Full solution

|x-1|<\frac{6}{x}

Consider Domain: First, we need to consider the domain of the inequality. Since we have a fraction with $x$ in the denominator, $x$ cannot be $0$ .
Split into Cases: Now, let's split the inequality into two cases, because the absolute value $|x-1|$ can be positive or negative. $\newline$ Case $1$ : $x - 1 < \frac{6}{x}$ when $x > 1$ $\newline$ Case $2$ : $-(x - 1) < \frac{6}{x}$ when $x < 1$
Solve Case $1$ : Let's solve Case $1$ : $x - 1 < \frac{6}{x}$ $\newline$ Multiply both sides by $x$ to get rid of the fraction: $x(x - 1) < 6$
Distribute $x$ : Now, distribute $x$ on the left side: $x^2 - x < 6$
Form Quadratic Inequality: Move all terms to one side to form a quadratic inequality: $x^2 - x - 6 < 0$
Factor Quadratic: Factor the quadratic: $(x - 3)(x + 2) < 0$
Find Intervals: Find the intervals where the inequality is true. The critical points are $x = 3$ and $x = -2$ . Test points in the intervals $(-\infty, -2)$ , $(-2, 3)$ , and $(3, \infty)$ .
Test Intervals: For $x$ in $(-\infty, -2)$ , choose $x = -3$ : $(-3 - 3)(-3 + 2) < 0$ , which is true. For $x$ in $(-2, 3)$ , choose $x = 0$ : $(0 - 3)(0 + 2) < 0$ , which is false. For $x$ in $(3, \infty)$ , choose $(-\infty, -2)$ $0$ : $(-\infty, -2)$ $1$ , which is false. So, the solution for Case $1$ is $x$ in $(-\infty, -2)$ $3$ , but we must remember $(-\infty, -2)$ $4$ , so we only take $(3, \infty)$ .
Solution for Case $1$ : Now, let's solve Case $2$ : $- (x - 1) < \frac{6}{x}$ This simplifies to $-x + 1 < \frac{6}{x}$
Solve Case $2$ : Multiply both sides by $x$ to get: $-x^2 + x < 6$