Concept:
Inequations:
- Adding (or subtracting) the same quantities to both sides of the inequality, does no affect the order of inequality.
If a > b, then a ± x > b ± x, ∀ x ∈ R.
- Multiplying (or dividing) by a negative quantity, reverses the order of inequality.
If a > b, then ax > bx, ∀ x > 0.
If a > b, then ax < bx, ∀ x < 0.
Calculation:
It is given that \(\rm \frac{x}{2x+1}\ge \frac{1}{4}\). We have the following two cases:
CASE 1: 2x + 1 > 0 ⇒ \(\rm x>-\frac{1}2\).
Multiplying both sides of \(\rm \frac{x}{2x+1}\ge \frac{1}{4}\) by 4(2x + 1), gives us:
⇒ 4x ≥ 2x + 1
Subtracting 2x from both sides, we get:
⇒ 2x ≥ 1
Dividing both sides by 2, we get:
⇒ \(\rm x\geq\frac{1}2\)
Combining this with our assumption 2x + 1 > 0, we get:
\(\rm x>-\frac{1}2\) AND \(\rm x\geq\frac{1}2\)
⇒ \(\rm x\geq\frac{1}2\)
CASE 2: 2x + 1 < 0 ⇒ \(\rm x<-\frac{1}2\).
Multiplying both sides of \(\rm \frac{x}{2x+1}\ge \frac{1}{4}\) by 4(2x + 1) (which is -ve in this case), will give us:
⇒ 4x ≤ 2x + 1
Subtracting 2x from both sides, we get:
⇒ 2x ≤ 1
⇒ \(\rm x\leq\frac{1}2\)
Combining this with our assumption, we get:
\(\rm x<-\frac{1}2\) AND \(\rm x\leq\frac{1}2\)
⇒ \(\rm x<-\frac{1}2\)
Considering both the cases together, we can say that \(\rm x\geq\frac{1}2\) OR \(\rm x<-\frac{1}2\).
⇒ \(\rm x \in \left(-\infty , - \frac{1}{2}\right)\cup \left(\frac{1}{2}, \infty\right)\).
Additional Information
Some rules of inequalities:
- If a > b, then a ± x > b ± x, ∀ x ∈ R.
- If a > b, then ax > bx, ∀ x > 0.
- If a > b, then ax < bx, ∀ x < 0.
- If 0 > a > b OR b < a < 0, then \(\rm \frac1a<\frac1b\).
- If a > b and x > y, then a + x > b + y. Nothing specific can be said about (a - x and b - y) or (ax and by).