\displaystyle \textbf{Question 1: }\text{Solve }12x<50\text{ when }
\displaystyle \text{i) }x\in R\hspace{1.0cm}\text{ii) }x\in Z\hspace{1.0cm}\text{iii) }x\in N
\displaystyle \text{Answer:}
\displaystyle \text{Given }12x<50
\displaystyle \Rightarrow x<\frac{50}{12}=\frac{25}{6}
\displaystyle \text{i) If }x\in R,
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{25}{6}\right).
\displaystyle \text{ii) If }x\in Z,
\displaystyle \therefore \text{The solution set is }\{\ldots,-3,-2,-1,0,1,2,3,4\}.
\displaystyle \text{iii) If }x\in N,
\displaystyle \therefore \text{The solution set is }\{1,2,3,4\}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve }-4x>30\text{ when}
\displaystyle \text{i) }x\in R\hspace{1.0cm}\text{ii) }x\in Z\hspace{1.0cm}\text{iii) }x\in N
\displaystyle \text{Answer:}
\displaystyle \text{Given }-4x>30
\displaystyle \Rightarrow x<-\frac{30}{4}=-\frac{15}{2}
\displaystyle \text{i) If }x\in R,
\displaystyle \therefore \text{The solution set is }\left(-\infty,-\frac{15}{2}\right).
\displaystyle \text{ii) If }x\in Z,
\displaystyle \therefore \text{The solution set is }\{\ldots,-9,-8\}.
\displaystyle \text{iii) If }x\in N,
\displaystyle \therefore \text{The solution set is }\phi,\text{ which is the null set.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve }4x-2<8\text{ when}
\displaystyle \text{i) }x\in R\hspace{1.0cm}\text{ii) }x\in Z\hspace{1.0cm}\text{iii) }x\in N
\displaystyle \text{Answer:}
\displaystyle \text{Given }4x-2<8
\displaystyle \Rightarrow 4x<10
\displaystyle \Rightarrow x<\frac{5}{2}
\displaystyle \text{i) If }x\in R,
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{5}{2}\right).
\displaystyle \text{ii) If }x\in Z,
\displaystyle \therefore \text{The solution set is }\{\ldots,-3,-2,-1,0,1,2\}.
\displaystyle \text{iii) If }x\in N,
\displaystyle \therefore \text{The solution set is }\{1,2\}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Solve }3x-7>x+1.
\displaystyle \text{Answer:}
\displaystyle \text{Given }3x-7>x+1
\displaystyle \Rightarrow 2x>8
\displaystyle \Rightarrow x>4
\displaystyle \therefore \text{The solution set is }(4,\infty).
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Solve }x+5>4x-10.
\displaystyle \text{Answer:}
\displaystyle \text{Given }x+5>4x-10
\displaystyle \Rightarrow -3x>-15
\displaystyle \Rightarrow x<5
\displaystyle \therefore \text{The solution set is }(-\infty,5).
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Solve }3x+9\geq-x+19.
\displaystyle \text{Answer:}
\displaystyle \text{Given }3x+9\geq-x+19
\displaystyle \Rightarrow 4x\geq10
\displaystyle \Rightarrow x\geq\frac{5}{2}
\displaystyle \therefore \text{The solution set is }\left[\frac{5}{2},\infty\right).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Solve }2(3-x)\geq\frac{x}{5}+4.
\displaystyle \text{Answer:}
\displaystyle 2(3-x)\geq\frac{x}{5}+4
\displaystyle \Rightarrow 6-2x\geq\frac{x}{5}+4
\displaystyle \Rightarrow \frac{x}{5}+2x\leq2
\displaystyle \Rightarrow \frac{11x}{5}\leq2
\displaystyle \Rightarrow x\leq\frac{10}{11}
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{10}{11}\right].
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Solve }\frac{3x-2}{5}\leq\frac{4x-3}{2}.
\displaystyle \text{Answer:}
\displaystyle \frac{3x-2}{5}\leq\frac{4x-3}{2}
\displaystyle \Rightarrow 2(3x-2)\leq5(4x-3)
\displaystyle \Rightarrow 6x-4\leq20x-15
\displaystyle \Rightarrow 11\leq14x
\displaystyle \Rightarrow x\geq\frac{11}{14}
\displaystyle \therefore \text{The solution set is }\left[\frac{11}{14},\infty\right).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Solve }-(x-3)+4<5-2x.
\displaystyle \text{Answer:}
\displaystyle -(x-3)+4<5-2x
\displaystyle \Rightarrow -x+3+4<5-2x
\displaystyle \Rightarrow -x+7<5-2x
\displaystyle \Rightarrow x+7<5
\displaystyle \Rightarrow x<-2
\displaystyle \therefore \text{The solution set is }(-\infty,-2).
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Solve }\frac{x}{5}<\frac{3x-2}{4}-\frac{5x-3}{5}.
\displaystyle \text{Answer:}
\displaystyle \frac{x}{5}<\frac{3x-2}{4}-\frac{5x-3}{5}
\displaystyle \Rightarrow \frac{x}{5}<\frac{5(3x-2)-4(5x-3)}{20}
\displaystyle \Rightarrow \frac{x}{5}<\frac{15x-10-20x+12}{20}
\displaystyle \Rightarrow \frac{x}{5}<\frac{-5x+2}{20}
\displaystyle \Rightarrow 4x<-5x+2
\displaystyle \Rightarrow 9x<2
\displaystyle \Rightarrow x<\frac{2}{9}
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{2}{9}\right).
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Solve }\frac{2(x-1)}{5}\leq\frac{3(2+x)}{7}.
\displaystyle \text{Answer:}
\displaystyle \frac{2(x-1)}{5}\leq\frac{3(2+x)}{7}
\displaystyle \Rightarrow \frac{2x-2}{5}\leq\frac{6+3x}{7}
\displaystyle \Rightarrow 7(2x-2)\leq5(6+3x)
\displaystyle \Rightarrow 14x-14\leq30+15x
\displaystyle \Rightarrow -44\leq x
\displaystyle \Rightarrow x\geq-44
\displaystyle \therefore \text{The solution set is }[-44,\infty).
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Solve }\frac{5x}{2}+\frac{3x}{4}\geq\frac{39}{4}.
\displaystyle \text{Answer:}
\displaystyle \frac{5x}{2}+\frac{3x}{4}\geq\frac{39}{4}
\displaystyle \Rightarrow 10x+3x\geq39
\displaystyle \Rightarrow 13x\geq39
\displaystyle \Rightarrow x\geq3
\displaystyle \therefore \text{The solution set is }[3,\infty).
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Solve }\frac{x-1}{3}+4<\frac{x-5}{5}-2.
\displaystyle \text{Answer:}
\displaystyle \frac{x-1}{3}+4<\frac{x-5}{5}-2
\displaystyle \Rightarrow \frac{x-1+12}{3}<\frac{x-5-10}{5}
\displaystyle \Rightarrow \frac{x+11}{3}<\frac{x-15}{5}
\displaystyle \Rightarrow 5(x+11)<3(x-15)
\displaystyle \Rightarrow 5x+55<3x-45
\displaystyle \Rightarrow 2x<-100
\displaystyle \Rightarrow x<-50
\displaystyle \therefore \text{The solution set is }(-\infty,-50).
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Solve }\frac{2x+3}{4}-3<\frac{x-4}{3}-2.
\displaystyle \text{Answer:}
\displaystyle \frac{2x+3}{4}-3<\frac{x-4}{3}-2
\displaystyle \Rightarrow \frac{2x+3-12}{4}<\frac{x-4-6}{3}
\displaystyle \Rightarrow \frac{2x-9}{4}<\frac{x-10}{3}
\displaystyle \Rightarrow 3(2x-9)<4(x-10)
\displaystyle \Rightarrow 6x-27<4x-40
\displaystyle \Rightarrow 2x<-13
\displaystyle \Rightarrow x<-\frac{13}{2}
\displaystyle \therefore \text{The solution set is }\left(-\infty,-\frac{13}{2}\right).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Solve }\frac{5-2x}{3}<\frac{x}{6}-5.
\displaystyle \text{Answer:}
\displaystyle \frac{5-2x}{3}<\frac{x}{6}-5
\displaystyle \Rightarrow 2(5-2x)<x-30
\displaystyle \Rightarrow 10-4x<x-30
\displaystyle \Rightarrow 40<5x
\displaystyle \Rightarrow x>8
\displaystyle \therefore \text{The solution set is }(8,\infty).
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Solve }\frac{4+2x}{3}\geq\frac{x}{2}-3.
\displaystyle \text{Answer:}
\displaystyle \frac{4+2x}{3}\geq\frac{x}{2}-3
\displaystyle \Rightarrow 2(4+2x)\geq3x-18
\displaystyle \Rightarrow 8+4x\geq3x-18
\displaystyle \Rightarrow x\geq-26
\displaystyle \therefore \text{The solution set is }[-26,\infty).
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Solve }\frac{2x+3}{5}-2<\frac{3(x-2)}{5}.
\displaystyle \text{Answer:}
\displaystyle \frac{2x+3}{5}-2<\frac{3(x-2)}{5}
\displaystyle \Rightarrow 2x+3-10<3x-6
\displaystyle \Rightarrow 2x-7<3x-6
\displaystyle \Rightarrow -1<x
\displaystyle \Rightarrow x>-1
\displaystyle \therefore \text{The solution set is }(-1,\infty).
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Solve }x-2\leq\frac{5x+8}{3}.
\displaystyle \text{Answer:}
\displaystyle x-2\leq\frac{5x+8}{3}
\displaystyle \Rightarrow 3x-6\leq5x+8
\displaystyle \Rightarrow -14\leq2x
\displaystyle \Rightarrow x\geq-7
\displaystyle \therefore \text{The solution set is }[-7,\infty).
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Solve }\frac{6x-5}{4x+1}<0.
\displaystyle \text{Answer:}
\displaystyle \frac{6x-5}{4x+1}<0
\displaystyle \text{For the fraction to be negative, the numerator and denominator must have opposite signs.}
\displaystyle \text{Case I: }6x-5>0\text{ and }4x+1<0
\displaystyle \Rightarrow x>\frac{5}{6}\text{ and }x<-\frac{1}{4}
\displaystyle \text{This is not possible.}
\displaystyle \text{Case II: }6x-5<0\text{ and }4x+1>0
\displaystyle \Rightarrow x<\frac{5}{6}\text{ and }x>-\frac{1}{4}
\displaystyle \Rightarrow -\frac{1}{4}<x<\frac{5}{6}
\displaystyle \therefore \text{The solution set is }\left(-\frac{1}{4},\frac{5}{6}\right).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Solve }\frac{2x-3}{3x-7}>0.
\displaystyle \text{Answer:}
\displaystyle \frac{2x-3}{3x-7}>0
\displaystyle \text{For the fraction to be positive, the numerator and denominator must have the same sign.}
\displaystyle \text{Case I: }2x-3>0\text{ and }3x-7>0
\displaystyle \Rightarrow x>\frac{3}{2}\text{ and }x>\frac{7}{3}
\displaystyle \Rightarrow x>\frac{7}{3}
\displaystyle \text{Case II: }2x-3<0\text{ and }3x-7<0
\displaystyle \Rightarrow x<\frac{3}{2}\text{ and }x<\frac{7}{3}
\displaystyle \Rightarrow x<\frac{3}{2}
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{3}{2}\right)\cup\left(\frac{7}{3},\infty\right).
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Solve }\frac{3}{x-2}<1.
\displaystyle \text{Answer:}
\displaystyle \frac{3}{x-2}<1
\displaystyle \Rightarrow \frac{3}{x-2}-1<0
\displaystyle \Rightarrow \frac{3-(x-2)}{x-2}<0
\displaystyle \Rightarrow \frac{5-x}{x-2}<0
\displaystyle \text{For the fraction to be negative, the numerator and denominator must have opposite signs.}
\displaystyle \text{Case I: }5-x>0\text{ and }x-2<0
\displaystyle \Rightarrow x<5\text{ and }x<2
\displaystyle \Rightarrow x<2
\displaystyle \text{Case II: }5-x<0\text{ and }x-2>0
\displaystyle \Rightarrow x>5\text{ and }x>2
\displaystyle \Rightarrow x>5
\displaystyle \therefore \text{The solution set is }(-\infty,2)\cup(5,\infty).
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Solve }\frac{1}{x-1}\leq2.
\displaystyle \text{Answer:}
\displaystyle \frac{1}{x-1}\leq2
\displaystyle \Rightarrow \frac{1}{x-1}-2\leq0
\displaystyle \Rightarrow \frac{1-2(x-1)}{x-1}\leq0
\displaystyle \Rightarrow \frac{3-2x}{x-1}\leq0
\displaystyle \text{For the fraction to be non-positive, the numerator and denominator must have opposite signs,}
\displaystyle \text{or the numerator must be zero.}
\displaystyle \text{Case I: }3-2x\geq0\text{ and }x-1<0
\displaystyle \Rightarrow x\leq\frac{3}{2}\text{ and }x<1
\displaystyle \Rightarrow x<1
\displaystyle \text{Case II: }3-2x\leq0\text{ and }x-1>0
\displaystyle \Rightarrow x\geq\frac{3}{2}\text{ and }x>1
\displaystyle \Rightarrow x\geq\frac{3}{2}
\displaystyle \therefore \text{The solution set is }(-\infty,1)\cup\left[\frac{3}{2},\infty\right).
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Solve }\frac{4x+3}{2x-5}<6.
\displaystyle \text{Answer:}
\displaystyle \frac{4x+3}{2x-5}<6
\displaystyle \Rightarrow \frac{4x+3}{2x-5}-6<0
\displaystyle \Rightarrow \frac{4x+3-6(2x-5)}{2x-5}<0
\displaystyle \Rightarrow \frac{-8x+33}{2x-5}<0
\displaystyle \Rightarrow \frac{8x-33}{2x-5}>0
\displaystyle \text{For the fraction to be positive, the numerator and denominator must have the same sign.}
\displaystyle \text{Case I: }8x-33<0\text{ and }2x-5<0
\displaystyle \Rightarrow x<\frac{33}{8}\text{ and }x<\frac{5}{2}
\displaystyle \Rightarrow x<\frac{5}{2}
\displaystyle \text{Case II: }8x-33>0\text{ and }2x-5>0
\displaystyle \Rightarrow x>\frac{33}{8}\text{ and }x>\frac{5}{2}
\displaystyle \Rightarrow x>\frac{33}{8}
\displaystyle \therefore \text{The solution set is }\left(-\infty,\frac{5}{2}\right)\cup\left(\frac{33}{8},\infty\right).
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Solve }\frac{5x-6}{x+6}<1.
\displaystyle \text{Answer:}
\displaystyle \frac{5x-6}{x+6}<1
\displaystyle \Rightarrow \frac{5x-6}{x+6}-1<0
\displaystyle \Rightarrow \frac{5x-6-(x+6)}{x+6}<0
\displaystyle \Rightarrow \frac{4x-12}{x+6}<0
\displaystyle \text{For the fraction to be negative, the numerator and denominator must have opposite signs.}
\displaystyle \text{Case I: }4x-12<0\text{ and }x+6>0
\displaystyle \Rightarrow x<3\text{ and }x>-6
\displaystyle \Rightarrow -6<x<3
\displaystyle \text{Case II: }4x-12>0\text{ and }x+6<0
\displaystyle \Rightarrow x>3\text{ and }x<-6
\displaystyle \text{This is not possible.}
\displaystyle \therefore \text{The solution set is }(-6,3).
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Solve }\frac{5x+8}{4-x}<2.
\displaystyle \text{Answer:}
\displaystyle \frac{5x+8}{4-x}<2
\displaystyle \Rightarrow \frac{5x+8}{4-x}-2<0
\displaystyle \Rightarrow \frac{5x+8-2(4-x)}{4-x}<0
\displaystyle \Rightarrow \frac{7x}{4-x}<0
\displaystyle \text{For the fraction to be negative, the numerator and denominator must have opposite signs.}
\displaystyle \text{Case I: }7x>0\text{ and }4-x<0
\displaystyle \Rightarrow x>0\text{ and }x>4
\displaystyle \Rightarrow x>4
\displaystyle \text{Case II: }7x<0\text{ and }4-x>0
\displaystyle \Rightarrow x<0\text{ and }x<4
\displaystyle \Rightarrow x<0
\displaystyle \therefore \text{The solution set is }(-\infty,0)\cup(4,\infty).
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Solve }\frac{x-1}{x+3}>2.
\displaystyle \text{Answer:}
\displaystyle \frac{x-1}{x+3}>2
\displaystyle \Rightarrow \frac{x-1}{x+3}-2>0
\displaystyle \Rightarrow \frac{x-1-2(x+3)}{x+3}>0
\displaystyle \Rightarrow \frac{-x-7}{x+3}>0
\displaystyle \Rightarrow \frac{-(x+7)}{x+3}>0
\displaystyle \text{For the fraction to be positive, the numerator and denominator must have the same sign.}
\displaystyle \text{Case I: }-(x+7)>0\text{ and }x+3>0
\displaystyle \Rightarrow x<-7\text{ and }x>-3
\displaystyle \text{This is not possible.}
\displaystyle \text{Case II: }-(x+7)<0\text{ and }x+3<0
\displaystyle \Rightarrow x>-7\text{ and }x<-3
\displaystyle \Rightarrow -7<x<-3
\displaystyle \therefore \text{The solution set is }(-7,-3).
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Solve }\frac{7x-5}{8x+3}>4.
\displaystyle \text{Answer:}
\displaystyle \frac{7x-5}{8x+3}>4
\displaystyle \Rightarrow \frac{7x-5}{8x+3}-4>0
\displaystyle \Rightarrow \frac{7x-5-4(8x+3)}{8x+3}>0
\displaystyle \Rightarrow \frac{-25x-17}{8x+3}>0
\displaystyle \Rightarrow \frac{-(25x+17)}{8x+3}>0
\displaystyle \text{For the fraction to be positive, the numerator and denominator must have the same sign.}
\displaystyle \text{Case I: }-(25x+17)>0\text{ and }8x+3>0
\displaystyle \Rightarrow x<-\frac{17}{25}\text{ and }x>-\frac{3}{8}
\displaystyle \text{This is not possible.}
\displaystyle \text{Case II: }-(25x+17)<0\text{ and }8x+3<0
\displaystyle \Rightarrow x>-\frac{17}{25}\text{ and }x<-\frac{3}{8}
\displaystyle \Rightarrow -\frac{17}{25}<x<-\frac{3}{8}
\displaystyle \therefore \text{The solution set is }\left(-\frac{17}{25},-\frac{3}{8}\right).
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Solve }\frac{x}{x-5}>\frac{1}{2}.
\displaystyle \text{Answer:}
\displaystyle \frac{x}{x-5}>\frac{1}{2}
\displaystyle \Rightarrow \frac{x}{x-5}-\frac{1}{2}>0
\displaystyle \Rightarrow \frac{2x-(x-5)}{2(x-5)}>0
\displaystyle \Rightarrow \frac{x+5}{2x-10}>0
\displaystyle \text{For the fraction to be positive, the numerator and denominator must have the same sign.}
\displaystyle \text{Case I: }x+5>0\text{ and }2x-10>0
\displaystyle \Rightarrow x>-5\text{ and }x>5
\displaystyle \Rightarrow x>5
\displaystyle \text{Case II: }x+5<0\text{ and }2x-10<0
\displaystyle \Rightarrow x<-5\text{ and }x<5
\displaystyle \Rightarrow x<-5
\displaystyle \therefore \text{The solution set is }(-\infty,-5)\cup(5,\infty).
\displaystyle \\

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