2210.32 – Fractional Inequality

Tags:Problem Set 2Algebra/TrigonometryAlgebra

If xx is a number such that 1x<2\frac{1}{x}<2 and 1x>3\frac{1}{x}>-3, then which of the following is true?

a. 13<x<12-\frac{1}{3} < x < \frac{1}{2},

b. 12<x<3-\frac{1}{2} < x < 3,

c. x>12,x > \frac{1}{2},

d. x>12x > \frac{1}{2} or 13<x<0-\frac{1}{3} < x<0,

e. x>12x > \frac{1}{2} or x<13.x < -\frac{1}{3}.


We have 3<1x<2-3< \frac{1}{x}< 2. We need to consider two cases, xx positive and xx negative, because when we multiply both sides of an inequality by a negative number (and we feel in our bones that we’re going to do this), the sense of the inequality is reversed.

If xx < 0, then x<12x < \frac{1}{2} is always true. From 1x>3\frac{1}{x}>-3 we get: 3x>1-3x > 1 so the stronger condition x<13x < -\frac{1}{3} holds.

If x>0x > 0, then 13<x-\frac{1}{3} < x is always true. From 1x<2\frac{1}{x}<2, we get: 1<2x1 < 2x so the stronger condition x>12x > \frac{1}{2} holds.

The two cases, xx positive and xx negative, lead to the two possibilities listed in (e).