Chapter 6: Linear Inequalities Exercise 6.1

Ans: Let us rewrite the given inequality, we get

$\frac{\text{3}\left( \text{x-2} \right)}{\text{5}}\leq\frac{\text{5}\left( \text{2-x} \right)}{\text{3}} $ 

$ \Rightarrow \text{9}\left( \text{x-2} \right)\leq\text{25}\left( \text{2-x} \right)$ 

$ \Rightarrow \text{9x-18}\leq\text{50-25x} $ 

$ \Rightarrow \text{9x+25x-18}\leq\text{50-25x+25x} $ 

$ \Rightarrow \text{34x-18}\leq\text{50} $ 

$ \Rightarrow \text{34x-18+18}\leq\text{50+18} $ 

$ \Rightarrow \text{34x}\leq\text{68} $ 

$ \Rightarrow \text{x}\leq\frac{\text{68}}{\text{34}} $ 

$ \Rightarrow \text{x}\leq\text{2} $

Therefore, we can say that the numbers less than or equal to $2$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

Ans: Let us rewrite the given inequality, we get

$\frac{\text{1}}{\text{2}}\left( \frac{\text{3x}}{\text{5}}\text{+4} \right)\geq\frac{\text{1}}{\text{3}}\left( \text{x-6} \right) $ 

$ \Rightarrow \text{3}\left( \frac{\text{3x}}{\text{5}}\text{+4} \right)\geq\text{2}\left( \text{x-6} \right) $ 

$ \Rightarrow \frac{\text{9}}{\text{5}}\text{x+12}\geq\text{2x-12} $ 

$ \Rightarrow \frac{\text{9}}{\text{5}}\text{x+12-12}\geq\text{2x-12-12} $ 

$ \Rightarrow \frac{\text{9}}{\text{5}}\text{x}\geq\text{2x-24} $ 

$ \Rightarrow \text{9x}\geq\text{10x-120} $ 

$ \Rightarrow \text{9x-10x}\geq\text{10x-10x-120} $ 

$ \Rightarrow \text{-x}\geq\text{-120} $ 

$ \Rightarrow \text{x}\leq\text{120} $

Therefore, we can say that the numbers less than or equal to $120$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,120 \right]$.

Ans: Let us rewrite the given inequality, we get

$\text{2}\left( \text{2x+3} \right)\text{-10}<\text{6}\left( \text{x-2} \right) $ 

$ \Rightarrow \text{4x+6-10}<\text{6x-12} $ 

$ \Rightarrow \text{4x-4}<\text{6x-12} $ 

$ \Rightarrow \text{4x-4+4}<\text{6x-12+4} $ 

$ \Rightarrow \text{4x}<\text{6x-8} $ 

$ \Rightarrow \text{4x-6x}<\text{6x-6x-8} $ 

$ \Rightarrow \text{-2x}<\text{-8} $ 

$ \Rightarrow \text{x}>\text{4}$

Therefore, we can say that the numbers greater than $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( 4,\infty\right)$.

Ans: Let us rewrite the given inequality, we get

$\text{37-}\left( \text{3x+5} \right)\geq\text{9x-8}\left( \text{x-3} \right) $ 

$ \Rightarrow \text{37-3x-5}\geq\text{9x-8x+24} $ 

$ \Rightarrow \text{-3x+32}\geq\text{x+24} $ 

$ \Rightarrow \text{-3x-x+32}\geq\text{x-x+24} $ 

$ \Rightarrow \text{-4x+32}\geq\text{24} $ 

$ \Rightarrow \text{-4x+32-32}\geq\text{24-32} $ 

$ \Rightarrow \text{-4x\geq-8} $ 

$ \Rightarrow \text{x\leq2}$

Therefore, we can say that the numbers less than or equal to $2$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

Ans: Let us rewrite the given inequality, we get

$\frac{\text{x}}{\text{4}}<\frac{\left( \text{5x-2} \right)}{\text{3}}\text{-}\frac{\left( \text{7x-3} \right)}{\text{5}} $ 

$ \Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{5}\left( \text{5x-2} \right)\text{-3}\left( \text{7x-3} \right)}{\text{15}} $ 

$ \Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{25x-10-21x+9}}{\text{15}} $ 

$ \Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{4x-1}}{\text{15}} $ 

$ \Rightarrow \text{15x}<\text{4}\left( \text{4x-1} \right) $ 

$ \Rightarrow \text{15x-16x}<\text{16x-16x-4} $ 

$ \Rightarrow \text{-x}<\text{-4} $ 

$\text{x}>\text{4}$

Therefore, we can say that the numbers greater $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( 4,\infty\right)$.

Ans: Let us rewrite the given inequality, we get

$\frac{\left( \text{2x-1} \right)}{\text{3}}\geq\frac{\left( \text{3x-2} \right)}{\text{4}}\text{-}\frac{\left( \text{2-x} \right)}{\text{5}} $ 

$ \Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\geq\frac{\text{5}\left( \text{3x-2} \right)\text{-4}\left( \text{2-x} \right)}{\text{20}} $ 

$ \Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\text{=}\frac{\text{15x-10-8+4x}}{\text{20}} $ 

$ \Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\geq\frac{\text{19x-18}}{\text{20}} $ 

$ \Rightarrow \text{20}\left( \text{2x-1} \right)\geq\text{3}\left( \text{19x-18} \right) $ 

$ \Rightarrow \text{40x-20}\geq\text{57x-54} $ 

$ \Rightarrow \text{-20+54}\geq\text{57x-40x} $ 

$ \Rightarrow \text{34}\geq\text{17x} $ 

$ \Rightarrow \text{2}\geq\text{x}$

Therefore, we can say that the numbers less than or equal to $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

Ans: Let us rewrite the given inequality, we get

$\text{3x-2}<\text{2x+1} $ 

$ \Rightarrow \text{3x-2x}<\text{1+2} $ 

$ \Rightarrow \text{x}<\text{3}$

Therefore, the graphical representation for the inequality in number line is:

Ans: Let us rewrite the given inequality, we get

$\text{5x-3}\geq\text{3x-5} $ 

$ \Rightarrow \text{5x-3x}\geq\text{-5+3} $ 

$ \Rightarrow \text{2x}\geq\text{-2} $ 

$ \Rightarrow \frac{\text{2x}}{\text{2}}\geq\frac{\text{-2}}{\text{2}} $ 

$ \Rightarrow \text{x}\geq\text{-1}$

Therefore, the graphical representation for the inequality in number line is:

Ans: Let us rewrite the given inequality, we get

$\text{3}\left( \text{1-x} \right)<\text{2}\left( \text{x+4} \right) $ 

$ \Rightarrow 3-3\text{x}<\text{2x+8} $ 

$ \Rightarrow \text{3-8}<\text{2x+3x} $ 

$ \Rightarrow \text{-5}<\text{5x} $ 

$ \Rightarrow \frac{-5}{5}<\frac{5\text{x}}{5} $ 

$ \Rightarrow -1<\text{x} $

Therefore, the graphical representation for the inequality in number line is:

Ans: Let us rewrite the given inequality, we get

$\frac{\text{x}}{\text{2}}\geq\frac{\left( \text{5x-2} \right)}{\text{3}}\text{-}\frac{\left( \text{7x-3} \right)}{\text{5}} $ 

$ \Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{5}\left( \text{5x-2} \right)\text{-3}\left( \text{7x-3} \right)}{\text{15}} $ 

$ \Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{25x-10-21x+9}}{\text{15}}$ 

$ \Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{4x-1}}{\text{15}} $ 

$ \Rightarrow \text{15x}\geq\text{2}\left( \text{4x-1} \right) $ 

$ \Rightarrow \text{15x}\geq\text{8x-2} $ 

$ \Rightarrow \text{15x-8x}\geq\text{8x-2-8x} $ 

$ \Rightarrow \text{7x}\geq\text{-2} $ 

$ \Rightarrow \text{x}\geq\text{-}\frac{\text{2}}{\text{7}}$

Therefore, the graphical representation for the inequality in number line is: