Chapter 6: Linear Inequalities Exercise 6.1

Ans: We are given $\text{24x}<\text{100}$.

$ 24\text{x}<100 $

$ \Rightarrow \frac{\text{24x}}{\text{24}}\text{}<\frac{\text{100}}{\text{24}} $

$ \Rightarrow \text{x}<\frac{25}{6} $

Therefore, we can say that only $1,2,3,4$ are only natural numbers less than $\frac{25}{6}$.

Therefore, the solution set is $\left\{ 1,2,3,4 \right\}$

Ans: We are given $\text{24x}<\text{100}$.

$ 24\text{x}<100 $

$\Rightarrow \frac{\text{24x}}{\text{24}}<\frac{\text{100}}{\text{24}} $ 

$ \Rightarrow \text{x}<\frac{\text{25}}{\text{6}}$

Therefore, we can say that only $..................-3,-2,-1,0,1,2,3,4.$ the integers less than $\frac{25}{6}$.

So, the solution set is $\left\{ ...............-3,-2,-1,0,1,2,3,4 \right\}$.

Ans: We are given $\text{-12x}>\text{30}$.

$ \text{-12x}>\text{30}$ 

$ \Rightarrow \frac{-12}{12}\text{x}>\frac{30}{12} $ 

$ \Rightarrow -\text{x}>\frac{5}{2} $ 

$ \Rightarrow \text{x}<-\frac{5}{2} $

Therefore, we can say that there is no natural number less than $\left( -\frac{5}{2} \right)$

So, there is no solution set for the given condition.

(ii) $\mathrm{x}$ is an integer.

We are given $\text{-12x}>\text{30}$.

$\text{-12x}>\text{30} $ 

$ \Rightarrow \frac{-12}{12}\text{x}>\frac{30}{12} $ 

$ \Rightarrow -\text{x}>-\frac{5}{2} $ 

$ \Rightarrow \text{x}<\frac{5}{2}$ 

Therefore, we can say that $.........-5,-4,-3$ are the numbers that satisfy the given inequality.

Hence, the solution set is $\left\{ .....-5,-4,-3 \right\}$.

Ans: We are given $\text{5x-3}<7$.

$ \text{5x-3}<7 $ 

$ \Rightarrow \text{5x-3+3}<7+3 $ 

$ \Rightarrow \text{5x}<10 $ 

$ \Rightarrow \frac{\text{5}}{\text{5}}\text{x}<\frac{10}{5} $ 

$ \Rightarrow \text{x}<2$

Therefore, we can say that $........-4,-3,-2,-1,0,1$ are the integers that satisfy the given inequality.

So, the solution set is $\left\{ ........-4,-3,-2,-1,0,1 \right\}$

(ii) $\mathrm{x}$ is a real number.

We are given $\text{5x-3}<7$.

$\text{5x-3}<7 $ 

$ \Rightarrow \text{5x-3+3}<7+3 $ 

$ \Rightarrow \text{5x}<10 $ 

$ \Rightarrow \frac{\text{5}}{\text{5}}\text{x}<\frac{10}{5} $ 

$ \Rightarrow \text{x}<2$

Therefore, we can say that the real numbers which are less than $2$ will satisfy the inequality.

Hence, the solution set is $\text{x}\in \left( -\infty ,2 \right)$.

Ans: We are given $\text{3x+8}>2$.

$3\text{x}+8>2 $ 

$ \Rightarrow 3\text{x}+8-8>2-8 $ 

$ \Rightarrow 3\text{x}>-6 $ 

$ \Rightarrow \frac{3}{3}\text{x}>\frac{-6}{3} $ 

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

Therefore, we can say that $-1,0,1,2,3.........$ are the integers that satisfy the given inequality.

So, the solution set is $\left\{ -1,0,1,2,3......... \right\}$

(ii) $\mathrm{x}$ is a real number.

$\text{3x+8}>\text{2} $ 

$ \Rightarrow 3\text{x}+8-8>2-8 $ 

$ \Rightarrow 3\text{x}>-6 $ 

$ \Rightarrow \frac{3}{3}\text{x}>\frac{-6}{3} $ 

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

Therefore, we can say that the real numbers which are greater than $-2$ will satisfy the inequality.

Hence, the solution set is $\text{x}\in \left( -2,\infty\right)$.

Ans: Let us rewrite the given inequality, we get

$\text{4x+3}<\text{5x+7} $ 

$ \Rightarrow \text{4x+3-7}<\text{5x+7-7} $ 

$ \Rightarrow \text{4x-4}<\text{5x} $ 

$ \Rightarrow \text{4x-4-4x}<\text{5x-4x} $ 

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

Therefore, we can say that the numbers greater than $\text{-4}$ will be the solutions that satisfy 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

$3\text{x}-7>5x-1 $ 

$ \Rightarrow 3\text{x}-7\text{+75x-1+7} $ 

$ \Rightarrow 3\text{x}>\text{5x+6} $ 

$ \Rightarrow 3\text{x}-5\text{x}>5\text{x}-5\text{x}+6 $ 

$ \Rightarrow -2\text{x}>6 $ 

$ \Rightarrow \frac{-2}{-2}\text{x}<-\frac{6}{-2} $ 

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

Therefore, we can say that the numbers less than $\text{-3}$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-3 \right)$.

Ans: Let us rewrite the given inequality, we get

$\text{3}\left( \text{x-1} \right)\leq\text{2}\left( \text{x-3} \right) $ 

$ \Rightarrow \text{3x-3}\leq\text{2x-6} $ 

$ \Rightarrow \text{3x-3+3}\leq\text{2x-6+3} $ 

$ \Rightarrow \text{3x}\leq\text{2x-3} $ 

$ \Rightarrow 3\text{x-2x}\leq\text{2x-2x-3} $ 

$ \Rightarrow \text{x}\leq\text{-3} $

Therefore, we can say that the numbers less than or equal to $\text{-3}$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-3 \right]$.

Ans: Let us rewrite the given inequality, we get

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

$ \Rightarrow \text{6-3x}\geq\text{2-2x} $ 

$ \Rightarrow \text{6-6-3x}\geq\text{2-6-2x} $ 

$ \Rightarrow \text{-3x}\geq\text{-4-2x} $ 

$ \Rightarrow \text{-3x+2x}\geq\text{-4-2x+2x} $ 

$ \Rightarrow \text{-x}\geq\text{-4} $ 

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

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 $\left( -\infty ,4 \right]$.

Ans: Let us rewrite the given inequality, we get

$\text{x+}\frac{\text{x}}{\text{2}}\text{+}\frac{\text{x}}{\text{3}}<\text{11}$ 

$ \Rightarrow \text{x}\left( \text{1+}\frac{\text{1}}{\text{2}}\text{+}\frac{\text{1}}{\text{3}} \right)<\text{11} $ 

$ \Rightarrow \text{x}\left( \frac{\text{6+3+2}}{\text{6}} \right)<\text{11}$ 

$ \Rightarrow \text{x}\left( \frac{\text{11}}{\text{6}} \right)<\text{11} $ 

$ \Rightarrow \text{x}\left( \frac{\text{11}}{\text{6}} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{6}}{\text{11}} \right)\text{11 }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{11}} $ 

$ \Rightarrow \text{x}<\text{6} $

Therefore, we can say that the numbers less than $6$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,6 \right)$.

Ans: Let us rewrite the given inequality, we get

$\frac{\text{x}}{\text{3}}>\frac{\text{x}}{\text{2}}\text{+1} $ 

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

$ \Rightarrow \text{x}\left( \frac{\text{1}}{\text{3}}\text{-}\frac{\text{1}}{\text{2}} \right)>\text{1} $ 

$ \Rightarrow \text{x}\left( \frac{\text{-1}}{6} \right)>\text{1} $ 

$ \Rightarrow \text{x}\left( \frac{\text{-1}}{6} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{-6}}{1} \right)<\text{-6} $ 

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

Therefore, we can say that the numbers less than $-6$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-6 \right)$.