Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

Solve Each of the Following System of Equations in R. 4 X + 1 ≤ 3 ≤ 6 X + 1 , X > 0 - Mathematics

Solve each of the following system of equations in R. $\frac{4}{x + 1} \leq 3 \leq \frac{6}{x + 1}, x > 0$

Solution

$\frac{4}{x + 1} \leq 3 \leq \frac{6}{x + 1}, x > 0$
$\Rightarrow \frac{4}{x + 1} \leq 3 and 3 \leq \frac{6}{x + 1}$
$\text{ Now },$
$\frac{4}{x + 1} \leq 3$
$\Rightarrow \frac{4}{x + 1} - 3 \leq 0$
$\Rightarrow \frac{4 - 3x - 3}{x + 1} \leq 0$
$\Rightarrow \frac{1 - 3x}{x + 1} \leq 0$
$\Rightarrow \frac{3x - 1}{x + 1} \geq 0$
$\Rightarrow x \in \left( - \infty , - 1 \right) \cup [\frac{1}{3}, \infty )$

Thus, the solution set of the inequation is $\left( - \infty , - 1 \right) \cup [\frac{1}{3}, \infty )$

$\text{ And }$
$\frac{6}{x + 1} \geq 3$
$\Rightarrow \frac{6}{x + 1} - 3 \geq 0$
$\Rightarrow \frac{6 - 3x - 3}{x + 1} \geq 0$
$\Rightarrow \frac{3 - 3x}{x + 1} \geq 0$
$\Rightarrow \frac{3x - 3}{x + 1} \leq 0$
$\Rightarrow x \in ( - 1, 1]$

Thus, the solution set of the inequation is $( - 1, 1]$

The common values of x in both the inequation is $\left[ \frac{1}{3}, 1 \right]$

Hence, the solution set of both the inequation is $\left[ \frac{1}{3}, 1 \right]$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 15 Linear Inequations
Exercise 15.2 | Q 21 | Page 16