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

Text

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

Solution

Move everything to the left over the common denominator $(2+x)(1+x)$: $\frac{1}{2+x}-\frac{4}{1+x} = \frac{(1+x) - 4(2+x)}{(2+x)(1+x)} = \frac{-3x-7}{(2+x)(1+x)} \le 0.$ Changing sign (and direction): $\dfrac{3x+7}{(2+x)(1+x)} \ge 0$. Key points: $x=-\tfrac73$ (numerator, included), $x=-2,\ x=-1$ (denominator, excluded). Sign study:

• $-\tfrac73 \le x < -2$: positive;
• $x > -1$: positive. $\boxed{-\tfrac{7}{3} \le x < -2 \ \vee\ x > -1}$