x/(2x)<9

Solution

\frac{x}{2 x} < 9

x < 0 or x > 0

Interval Notation

(- \infty, 0) \cup (0, \infty)

Solution steps

\frac{x}{2 x} < 9

Rewrite in standard form

1 > 0

True for all x

Find singularity points

Find the zeros of the denominator

2 x : x = 0

Since the inequality is not defined for

x = 0

x < 0 or x > 0

Combine the intervals

True for all x and x < 0 or x > 0

Merge Overlapping Intervals

x < 0 or x > 0

\frac{x + 1}{( x - 1 )} \leq 3

5 x^{3} - 17 x^{2} + 17 x - 5 \leq 0

(x + 2)^{2} - (x - 2)^{2} > 8 x - 2

\frac{( x ^{3} + 1 )}{( 7 x ^{2} - 1 )} \leq 0

\frac{1}{( x ^{2} )} \leq 4