Carcass wrote:
Real numbers x, y, and z satisfy the inequalities \(0<x<1\), \(-1<y<0\), and \(1<z<2\). Which of the following numbers is necessarily positive?
(A) \(y+x^2\)
(B) \(y+xz\)
(C) \(y+y^2\)
(D) \(y+2y^2\)
(E) \(y+z\)
Given:
\(-1<y<0\)
\(1<z<2\)
Since the inequality symbols are
facing the same direction, we can
add the two inequalities to get: \(0<y+z<2\)
So, it must be the case that \(y+z\) is positive.
Answer: E
_________________
Brent Hanneson - founder of Greenlight Test Prep