Class 8: Linear Inequations

$\displaystyle \textbf{Question 1: }\text{If }x\in\{-3,-2,-1,0,1,2,3\}, \text{ find the solution set of each of the following:}$
$\displaystyle \text{(i) }x+2<1\qquad \text{(ii) }2x-1<4\qquad \text{(iii) }\frac{2}{3}x<1$
$\displaystyle \text{(iv) }1-x>0\qquad \text{(v) }3-5x<-1\qquad \text{(vi) }2-3x>1$
$\displaystyle \text{(vii) }-6\geq2x-4\qquad \text{(viii) }3x-5\geq-12\qquad \text{(ix) }14-2x<6$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }x+2<1$
$\displaystyle \Rightarrow x<1-2$
$\displaystyle \Rightarrow x<-1$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2\}.$

$\displaystyle \text{(ii) }2x-1<4$
$\displaystyle \Rightarrow 2x<5$
$\displaystyle \Rightarrow x<\frac{5}{2}$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1,0,1,2\}.$

$\displaystyle \text{(iii) }\frac{2}{3}x<1$
$\displaystyle \Rightarrow x<\frac{3}{2}$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1,0,1\}.$

$\displaystyle \text{(iv) }1-x>0$
$\displaystyle \Rightarrow -x>-1$
$\displaystyle \Rightarrow x<1$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1,0\}.$

$\displaystyle \text{(v) }3-5x<-1$
$\displaystyle \Rightarrow -5x<-4$
$\displaystyle \Rightarrow x>\frac{4}{5}$
$\displaystyle \therefore \text{ Solution Set}=\{1,2,3\}.$

$\displaystyle \text{(vi) }2-3x>1$
$\displaystyle \Rightarrow -3x>-1$
$\displaystyle \Rightarrow x<\frac{1}{3}$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1,0\}.$

$\displaystyle \text{(vii) }-6\geq2x-4$
$\displaystyle \Rightarrow -2\geq2x$
$\displaystyle \Rightarrow -1\geq x$
$\displaystyle \Rightarrow x\leq-1$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1\}.$

$\displaystyle \text{(viii) }3x-5\geq-12$
$\displaystyle \Rightarrow 3x\geq-7$
$\displaystyle \Rightarrow x\geq-\frac{7}{3}$
$\displaystyle \therefore \text{ Solution Set}=\{-2,-1,0,1,2,3\}.$

$\displaystyle \text{(ix) }14-2x<6$
$\displaystyle \Rightarrow -2x<-8$
$\displaystyle \Rightarrow x>4$
$\displaystyle \therefore \text{ Solution Set}=\varnothing.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }x\in\mathbb{N}, \text{ find the solution set of each of the following inequations,}$
$\displaystyle \text{where }\mathbb{N}=\{1,2,3,\ldots\}:$
$\displaystyle \text{(i) }3x-8<0\qquad \text{(ii) }7x+3\leq17\qquad \text{(iii) }5-x>1$
$\displaystyle \text{(iv) }1-3x>-4\qquad \text{(v) }\frac{3}{2}-\frac{x}{2}>-1\qquad \text{(vi) }-\frac14\leq\frac12-\frac{x}{3}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }3x-8<0$
$\displaystyle \Rightarrow 3x<8$
$\displaystyle \Rightarrow x<\frac{8}{3}$
$\displaystyle \therefore \text{ Solution Set}=\{1,2\}.$

$\displaystyle \text{(ii) }7x+3\leq17$
$\displaystyle \Rightarrow 7x\leq14$
$\displaystyle \Rightarrow x\leq2$
$\displaystyle \therefore \text{ Solution Set}=\{1,2\}.$

$\displaystyle \text{(iii) }5-x>1$
$\displaystyle \Rightarrow -x>-4$
$\displaystyle \Rightarrow x<4$
$\displaystyle \therefore \text{ Solution Set}=\{1,2,3\}.$

$\displaystyle \text{(iv) }1-3x>-4$
$\displaystyle \Rightarrow -3x>-5$
$\displaystyle \Rightarrow x<\frac{5}{3}$
$\displaystyle \therefore \text{ Solution Set}=\{1\}.$

$\displaystyle \text{(v) }\frac{3}{2}-\frac{x}{2}>-1$
$\displaystyle \Rightarrow -\frac{x}{2}>-\frac{5}{2}$
$\displaystyle \Rightarrow x<5$
$\displaystyle \therefore \text{ Solution Set}=\{1,2,3,4\}.$

$\displaystyle \text{(vi) }-\frac14\leq\frac12-\frac{x}{3}$
$\displaystyle \Rightarrow -\frac14-\frac12\leq-\frac{x}{3}$
$\displaystyle \Rightarrow -\frac34\leq-\frac{x}{3}$
$\displaystyle \Rightarrow \frac34\geq\frac{x}{3}$
$\displaystyle \Rightarrow x\leq\frac94$
$\displaystyle \therefore \text{ Solution Set}=\{1,2\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If }x\in\mathbb{Z}, \text{ find the solution set of each of the following inequations,}$
$\displaystyle \text{where }\mathbb{Z}=\{\ldots,-3,-2,-1,0,1,2,3,\ldots\}:$
$\displaystyle \text{(i) }9x-7\leq25+3x\qquad \text{(ii) }-17<9x-8\qquad \text{(iii) }-4(x+5)>10$
$\displaystyle \text{(iv) }4-3x<13+x\qquad \text{(v) }5-4x<10-x\qquad \text{(vi) }10-2(1+4x)<20$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }9x-7\leq25+3x$
$\displaystyle \Rightarrow 9x-3x\leq25+7$
$\displaystyle \Rightarrow 6x\leq32$
$\displaystyle \Rightarrow x\leq\frac{16}{3}$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\leq5.$
$\displaystyle \therefore \text{ Solution Set} =\{\ldots,-3,-2,-1,0,1,2,3,4,5\}.$

$\displaystyle \text{(ii) }-17<9x-8$
$\displaystyle \Rightarrow -9<9x$
$\displaystyle \Rightarrow -1<x$
$\displaystyle \Rightarrow x>-1$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\geq0.$
$\displaystyle \therefore \text{ Solution Set}=\{0,1,2,3,\ldots\}.$

$\displaystyle \text{(iii) }-4(x+5)>10$
$\displaystyle \Rightarrow -4x-20>10$
$\displaystyle \Rightarrow -4x>30$
$\displaystyle \Rightarrow x<-\frac{15}{2}$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\leq-8.$
$\displaystyle \therefore \text{ Solution Set} =\{\ldots,-11,-10,-9,-8\}.$

$\displaystyle \text{(iv) }4-3x<13+x$
$\displaystyle \Rightarrow -3x-x<13-4$
$\displaystyle \Rightarrow -4x<9$
$\displaystyle \Rightarrow x>-\frac{9}{4}$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\geq-2.$
$\displaystyle \therefore \text{ Solution Set} =\{-2,-1,0,1,2,3,\ldots\}.$

$\displaystyle \text{(v) }5-4x<10-x$
$\displaystyle \Rightarrow -4x+x<10-5$
$\displaystyle \Rightarrow -3x<5$
$\displaystyle \Rightarrow x>-\frac{5}{3}$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\geq-1.$
$\displaystyle \therefore \text{ Solution Set} =\{-1,0,1,2,3,\ldots\}.$

$\displaystyle \text{(vi) }10-2(1+4x)<20$
$\displaystyle \Rightarrow 10-2-8x<20$
$\displaystyle \Rightarrow 8-8x<20$
$\displaystyle \Rightarrow -8x<12$
$\displaystyle \Rightarrow x>-\frac{3}{2}$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ we have }x\geq-1.$
$\displaystyle \therefore \text{ Solution Set} =\{-1,0,1,2,3,\ldots\}.$ $\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the solution set of each of the following} \text{ inequations and represent it on a real number line.}$
$\displaystyle \text{(i) }1-4x\geq-1,\quad x\in\mathbb{N} \qquad \text{(ii) }-3\leq4x+1<9,\quad x\in\mathbb{N}$
$\displaystyle \text{(iii) }0<2x-5<5,\quad x\in\mathbb{W}, \quad \mathbb{W}=\{0,1,2,3,\ldots\}$
$\displaystyle \text{(iv) }-3<\frac{x}{2}-1<1,\quad x\in\mathbb{Z} \qquad \text{(v) }-4<\frac{2x}{5}+1<-3,\quad x\in\mathbb{Z}$
$\displaystyle \text{(vi) }3+\frac{x}{4}<\frac{2x}{3}+5,\quad x\in\mathbb{R}$
$\displaystyle \text{(vii) }\frac{3x+1}{4}\leq\frac{5x-2}{3}, \quad x\in\mathbb{R}$
$\displaystyle \text{(viii) }\frac{1}{3}(4x-1)+3 \leq4+\frac{2}{5}(6x+2)+\frac{4}{5},\quad x\in\mathbb{R}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }1-4x\geq-1$
$\displaystyle \Rightarrow -4x\geq-2$
$\displaystyle \Rightarrow x\leq\frac{1}{2}$
$\displaystyle \text{Since }x\in\mathbb{N}=\{1,2,3,\ldots\}, \text{ there is no possible value of }x.$
$\displaystyle \therefore \text{ Solution Set}=\varnothing.$

$\displaystyle \text{(ii) }-3\leq4x+1<9$
$\displaystyle \Rightarrow -4\leq4x<8$
$\displaystyle \Rightarrow -1\leq x<2$
$\displaystyle \text{Since }x\in\mathbb{N},\text{ the only possible value is }x=1.$
$\displaystyle \therefore \text{ Solution Set}=\{1\}.$

$\displaystyle \text{(iii) }0<2x-5<5$
$\displaystyle \Rightarrow 5<2x<10$
$\displaystyle \Rightarrow \frac{5}{2}<x<5$
$\displaystyle \text{Since }x\in\mathbb{W},\text{ the possible values are }x=3,4.$
$\displaystyle \therefore \text{ Solution Set}=\{3,4\}.$

$\displaystyle \text{(iv) }-3<\frac{x}{2}-1<1$
$\displaystyle \Rightarrow -2<\frac{x}{2}<2$
$\displaystyle \Rightarrow -4<x<4$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ the possible integer values are}$
$\displaystyle -3,-2,-1,0,1,2,3.$
$\displaystyle \therefore \text{ Solution Set}=\{-3,-2,-1,0,1,2,3\}.$

$\displaystyle \text{(v) }-4<\frac{2x}{5}+1<-3$
$\displaystyle \Rightarrow -5<\frac{2x}{5}<-4$
$\displaystyle \Rightarrow -25<2x<-20$
$\displaystyle \Rightarrow -\frac{25}{2}<x<-10$
$\displaystyle \text{Since }x\in\mathbb{Z},\text{ the possible values are }x=-12,-11.$
$\displaystyle \therefore \text{ Solution Set}=\{-12,-11\}.$

$\displaystyle \text{(vi) }3+\frac{x}{4}<\frac{2x}{3}+5$
$\displaystyle \Rightarrow \frac{x}{4}<\frac{2x}{3}+2$
$\displaystyle \Rightarrow \frac{x}{4}-\frac{2x}{3}<2$
$\displaystyle \Rightarrow -\frac{5x}{12}<2$
$\displaystyle \Rightarrow -5x<24$
$\displaystyle \Rightarrow x>-\frac{24}{5}$
$\displaystyle \therefore \text{ Solution Set} =\left\{x\in\mathbb{R}:x>-\frac{24}{5}\right\}.$

$\displaystyle \text{(vii) }\frac{3x+1}{4}\leq\frac{5x-2}{3}$
$\displaystyle \Rightarrow 3(3x+1)\leq4(5x-2)$
$\displaystyle \Rightarrow 9x+3\leq20x-8$
$\displaystyle \Rightarrow 11\leq11x$
$\displaystyle \Rightarrow x\geq1$
$\displaystyle \therefore \text{ Solution Set} =\{x\in\mathbb{R}:x\geq1\}.$

$\displaystyle \text{(viii) }\frac{1}{3}(4x-1)+3 \leq4+\frac{2}{5}(6x+2)+\frac{4}{5}$
$\displaystyle \Rightarrow \frac{4x}{3}-\frac{1}{3}+3 \leq4+\frac{12x}{5}+\frac{4}{5}+\frac{4}{5}$
$\displaystyle \Rightarrow \frac{4x}{3}+\frac{8}{3} \leq\frac{12x}{5}+\frac{28}{5}$
$\displaystyle \Rightarrow 20x+40\leq36x+84$
$\displaystyle \Rightarrow -44\leq16x$
$\displaystyle \Rightarrow x\geq-\frac{11}{4}$
$\displaystyle \therefore \text{ Solution Set} =\left\{x\in\mathbb{R}:x\geq-\frac{11}{4}\right\}.$