Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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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. \[\frac{4}{x + 1} \leq 3 \leq \frac{6}{x + 1}, x > 0\]

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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?
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APPEARS IN

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