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Model Paper 7 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 75 Marks
Question
Solve the following system of linear inequalities. 
$\begin{array}{l}2(2 x +3)-10<6( x -2) \\ \text { and } \frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3}\end{array}$
✓

Answer

The given system of linear inequalities is
|$\begin{array}{l}2(2 x+3)-10<6(x-2) \ldots \text { (i) } \\ \text { and } \frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3} \ldots \text { (ii) }\end{array}$
From inequality (i), we get
2(2x + 3) - 10 < 6(x - 2)
$\begin{array}{l}\Rightarrow 4 x +6-10<6 x -12 \\ \Rightarrow 4 x -4<6 x -12 \\ \Rightarrow
4 x -4+4<6 x -12+4 \text { [adding } 4 \text { on both sides] } \\ \Rightarrow 4 x <6 x -8 \\ \Rightarrow 4 x -6 x <6 x -8-6 x \text { [subtracting } 6 x \text { from both sides] } \\ \Rightarrow-2 x <-8 \\ \Rightarrow 2 x >8 \text { [dividing both sides by }-1 \text { and then inequality sign will change] } \\ \Rightarrow \quad \frac{2 x}{2}>\frac{8}{2} \text { [dividing both sides by } 2 \text { ] }\end{array}$
$\therefore \quad x>4 \ldots$ (iii)
Thus, any value of x greater than 4 satisfies the inequality.
$\therefore$ Solution set is $x \in(4, \infty)$
The representation of solution of inequality (i) is 
Image
From inequality (ii), we get
$\frac{2 x-3}{4}+6 \geq 2+\frac{4 x}{3} \Rightarrow \frac{2 x-3+24}{4} \geq \frac{6+4 x}{3}$
$\begin{array}{l}\Rightarrow \quad \frac{2 x+21}{4} \geq \frac{6+4 x}{3} \Rightarrow 3(2 x+21) \geq 4(6+4 x) \\ \Rightarrow \quad 6 x+63 \geq 24+16 x \\ \Rightarrow \quad-10 x \geq-39 \Rightarrow 10 x \leq 39 \\ \Rightarrow \quad \frac{10 x}{10} \leq \frac{39}{10} \\ \Rightarrow \quad x<3.9\end{array}$
From Eqs. (iii) and (iv), it is clear, that there is no common value of x, which satisfies both inequalities (iii) and (iv). Hence, the given system of inequalities has no solution. 

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