Question 12 Marks
Solve for x, the inequalities Exercises.
$\frac{4}{\text{x+1}}\leq3\leq\frac{6}{\text{x+1}}, (\text{x}>0)$
$\frac{4}{\text{x+1}}\leq3\leq\frac{6}{\text{x+1}}, (\text{x}>0)$
Answer
View full question & answer→$\frac{4}{\text{x+1}}\leq3\leq\frac{6}{\text{x+1}}, (\text{x}>0)$
$\frac{4}{\text{x+1}}\leq3$ and $3\leq\frac{6}{\text{x}+1}, \text{x}>0$
$\Rightarrow4\leq3({\text{x+1}})$ and $3({\text{x+1}})\leq6,\text{x}<0$
$\Rightarrow\frac{1}{3}\leq\text{x}$ and $\text{x}\leq1, \text{x}>0$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq1$
$\frac{4}{\text{x+1}}\leq3$ and $3\leq\frac{6}{\text{x}+1}, \text{x}>0$
$\Rightarrow4\leq3({\text{x+1}})$ and $3({\text{x+1}})\leq6,\text{x}<0$
$\Rightarrow\frac{1}{3}\leq\text{x}$ and $\text{x}\leq1, \text{x}>0$
$\Rightarrow\frac{1}{3}\leq\text{x}\leq1$