\begin{array}{l}
\text {Solve the following equations and verify your answer: } \\ \\
\text {In this exercise we will use cross multiplication and transposition to solve the below problems. } \\ \\\text {Q1. } \quad \frac {(2 x-3)}{(3 x+2)}= – \frac{2}{3} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {(2 x-3)}{(3 x+2)}= – \frac{2}{3} \\
\Rightarrow \quad 3(2 x-3)=-2(3 x+2) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 6 x-9=-6 x-4 \\
\Rightarrow \quad 6 x+6 x=9-4 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad 12 x=5 \\
\Rightarrow \quad x=\frac{5}{12} \\\text {On substituting } x=\frac{5}{12} \text { on both side of equation, we get} \\LHS = \frac {(2 x-3)}{(3 x+2)} \\
=\frac {[2(\frac{5}{12}) -3]}{[3 (\frac{5}{12})+2]} \\=\frac {[\frac{5-18}{6}]}{[\frac{5+8}{4}]} \\
=\frac {[\frac{-13}{6}]}{[\frac{13}{4}]} \\
=- \frac{4}{6} \\
=- \frac{2}{3} = RHS \\\text {Hence verified.} \\\end{array}\begin{array}{l}
\text {Q2. } \quad \frac{2-y}{y+7}=\frac{3}{5} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \frac{2-y}{y+7}=\frac{3}{5} \\
\Rightarrow \quad (2-y) 5=3(y+7) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 10-5 y=3 y+21 \\
\Rightarrow \quad -5 y-3 y=21-10 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad -8 y=11 \\
\Rightarrow \quad y= -\frac{11}{8} \\\text {On substituting } y= -\frac{11}{8} \text { on both side of equation, we get} \\LHS = \frac{2-(-\frac{11}{8})}{\frac{-11}{8}+7} \\
=\frac {\frac{16+11}{8}}{\frac{-11+56}{8}} \\
=\frac{27}{45} \\
=\frac{3}{5} = RHS \\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q3. } \quad \frac{5 x-7}{3 x}=2 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{5 x-7}{3 x}=2 \\\Rightarrow \quad 5 x-7=2 \times 3 x \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 5 x-7=6 x \\
\Rightarrow \quad 5 x-6 x=7 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad-x=7 \\
\Rightarrow \quad x=-7 \\\text {On substituting } x=-7 \text { on both side of equation, we get} \\
LHS =\frac{5 x-7}{3 x} \\
=\frac{5(-7)-7}{3(-7)} \\
=\frac{-35-7}{-21} \\
=\frac{42}{21} \\
=2 = RHS \\
\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q4.} \quad (3 x+5) /(2 x+7)=4 \\
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Sol. Given equation is } \quad \frac{3 x+5}{2 x+7}=4 \\
\Rightarrow \quad 3 x+5=4(2 x+7) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 3 x+5=8 x+28 \\
\Rightarrow \quad 3 x-8 x=28-5 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad -5 x=23 \\
\Rightarrow \quad x=-\frac{23}{5} \\\text {On substituting } x=-\frac{23}{5} \text { on both side of equation, we get} \\
LHS =\frac{3(-\frac{23}{5})+5}{2(\frac{-23}{5})+7} \\
=\frac{\frac{-69+25}{5}}{\frac{-46+35}{5}} \\
=\frac{-44}{-11} \\
=4 = RHS \\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q5.} \quad \frac{2 y+5}{y+4}=1 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{2 y+5}{y+4}=1 \\
\Rightarrow \quad 2 y+5=y+4 \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 2 y-y=4-5 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad y=-1 \\\text {On substituting } y=-1 \text { on both side of equation, we get} \\
LHS =\frac{2(-1)+5}{-1+4} \\
=\frac{-2+5}{3}= 1 = RHS \\
\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q6.} \quad \frac{2 x+1}{3 x-2}=\frac{5}{9} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{2 x+1}{3 x-2}=\frac{5}{9} \\
\Rightarrow \quad (2 x+1) 9=5(3 x-2) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 18 x+9=15 x-10 \\
\Rightarrow \quad \quad 3 x=-10-9 \quad \quad [\text {Using transposition}] \\
\Rightarrow \quad x=\frac{-19}{3} \\\text {On substituting } x=\frac{-19}{3} \text { on both side of equation, we get} \\
LHS = \frac{2 x+1}{3 x-2}
=\frac{2(\frac{-19}{3})+1}{3(\frac{-19}{3})-2} \\
=\frac{\frac{-38+3}{3}}{\frac{-57-6}{3}} \\
=\frac{-35}{-63} \\
=\frac{5}{9} = RHS\\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q7. } \quad \frac{1-9 y}{19-3 y}=\frac{5}{8} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{1-9 y}{19-3 y}=\frac{5}{8} \\
\Rightarrow \quad 8(1-9 y)=5(19-3 y) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 8-72 y=95-15 y \\
\Rightarrow \quad -72 y+15 y=95-8 \\
\Rightarrow \quad -57 y=87 \\
\Rightarrow \quad y=-\frac{87}{57} \\
\Rightarrow \quad y=-\frac{29}{19} \\\text {On substituting } y=-\frac{29}{19} \text { on both side of equation, we get} \\
LHS = \frac{1-9 y}{19-3 y}=\frac{1-9(\frac{-29}{19})}{19-3(-\frac{29}{19})} \\
=\frac{19+9(29)}{19(19)+3(29)} \\=\frac{19+261}{361+87} \\
=\frac{280}{448} \\
=\frac{5}{8} = RHS \\
\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q8. } \quad \frac{2 x}{3 x+1}=-3 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{2 x}{3 x+1}=-3 \\
\Rightarrow \quad 2 x=-3(3 x+1) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 2 x=-9 x-3 \\
\Rightarrow \quad 2 x+9 x=-3 \\
\Rightarrow \quad 11 x=-3 \\
\Rightarrow \quad x=\frac{-3}{11} \\\text {On substituting } x=\frac{-3}{11} \text { on both side of equation, we get} \\
=\frac{2(-\frac{3}{11})}{3(-\frac{3}{11})+1} \\
=\frac{-\frac{-6}{11}}{\frac{-9+11}{11}} \\
=\frac{-6}{2} \\
=-3 = RHS \\
\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q9. } \quad \frac{y-(7-8 y)}{9 y-(3+4 y)}=\frac{2}{3} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{y-(7-8 y)}{9 y-(3+4 y)}=\frac{2}{3}\\\Rightarrow \quad \frac {9 y-7}{5 y-3}=\frac{2}{3} \\
\Rightarrow \quad 3(9 y-7)=2(5 y-3) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 27 y-21=10 y-6 \\
\Rightarrow \quad 27 y-10 y=21-6 \\
\Rightarrow \quad 17 y=15 \\
\Rightarrow \quad y=\frac{15}{17} \\\text {On substituting } y=\frac{15}{17} \text { on both side of equation, we get} \\LHS = \frac{y-(7-8 y)}{9 y-(3+4 y)} \\
=\frac {9 y-7}{5 y-3} \\
=\frac{9(\frac{15}{17})-7}{5(\frac{15}{17})-3} \\
=\frac{9(15)-7(17)}{5(15)-3(17)} \\=\frac{135 -119)}{75 -51} \\
=\frac{16}{24} \\
=\frac{2}{3} =RHS \\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q10. } \quad \frac{6}{2 x-(3-4 x)}=\frac{2}{3} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{6}{2 x-(3-4 x)}=\frac{2}{3} \\
\Rightarrow \quad 6(3)=2(6 x-3) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 18=12 x-6 \\
\Rightarrow \quad 12 x=18+6 \\
\Rightarrow \quad 12 x=24 \\
\Rightarrow \quad x=\frac{24}{12}=2 \\\text {On substituting } x=2 \text { on both side of equation, we get} \\LHS =\frac{6}{2 x-(3-4 x)} \\= \frac{6}{6 x-3} \\
= \frac{6}{6(2)-3} \\
= \frac{6}{9} \\
= \frac{2}{3} = RHS \\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q11. } \quad \frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{2}{3 x}-\frac{3}{2 x}=\frac{1}{12} \\\Rightarrow \quad \frac{4-9}{6 x}=\frac{1}{12} \\
\Rightarrow \quad (-5) \times 12 =6 x \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad x=-\frac{5 \times 12}{6} \\
\Rightarrow \quad x=-10 \\\text {On substituting } x=-10 \text { on both side of equation, we get} \\LHS =\frac{2}{3 x}-\frac{3}{2 x} \\
=\frac{4-9}{6 x} \\
=\frac{-5}{6 x} \\
=\frac{-5}{6(-10)} \\
= \frac{1}{12} = RHS \\\text {Hence verified.} \\
\end{array}\begin{array}{l}
\text {Q12. } \quad \frac{3 x+5}{4 x+2}=\frac{3 x+4}{4 x+7} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{3 x+5}{4 x+2}=\frac{3 x+4}{4 x+7} \\\Rightarrow \quad (3 x+5)(4 x+7)=(3 x+4)(4 x+2) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 12 x^{2}+21 x+20 x+35=12 x^{2}+6 x+16 x+8 \\
\Rightarrow \quad 12 x^{2}-12 x^{2}+21 x+20 x-6 x-16 x=-35+8 \\
\Rightarrow \quad 19 x=-27 \\
\Rightarrow \quad x=\frac{-27}{19} \\\text {On substituting } x=\frac{-27}{19} \text { on both side of equation, we get} \\LHS =
\frac{3(\frac{-27}{19})+5}{4(\frac{-27}{19})+2} \\= \frac{(\frac{-81+95}{19})}{(\frac{-108+38}{19})} \\
=\frac{-81+95}{-108+38} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
=\frac{14}{-70} \\
= -\frac{1}{5} \\RHS = \frac{3(\frac{-27}{19})+4}{4(-\frac{27}{19})+7} \\
= \frac{(\frac{-81+76}{19})}{(\frac{-108+133}{19})} \\
=\frac{-81+76}{-108+133} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
=\frac{-5}{25} \\
=-\frac{1}{5} \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q13. } \quad \frac{7 x-2}{5 x-1}=\frac{7 x+3}{5 x+4} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{7 x-2}{5 x-1}=\frac{7 x+3}{5 x+4} \\\Rightarrow \quad (7 x-2)(5 x+4)=(7 x+3)(5 x-1) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad (7 x-2)(5 x+4)- (7 x+3)(5 x-1) = 0 \\
\Rightarrow \quad 35 x^{2} + 28 x – 10 x – 8 – 35 x^{2} + 7 x-15 x+3=0 \\
\Rightarrow \quad 10 x-5=0 \\
\Rightarrow \quad 10 x=5 \\
\Rightarrow \quad x=\frac{5}{10} \\
\Rightarrow \quad x=\frac{1}{2} \\\text {On substituting } x=\frac{1}{2} \text { on both side of equation, we get} \\LHS = \frac{7(\frac{1}{2})-2}{5(\frac{1}{2})-1} \\
=\frac{\frac{7-4}{2}}{\frac{5-2}{2}} \\
=\frac{3}{3} \times \frac{2}{2} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
=1 \\
RHS = \frac{7(\frac{1}{2})+3}{\frac{5}{6}(\frac{1}{2})+4} \\
=\frac{\frac{7+6}{2}}{\frac{5+8}{2}} \\
=\frac{13}{13} \times \frac{2}{2} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
=1 \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q14. } \quad (\frac{x+1}{x+2})^{2}=\frac{x+2}{x+4} \\\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad (\frac{x+1}{x+2})^{2}=\frac{x+2}{x+4} \\\Rightarrow \quad (x+1)^{2}(x+4)=(x+2)^{2}(x+2) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad (x+1)^{2}(x+4) – (x+2)^{2}(x+2) =0 \\
\Rightarrow \quad (x^{2}+2 x+1)(x+4)-(x^{2}+4 x+4)(x+2)=0 \\
\Rightarrow \quad x^{3}+2 x^{2}+x+4 x^{2}+8 x+4-(x^{3}+4 x^{2}+4 x+2 x^{2}+8 x+8)=0 \\
\Rightarrow \quad x^{3}+2 x^{2}+x+4 x^{2}+8 x+4-x^{3}-4 x^{2}-4 x-2 x^{2}-8 x-8=0 \\
\Rightarrow \quad -3 x-4=0 \\
\Rightarrow \quad x=- \frac{4}{3} \\\text {On substituting } x=\frac{1}{2} \text { on both side of equation, we get} \\LHS =\frac{(\frac{-4}{3}+1)^{2}}{(\frac{-4}{3}+2)^{2}} \\
=\frac{(\frac{-4+3}{3})^{2}}{(\frac{-4+6}{3})^{2}} \\
=\frac{(-\frac{1}{3})^{2}}{(\frac{2}{3})^{2}} \\
=\frac{\frac{1}{9}}{\frac{4}{9}} \\
=\frac{1}{4} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\RHS =\frac{\frac{-4}{3}+2}{-\frac{4}{3}+4} \\
=\frac{\frac{-4+6}{3}}{\frac{-4+12}{3}} \\
=\frac{2}{8} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
=\frac{1}{4} \\
\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q15. } \quad (\frac{x+1}{x-4})^{2}=\frac{x+8}{x-2} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad (\frac{x+1}{x-4})^{2}=\frac{x+8}{x-2} \\\Rightarrow \quad (x+1)^{2}(x-2)=(x-4)^{2}(x+8) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad (x+1)^{2}(x-2) – (x-4)^{2}(x+8) = 0 \\
\Rightarrow \quad (x^{2}+2 x+1)(x-2)-(x^{2}-8 x+16)(x+8)=0 \\
\Rightarrow \quad x^{3}+2 x^{2}+x-2 x^{2}-4 x-2-(x^{3}-8 x^{2}+16 x+8 x^{2}-64 x+128)=0 \\
\Rightarrow \quad x^{3}+2 x^{2}+x-2 x^{2}-4 x-2-x^{3}+8 x^{2}-16 x-8 x^{2}+64 x-128=0 \\
\Rightarrow \quad 45 x-130=0 \\
\Rightarrow \quad x=\frac{130}{45} \\
\Rightarrow \quad x=\frac{26}{9} \\\text {On substituting } x=\frac{26}{9} \text { on both side of equation, we get} \\LHS =(\frac{\frac{26}{9}+1}{\frac{26}{9}-4})^{2}\\
=(\frac{\frac{26+9}{9}}{\frac{26-36}{9}})^{2}\\=(\frac{35}{-10})^{2} \\
=(\frac{7}{-2})^{2} \\
=\frac{49}{4} \\RHS =\frac{\frac{26}{9}+8}{\frac{26}{9}-2} \\
=\frac{\frac{26+72}{9}}{\frac{26-18}{9}} \\
=\frac{\frac{98}{9}}{\frac{8}{9}} \\
=\frac{49}{4} \quad \quad [\text {Using : }\frac{a}{b} \div \frac{c}{d}=\frac{a d}{b c}] \\
\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q16. } \quad \frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6} \\\Rightarrow \quad (9 x-7)(x+6)=(3 x-4)(3 x+5) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 9 x^{2}+54 x-7 x-42-(9 x^{2}+15 x-12 x-20)=0 \\
\Rightarrow \quad 44 x-22=0 \\
\Rightarrow \quad 44 x=22 \\
\Rightarrow \quad x=\frac {22}{44} \\
\Rightarrow \quad x=\frac {1}{2} \\\text {On substituting } x=\frac {1}{2} \text { on both side of equation, we get} \\LHS =\frac{9(\frac{1}{2})-7}{3(\frac{1}{2})+5} \\
=\frac{\frac{9-14}{2}}{\frac{3+10}{2}} \\
=\frac{-5}{13} \\RHS = \frac{3(\frac{1}{2})-4}{\frac{1}{2}+6} \\
=\frac{\frac{3-8}{2}}{\frac{1+12}{2}} \\
=\frac{-5}{13} \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q17. } \quad \frac {x+2}{x+5}= \frac {x}{x+6} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad {x+2}{x+5}= \frac {x}{x+6} \\\Rightarrow \quad (x+2)(x+6) =x(x+5) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad (x+2)(x+6)-x(x+5)=0 \\\Rightarrow \quad x^{2} + 8 x + 12 x^{2} – 5 x=0 \\
\Rightarrow \quad 3 x+12=0 \\
\Rightarrow \quad 3 x=-12 \\
\Rightarrow \quad x=-\frac{12}{3} \\
$\Rightarrow \quad x=-4 \\\text {On substituting } x=-4 \text { on both side of equation, we get} \\LHS = \frac {(-4+2)}{(-4+5)} \\
=\frac {-2}{1} = -2 \\RHS = \frac {-4}{(-4+6)} \\
=\frac {-4}{2} =-2 \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q18. } \quad \frac {2 x-(7-5 x)}{9 x-(3+4 x)} = \frac {7}{6} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {2 x-(7-5 x)}{9 x-(3+4 x)} = \frac {7}{6} \\\Rightarrow \quad \frac {(7 x-7)}{(5 x-3)}= \frac {7}{6} \\
\Rightarrow \quad 6(7 x-7)=7(5 x-3) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 42 x-42=35 x-21 \\
\Rightarrow \quad 42 x-35 x=-21+42 \\
\Rightarrow \quad 7 x=21 \\
\Rightarrow \quad x=\frac {21}{7} \\
\Rightarrow \quad x=3 \\\text {On substituting } x=3 \text { on both side of equation, we get} \\LHS = \frac {[7(3)-7]}{[5(3)-3]} \\
=\frac {(21-7)}{(15-3)} \\=\frac {14}{12} \\
=\frac {7}{6} = RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q19. } \quad \frac {(15(2-x)-5(x+6))}{(1-3 x)}=10 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {(15(2-x)-5(x+6))}{(1-3 x)}=10 \\\Rightarrow \quad 30-15 x -5x -30=10(1-3 x) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad -20x=10-30 x \\
\Rightarrow \quad -20x + 30 x =10 \\
\Rightarrow \quad 10 x=10 \\
\Rightarrow \quad x=\frac {10}{10} \\
\Rightarrow \quad x=1 \\
\text {On substituting } x=1 \text { on both side of equation, we get} \\LHS = \frac {[15(2-1)-5(1+6)]}{(1-3)} \\
= \frac {15-5(7)}{-2} \\
= \frac {15-35}{-2} \\
= \frac {-20}{-2} \\
= 10 = RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q20. } \quad \frac {(x+3)}{(x-3)} + \frac {(x+2)}{(x-2)} =2 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {(x+3)}{(x-3)} + \frac {(x+2)}{(x-2)} =2 \\\Rightarrow \quad \frac {((x+3)(x-2)+(x+2)(x-3))}{(x-3)(x-2)} = 2 \\
\Rightarrow \quad (x+3)(x-2)+(x+2)(x-3)=2(x-3)(x-2) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad x^{2}+3 x-2 x-6+x^{2}-3 x+2 x-6=2(x^{2}-3 x-2 x+6) \\
\Rightarrow \quad 2 x^{2}-12=2 x^{2}-10 x+12 \\
\Rightarrow \quad 2 x^{2}-2 x^{2}+10 x=12+12 \\
\Rightarrow \quad 10 x=24 \\
\Rightarrow \quad x= \frac {24}{10} \\
\Rightarrow \quad x=\frac {12}{5} \\
\text {On substituting } x=\frac {12}{5} \text { on both side of equation, we get} \\LHS = \frac {(x+3)}{(x-3)} + \frac {(x+2)}{(x-2)} \\=\frac {[(x+3)(x-2)+(x+2)(x-3)]}{(x-3)(x-2)} \\
=\frac {x^{2} + 3x – 2x – 6 + x^{2} – 3x + 2x – 6}{(x^{2}-2x -3x +6} \\
=\frac {2x^{2} – 12}{x^{2}-5x +6} \\
=\frac {2 (\frac {12}{5})^{2} – 12}{(\frac {12}{5})^{2}-5(\frac {12}{5}) +6} \\
=\frac {2 (\frac {144}{25}) – 12}{\frac {144}{25}-12 +6} \\
=\frac {\frac {2 \times 144 – 12 \times 25}{25}}{\frac {144 -6 \times 25}{25}} \\
=\frac {288 – 300}{144 -150} \\
=\frac {-12}{-6} \\
=2= RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q21. } \quad \frac {(x+2)(2 x-3)-2x^{2}+6}{x-5}=2 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {(x+2)(2 x-3)-2x^{2}+6}{x-5}=2 \\\Rightarrow \quad (x+2)(2 x-3)-2x^{2}+6=2(x-5) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad x^{2} -3x +4x -6 -2x^{2}+6=2x-10 \\
\Rightarrow \quad x -2x =-10 \\
\Rightarrow \quad -x =-10 \\
\Rightarrow \quad x= 10 \\\text {On substituting } x=10 \text { on both side of equation, we get} \\LHS = \frac {(x+2)(2 x-3)-2x^{2}+6}{x-5} \\
= \frac {x^{2} -3x +4x -6 -2x^{2}+6}{x-5} \\
= \frac {x}{x-5} \\
= \frac {10}{10-5} \\
=2 = RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q22. } \quad \frac {x^{2}-(x+1)(x+2)}{5 x+1}=6 \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {x^{2}-(x+1)(x+2)}{5 x+1}=6 \\\Rightarrow \quad x^{2}- x^{2}- 3x -2 =6(5 x+1) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad – 3x -2 =30 x+ 6 \\
\Rightarrow \quad -3x -30x = 6 + 2 \\
\Rightarrow \quad -33x =8\\
\Rightarrow \quad x =- \frac {8}{33} \\\text {On substituting } x =- \frac {8}{33} \text { on both side of equation, we get} \\LHS = \frac {x^{2}-(x+1)(x+2)}{5 x+1} \\
= \frac {x^{2}- x^{2}- 3x -2}{5 x+1} \\
= \frac {-3x -2}{5 x+1} \\= \frac {-3(\frac {-8}{33}) -2}{5(\frac {-8}{33})+1} \\= \frac {\frac {24-66}{33}}{\frac {-40+33}{33}} \\= \frac {\frac {-42}{33}}{\frac {-7}{33}} \\
= \frac {42}{7} \\
=6 = RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q23.} \quad \frac {(2 x+3)-(5 x-7)}{6 x+11} =- \frac {8}{3} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {(2 x+3)-(5 x-7)}{6 x+11} =- \frac {8}{3} \\\Rightarrow \quad 2x +3 -5x +7 =- \frac {8}{3} \times (6 x+11) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad -3x +10 =- 16x – \frac {88}{3} \\
\Rightarrow \quad -3x +16x =- 10 – \frac {88}{3} \\
\Rightarrow \quad 13x = \frac {-30-88}{3} \\
\Rightarrow \quad x = \frac {-118}{3 \times 13} \\
\Rightarrow \quad x = – \frac {118}{39} \\\text {On substituting } x = – \frac {118}{39} \text { on both side of equation, we get} \\LHS = \frac {(2 x+3)-(5 x-7)}{6 x+11} \\= \frac {-3x +10}{6 x+11} \\
= \frac {-3(\frac {-118}{39}) +10}{6(\frac {-118}{39})+11} \\
= \frac {\frac {118+130}{13}}{\frac {-2 \times 118 +143}{13}} \\
= \frac {248}{-93} \\
= -\frac {8}{3} \\ = RHS \\\text {As LHS = RHS, hence verified.} \\
\end{array}\begin{array}{l}
\text {Q24. } \quad \text {Find the positive value of x for which the given equation is satisfied: } \\
\text {(i) } \quad \frac {x^{2}-9}{5+x^{2}}=- \frac {5}{9} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {x^{2}-9}{5+x^{2}}=- \frac {5}{9} \\\Rightarrow \quad 9(x^{2}-9)=-5(5+x^{2}) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 9 x^{2}-81=-25-5 x^{2} \\
\Rightarrow \quad 9 x^{2}+5 x^{2}=-25+81 \\
\Rightarrow \quad 14 x^{2}=56 \\
\Rightarrow \quad x^{2}=56 / 14 \\
\Rightarrow \quad x^{2}=4 \\
\Rightarrow \quad x=\sqrt{4} \\
\Rightarrow \quad x= \pm 2 \\\text {Hence, 2 is the right value.} \\
\end{array}\begin{array}{l}
\text {(ii) } \quad \frac {y^{2}+4}{3 y^{2}+7} = \frac {1}{2} \\
\end{array}\begin{array}{l}
\text {Sol. Given equation is } \quad \frac {y^{2}+4}{3 y^{2}+7} = \frac {1}{2} \\\Rightarrow \quad 2(y^{2}+4)=1(3 y^{2}+7) \quad \quad [\text {By cross-multiplication}] \\
\Rightarrow \quad 2 y^{2}+8=3 y^{2}+7 \\
\Rightarrow \quad 3 y^{2}-2 y^{2}=8 -7 \\
\Rightarrow \quad y^{2}= 1 \\
\Rightarrow \quad y=\sqrt{1} \\
\Rightarrow \quad y= \pm 1 \\\text {Hence, 1 is the right value.} \\
\end{array}
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