JAMB 1998 · UME · Q17

If $x$ is a positive real number, find the range of values for which $\frac{1}{3x} + \frac12 > \frac{1}{4x}$ .

Worked solution (try it first)

Multiply every term by $12x$ , which is positive, so the sign stays: $4 + 6x > 3$ .
Subtract 4 from both sides: $6x > -1$ , so $x > -\frac16$ .
Every positive $x$ satisfies $x > -\frac16$ , so the answer is all $x > 0$ , option B.