Solve the inequality (2x - 1)/3 - 4/(x + 2) > (x - 3)/6 + 1.

Step-by-step solution

1. 1

   Note the domain and the plan. $x+2 \neq 0$, so $x \neq -2$. Because the variable sits in a denominator, the only safe method is to move everything to one side and sign-chart a single quotient - not to multiply across.

2. 2

   Combine the three polynomial pieces over the denominator 6. Ignoring the $\tfrac{4}{x+2}$ term for a moment, $\frac{2x-1}{3}-\frac{x-3}{6}-1 = \frac{2(2x-1)-(x-3)-6}{6} = \frac{4x-2-x+3-6}{6} = \frac{3x-5}{6}.$

3. 3

   Reduce the inequality to a single fraction. It is now $\dfrac{3x-5}{6}-\dfrac{4}{x+2}>0$. Using the common denominator $6(x+2)$: $\frac{(3x-5)(x+2)-24}{6(x+2)} > 0.$ Expanding the numerator: $3x^2+6x-5x-10-24 = 3x^2+x-34$, and since $6>0$ it can be dropped: $\frac{3x^2+x-34}{x+2} > 0.$

4. 4

   Find the numerator's roots. $3x^2+x-34$ has no integer factorisation (the pair would need product $-102$ and sum $1$), so use the formula: $x = \frac{-1 \pm \sqrt{1+408}}{6} = \frac{-1\pm\sqrt{409}}{6} \approx -3.5373 \ \text{and}\ 3.2040.$

5. 5

   Sign-chart across the three critical values. In order: $\tfrac{-1-\sqrt{409}}{6} \approx -3.5373$, then $-2$, then $\tfrac{-1+\sqrt{409}}{6} \approx 3.2040$. The upward parabola is positive outside its roots and negative between them, while $x+2$ flips at $-2$. That gives signs $-,\,+,\,-,\,+$ across the four intervals.

6. 6

   Read off the two positive intervals and check. $x \in \left(\frac{-1-\sqrt{409}}{6},\ -2\right) \cup \left(\frac{-1+\sqrt{409}}{6},\ \infty\right).$ Substituting into the original inequality: $x=-3$ gives left minus right $=+1.667>0$ (in), $x=-1$ gives $-5.333$ (out), $x=3$ gives $-0.133$ (out, just below the boundary $3.2040$), and $x=3.5$ gives $+0.189$ (in). All endpoints are excluded, since the inequality is strict and $x=-2$ is outside the domain.