# Solve Each of the Following System of Equations in R. 2 X − 3 4 − 2 ≥ 4 X 3 − 6 , 2 ( 2 X + 3 ) < 6 ( X − 2 ) + 10 - Mathematics

Solve each of the following system of equations in R.

$\frac{2x - 3}{4} - 2 \geq \frac{4x}{3} - 6, 2\left( 2x + 3 \right) < 6\left( x - 2 \right) + 10$

#### Solution

$\frac{2x - 3}{4} - 2 \geq \frac{4x}{3} - 6$
$\Rightarrow \frac{2x - 3}{4} - \frac{4x}{3} \geq - 6 + 2$
$\Rightarrow \frac{3\left( 2x - 3 \right) - 16x}{12} \geq - 4$
$\Rightarrow 6x - 9 - 16x \geq - 48$
$\Rightarrow - 10x \geq - 39$
$\Rightarrow 10x \leq 39 \left[ \text{ Multiplying both sides by } - 1 \right]$
$\Rightarrow x \leq \frac{39}{10}$
$\Rightarrow x \in ( - \infty , \frac{39}{10}] . . . (i)$
$\text{ Also }, 2\left( 2x + 3 \right) < 6\left( x - 2 \right) + 10$
$\Rightarrow 4x + 6 < 6x - 12 + 10$
$\Rightarrow 4x + 6 < 6x - 2$
$\Rightarrow 6x - 2 > 4x + 6$
$\Rightarrow 6x - 4x > 6 + 2$
$\Rightarrow 2x > 8$
$\Rightarrow x > 4$
$\Rightarrow x \in \left( 4, \infty \right) . . . (ii)$
$\text{ Hence, thesolution of the given set of inequalities is theintersection of } (i) \text{ and } (ii),$
$( - \infty , \frac{39}{10}] \cap \left( 4, \infty \right) = \varnothing$
$\text{ which is an empty set } .$
$\text{ Thus, there is no solution of the given set of inequations } .$

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 15 | Page 15