Solve the following linear inequations:

\displaystyle \textbf{Question 1: } \Big|x+\frac{1}{3}\Big|>\frac{8}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\Big|x+\frac{1}{3}\Big|>\frac{8}{3}.
\displaystyle \text{Case I: }x+\frac{1}{3}>\frac{8}{3}\Rightarrow x>\frac{8}{3}-\frac{1}{3}\Rightarrow x>\frac{7}{3}.
\displaystyle \text{Case II: }x+\frac{1}{3}<-\frac{8}{3}\Rightarrow x>-\frac{1}{3}-\frac{8}{3}\Rightarrow x<-3.
\displaystyle \therefore \text{The solution set is }\left(\frac{7}{3},\infty\right)\cup(-\infty,-3).
\displaystyle \\

\displaystyle \textbf{Question 2: } |4-x|+1<3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }|4-x|+1<3.
\displaystyle \Rightarrow |4-x|<2.
\displaystyle \Rightarrow -2<4-x<2.
\displaystyle \Rightarrow -6<-x<-2.
\displaystyle \Rightarrow 2<x<6.
\displaystyle \therefore \text{The solution set is }(2,6).
\displaystyle \\

\displaystyle \textbf{Question 3: } \Big|\frac{3x-4}{2}\Big|\leq\frac{5}{12}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\Big|\frac{3x-4}{2}\Big|\leq\frac{5}{12}.
\displaystyle \Rightarrow -\frac{5}{12}\leq\frac{3x-4}{2}\leq\frac{5}{12}.
\displaystyle \Rightarrow -5\leq18x-24\leq5.
\displaystyle \Rightarrow 19\leq18x\leq29.
\displaystyle \Rightarrow \frac{19}{18}\leq x\leq\frac{29}{18}.
\displaystyle \therefore \text{The solution set is }\left[\frac{19}{18},\frac{29}{18}\right].
\displaystyle \\

\displaystyle \textbf{Question 4: } \frac{|x-2|}{x-2}>0.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{|x-2|}{x-2}>0,\qquad x\neq2.
\displaystyle \text{Case I: When }x-2>0\Rightarrow x>2.
\displaystyle \therefore |x-2|=x-2.
\displaystyle \frac{|x-2|}{x-2}=\frac{x-2}{x-2}=1>0.
\displaystyle \therefore x\in(2,\infty).
\displaystyle \text{Case II: When }x-2<0\Rightarrow x<2.
\displaystyle \therefore |x-2|=-(x-2).
\displaystyle \frac{|x-2|}{x-2}=\frac{-(x-2)}{x-2}=-1<0.
\displaystyle \therefore \text{There is no solution in this case.}
\displaystyle \therefore \text{The solution set is }(2,\infty).
\displaystyle \\

\displaystyle \textbf{Question 5: } \frac{1}{|x|-3}<\frac{1}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{1}{|x|-3}<\frac{1}{2},\qquad x\neq\pm3.
\displaystyle \Rightarrow \frac{1}{|x|-3}-\frac{1}{2}<0.
\displaystyle \Rightarrow \frac{2-(|x|-3)}{2(|x|-3)}<0.
\displaystyle \Rightarrow \frac{5-|x|}{2(|x|-3)}<0.
\displaystyle \text{Case I: When }x\geq0,\ |x|=x.
\displaystyle \therefore \frac{5-x}{2(x-3)}<0.
\displaystyle \text{Case I(a): }5-x<0\text{ and }x-3>0.
\displaystyle \Rightarrow x>5\text{ and }x>3\Rightarrow x>5.
\displaystyle \text{Case I(b): }5-x>0\text{ and }x-3<0.
\displaystyle \Rightarrow x<5\text{ and }x<3\Rightarrow x<3.
\displaystyle \text{Since }x\geq0,\text{ the solution in Case I is }[0,3)\cup(5,\infty).
\displaystyle \text{Case II: When }x<0,\ |x|=-x.
\displaystyle \therefore \frac{5+x}{2(-x-3)}<0.
\displaystyle \Rightarrow \frac{5+x}{-2x-6}<0.
\displaystyle \text{Case II(a): }5+x<0\text{ and }-2x-6>0.
\displaystyle \Rightarrow x<-5\text{ and }x<-3\Rightarrow x<-5.
\displaystyle \text{Case II(b): }5+x>0\text{ and }-2x-6<0.
\displaystyle \Rightarrow x>-5\text{ and }x>-3\Rightarrow x>-3.
\displaystyle \text{Since }x<0,\text{ the solution in Case II is }(-\infty,-5)\cup(-3,0).
\displaystyle \therefore \text{The complete solution set is}
\displaystyle [0,3)\cup(5,\infty)\cup(-\infty,-5)\cup(-3,0).
\displaystyle \therefore \text{The solution set is }(-\infty,-5)\cup(-3,3)\cup(5,\infty).
\displaystyle \\

\displaystyle \textbf{Question 6: } \frac{|x+2|-x}{x}<2.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{|x+2|-x}{x}<2,\qquad x\neq0.
\displaystyle \Rightarrow \frac{|x+2|-x}{x}-2<0.
\displaystyle \Rightarrow \frac{|x+2|-3x}{x}<0.
\displaystyle \text{Case I: When }x+2\geq0\Rightarrow x\geq-2.
\displaystyle \therefore |x+2|=x+2.
\displaystyle \Rightarrow \frac{x+2-3x}{x}<0.
\displaystyle \Rightarrow \frac{2-2x}{x}<0.
\displaystyle \text{Case I(a): }2-2x<0\text{ and }x>0.
\displaystyle \Rightarrow x>1\text{ and }x>0\Rightarrow x>1.
\displaystyle \text{Case I(b): }2-2x>0\text{ and }x<0.
\displaystyle \Rightarrow x<1\text{ and }x<0\Rightarrow x<0.
\displaystyle \text{Since }x\geq-2,\text{ the solution in Case I is }[-2,0)\cup(1,\infty).
\displaystyle \text{Case II: When }x+2<0\Rightarrow x<-2.
\displaystyle \therefore |x+2|=-(x+2).
\displaystyle \Rightarrow \frac{-(x+2)-3x}{x}<0.
\displaystyle \Rightarrow \frac{-4x-2}{x}<0.
\displaystyle \text{For }x<-2,\ -4x-2>0\text{ and }x<0.
\displaystyle \therefore \frac{-4x-2}{x}<0\text{ for every }x<-2.
\displaystyle \therefore \text{The solution in Case II is }(-\infty,-2).
\displaystyle \therefore \text{The complete solution set is}
\displaystyle [-2,0)\cup(1,\infty)\cup(-\infty,-2).
\displaystyle \therefore \text{The solution set is }(-\infty,0)\cup(1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 7: } \Big|\frac{2x-1}{x-1}\Big|>2.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\Big|\frac{2x-1}{x-1}\Big|>2,\qquad x\neq1.
\displaystyle \text{Case I: }\frac{2x-1}{x-1}>2.
\displaystyle \Rightarrow \frac{2x-1}{x-1}-2>0.
\displaystyle \Rightarrow \frac{2x-1-2x+2}{x-1}>0.
\displaystyle \Rightarrow \frac{1}{x-1}>0.
\displaystyle \Rightarrow x-1>0\Rightarrow x>1.
\displaystyle \therefore \text{The solution in Case I is }(1,\infty).
\displaystyle \text{Case II: }\frac{2x-1}{x-1}<-2.
\displaystyle \Rightarrow \frac{2x-1}{x-1}+2<0.
\displaystyle \Rightarrow \frac{2x-1+2x-2}{x-1}<0.
\displaystyle \Rightarrow \frac{4x-3}{x-1}<0.
\displaystyle \text{Case II(a): }4x-3>0\text{ and }x-1<0.
\displaystyle \Rightarrow x>\frac{3}{4}\text{ and }x<1.
\displaystyle \Rightarrow \frac{3}{4}<x<1.
\displaystyle \text{Case II(b): }4x-3<0\text{ and }x-1>0.
\displaystyle \Rightarrow x<\frac{3}{4}\text{ and }x>1,
\displaystyle \text{which is not possible.}
\displaystyle \therefore \text{The solution in Case II is }\left(\frac{3}{4},1\right).
\displaystyle \therefore \text{The complete solution set is }(1,\infty)\cup\left(\frac{3}{4},1\right).
\displaystyle \\

\displaystyle \textbf{Question 8: } |x-1|+|x-2|+|x-3|\geq6.
\displaystyle \text{Answer:}
\displaystyle \text{Given }|x-1|+|x-2|+|x-3|\geq6.
\displaystyle \text{The critical points are }x=1,\ x=2\text{ and }x=3.
\displaystyle \text{Case I: When }x<1.
\displaystyle (1-x)+(2-x)+(3-x)\geq6.
\displaystyle \Rightarrow 6-3x\geq6.
\displaystyle \Rightarrow -3x\geq0\Rightarrow x\leq0.
\displaystyle \therefore \text{The solution in Case I is }(-\infty,0].
\displaystyle \text{Case II: When }1\leq x<2.
\displaystyle (x-1)+(2-x)+(3-x)\geq6.
\displaystyle \Rightarrow 4-x\geq6.
\displaystyle \Rightarrow x\leq-2.
\displaystyle \text{Since }1\leq x<2,\text{ there is no solution in this case.}
\displaystyle \text{Case III: When }2\leq x<3.
\displaystyle (x-1)+(x-2)+(3-x)\geq6.
\displaystyle \Rightarrow x\geq6.
\displaystyle \text{Since }2\leq x<3,\text{ there is no solution in this case.}
\displaystyle \text{Case IV: When }x\geq3.
\displaystyle (x-1)+(x-2)+(x-3)\geq6.
\displaystyle \Rightarrow 3x-6\geq6.
\displaystyle \Rightarrow 3x\geq12\Rightarrow x\geq4.
\displaystyle \therefore \text{The solution in Case IV is }[4,\infty).
\displaystyle \therefore \text{The solution set is }(-\infty,0]\cup[4,\infty).
\displaystyle \\

\displaystyle \textbf{Question 9: } \frac{|x-2|-1}{|x-2|-2}\leq0.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{|x-2|-1}{|x-2|-2}\leq0,\qquad x\neq0,\ 4.
\displaystyle \text{Case I: When }x-2\geq0\Rightarrow x\geq2.
\displaystyle \therefore |x-2|=x-2.
\displaystyle \Rightarrow \frac{x-2-1}{x-2-2}\leq0.
\displaystyle \Rightarrow \frac{x-3}{x-4}\leq0.
\displaystyle \text{The critical points are }x=3\text{ and }x=4.
\displaystyle \therefore 3\leq x<4.
\displaystyle \text{Since }x\geq2,\text{ the solution in Case I is }[3,4).
\displaystyle \text{Case II: When }x-2<0\Rightarrow x<2.
\displaystyle \therefore |x-2|=2-x.
\displaystyle \Rightarrow \frac{2-x-1}{2-x-2}\leq0.
\displaystyle \Rightarrow \frac{1-x}{-x}\leq0.
\displaystyle \Rightarrow \frac{x-1}{x}\leq0.
\displaystyle \text{The critical points are }x=0\text{ and }x=1.
\displaystyle \therefore 0<x\leq1.
\displaystyle \text{Since }x<2,\text{ the solution in Case II is }(0,1].
\displaystyle \therefore \text{The complete solution set is }(0,1]\cup[3,4).
\displaystyle \\

\displaystyle \textbf{Question 10: } \frac{1}{|x|-3}\leq\frac{1}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }\frac{1}{|x|-3}\leq\frac{1}{2},\qquad x\neq\pm3.
\displaystyle \Rightarrow \frac{1}{|x|-3}-\frac{1}{2}\leq0.
\displaystyle \Rightarrow \frac{2-(|x|-3)}{2(|x|-3)}\leq0.
\displaystyle \Rightarrow \frac{5-|x|}{2|x|-6}\leq0.
\displaystyle \text{Case I: When }x\geq0,\ |x|=x.
\displaystyle \therefore \frac{5-x}{2x-6}\leq0.
\displaystyle \text{Case I(a): }5-x\leq0\text{ and }2x-6>0.
\displaystyle \Rightarrow x\geq5\text{ and }x>3\Rightarrow x\geq5.
\displaystyle \text{Case I(b): }5-x\geq0\text{ and }2x-6<0.
\displaystyle \Rightarrow x\leq5\text{ and }x<3\Rightarrow x<3.
\displaystyle \text{Since }x\geq0,\text{ the solution in Case I is }[0,3)\cup[5,\infty).
\displaystyle \text{Case II: When }x<0,\ |x|=-x.
\displaystyle \therefore \frac{5+x}{-2x-6}\leq0.
\displaystyle \Rightarrow \frac{5+x}{2x+6}\geq0.
\displaystyle \text{Case II(a): }5+x\geq0\text{ and }2x+6>0.
\displaystyle \Rightarrow x\geq-5\text{ and }x>-3\Rightarrow x>-3.
\displaystyle \text{Since }x<0,\text{ this gives }-3<x<0.
\displaystyle \text{Case II(b): }5+x\leq0\text{ and }2x+6<0.
\displaystyle \Rightarrow x\leq-5\text{ and }x<-3\Rightarrow x\leq-5.
\displaystyle \therefore \text{The solution in Case II is }(-\infty,-5]\cup(-3,0).
\displaystyle \therefore \text{The complete solution set is}
\displaystyle [0,3)\cup[5,\infty)\cup(-\infty,-5]\cup(-3,0).
\displaystyle \therefore \text{The solution set is }(-\infty,-5]\cup(-3,3)\cup[5,\infty).
\displaystyle \\

\displaystyle \textbf{Question 11: } |x+1|+|x|>3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }|x+1|+|x|>3.
\displaystyle \text{The critical points are }x=-1\text{ and }x=0.
\displaystyle \text{Case I: When }x<-1.
\displaystyle \therefore |x+1|=-(x+1)\text{ and }|x|=-x.
\displaystyle \Rightarrow -(x+1)-x>3.
\displaystyle \Rightarrow -2x-1>3.
\displaystyle \Rightarrow -2x>4\Rightarrow x<-2.
\displaystyle \therefore \text{The solution in Case I is }(-\infty,-2).
\displaystyle \text{Case II: When }-1\leq x<0.
\displaystyle \therefore |x+1|=x+1\text{ and }|x|=-x.
\displaystyle \Rightarrow (x+1)-x>3.
\displaystyle \Rightarrow 1>3,
\displaystyle \text{which is not possible.}
\displaystyle \therefore \text{There is no solution in Case II.}
\displaystyle \text{Case III: When }x\geq0.
\displaystyle \therefore |x+1|=x+1\text{ and }|x|=x.
\displaystyle \Rightarrow (x+1)+x>3.
\displaystyle \Rightarrow 2x+1>3.
\displaystyle \Rightarrow 2x>2\Rightarrow x>1.
\displaystyle \therefore \text{The solution in Case III is }(1,\infty).
\displaystyle \therefore \text{The solution set is }(-\infty,-2)\cup(1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 12: } 1\leq|x-2|\leq3.
\displaystyle \text{Answer:}
\displaystyle \text{Given }1\leq|x-2|\leq3.
\displaystyle \text{First, consider }|x-2|\geq1.
\displaystyle \Rightarrow x-2\geq1\text{ or }x-2\leq-1.
\displaystyle \Rightarrow x\geq3\text{ or }x\leq1.
\displaystyle \therefore \text{The solution set is }(-\infty,1]\cup[3,\infty).
\displaystyle \text{Next, consider }|x-2|\leq3.
\displaystyle \Rightarrow -3\leq x-2\leq3.
\displaystyle \Rightarrow -1\leq x\leq5.
\displaystyle \therefore \text{The solution set is }[-1,5].
\displaystyle \text{Both inequalities must hold simultaneously.}
\displaystyle \therefore \text{The required solution set is}
\displaystyle \big((-\infty,1]\cup[3,\infty)\big)\cap[-1,5].
\displaystyle \therefore \text{The solution set is }[-1,1]\cup[3,5].
\displaystyle \\

\displaystyle \textbf{Question 13: } |3-4x|\geq9.
\displaystyle \text{Answer:}
\displaystyle \text{Given }|3-4x|\geq9.
\displaystyle \Rightarrow 3-4x\geq9\text{ or }3-4x\leq-9.
\displaystyle \Rightarrow -4x\geq6\text{ or }-4x\leq-12.
\displaystyle \Rightarrow x\leq-\frac{3}{2}\text{ or }x\geq3.
\displaystyle \therefore \text{The solution set is }\left(-\infty,-\frac{3}{2}\right]\cup[3,\infty).
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.