \right|\,\,\,\,\,\, \Leftrightarrow \,\,\,\,xy \ge 0$$
(Try some values for x and y to feel comfortable about it. In less than 3min you will realize the result is pretty evident!)
SOLUTION: (Now absolutely trivial!)
$$\left| {x + y} \right|\,\,\mathop < \limits^? \,\,\,\left| x \right| + \left| y \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{xy\,\,\mathop < \limits^? \,\,0}$$
$$\left( 1 \right)\,\,\,x < 0\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,\,y > 0\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
x < 0 \hfill \cr
y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,xy < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
