Fractional inequality (2x+1)/(2−x) + 3 ≤ 1 − (x+4)/(1−2x)

Text

Solve the following fractional inequality: $\frac{2x+1}{2-x}+3 \leq 1 - \frac{x+4}{1-2x}.$

Solution

Move everything to the left: $\dfrac{2x+1}{2-x}+2+\dfrac{x+4}{1-2x} \le 0$, over the common denominator $(2-x)(1-2x)$. Expanding the numerator: $(2x+1)(1-2x) + 2(2-x)(1-2x) + (x+4)(2-x) = (1-4x^2) + (4x^2-10x+4) + (-x^2-2x+8) = -x^2 - 12x + 13.$ Hence $\frac{-(x^2+12x-13)}{(2-x)(1-2x)} \le 0 \;\Longleftrightarrow\; \frac{(x+13)(x-1)}{(2-x)(1-2x)} \ge 0.$ Key points: $x=-13,\ x=1$ (numerator, included), $x=\tfrac12,\ x=2$ (denominator, excluded). Sign study:

• $x\le -13$: positive;
• $\tfrac12 < x \le 1$: positive;
• $x > 2$: positive. $\boxed{x \le -13 \ \vee\ \tfrac12 < x \le 1 \ \vee\ x > 2}$