Answer:
1. \[\frac{x}{2}-\frac{1}{5}\,=\frac{x}{3}+\frac{1}{4}\] We have \[\frac{x}{2}-\frac{1}{5}\,=\frac{x}{3}+\frac{1}{4}\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[\frac{x}{2}-\frac{x}{3}\,=\frac{1}{4}+\frac{1}{5}\] I Transposing \[\frac{x}{3}\] to LHS and \[-\frac{1}{5}\] to RHS \[\Rightarrow \] \[\frac{3x-2x}{6}=\frac{5+4}{20}\] \[\Rightarrow \] \[\frac{x}{6}=\frac{9}{20}\] \[\Rightarrow \] \[x=\frac{9}{20}\times 6\] I Multiplying both sides by 6 \[\Rightarrow \] \[x=\frac{27}{10}\] This is the required solution. Check: \[\text{LHS}=\frac{1}{2}\times \frac{27}{10}-\frac{1}{5}\] \[=\frac{27}{20}-\frac{1}{5}\] \[=\frac{27-4}{20}\] \[=\frac{23}{20}\] RHS \[=\frac{1}{3}\times \,\frac{27}{10}+\frac{1}{4}\] \[=\frac{9}{10}+\frac{1}{4}\] \[=\frac{18+5}{20}\] \[=\frac{23}{20}\] Therefore, LHS = RHS |as desired 2. \[\frac{n}{2}-\frac{3n}{4}\,+\frac{5n}{6}=21\] We have \[\frac{n}{2}-\frac{3n}{4}\,+\frac{5n}{6}=21\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[\frac{6n-9n+10n}{12}=21\] |LCM (2, 4, 6) = 12 \[\Rightarrow \] \[\frac{7n}{12}=21\] \[\Rightarrow \] \[n=21\times \frac{12}{7}=36\] | Multiplying both sides by This is the required solution. Check: \[\text{LHS}=\frac{1}{2}\times 36-\frac{3}{4}\] \[\times 36+\frac{5}{6}\,\times 36\] \[=18-27+30\] = 21 = RHS I as desired 3. \[x+7-\frac{8x}{3}\,=\frac{17}{6}-\frac{5x}{2}\] We have \[x+7-\frac{8x}{3}\,=\frac{17}{6}-\frac{5x}{2}\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[x-\frac{8x}{3}\,+\frac{5x}{2}=\frac{17}{6}-7\] |Transposing \[\frac{-5x}{2}\] to LHS and 7 to RHS \[\Rightarrow \] \[\frac{6x-16x+15x}{6}\,=\frac{17-42}{6}\] \[\Rightarrow \] \[\frac{5x}{6}=\frac{-25}{6}\] \[\Rightarrow \] \[x=\frac{-25}{6}\times \frac{6}{5}\] | Multiplying both sides by \[\frac{6}{5}\] \[\Rightarrow \] \[x=-5\] This is the required solution. Check: \[\text{LHS}=-5+7-\frac{8}{3}\,(-5)\] \[=-5+7+\frac{40}{3}\] \[=2+\frac{40}{3}\,=\frac{6+40}{3}\] \[=\frac{46}{3}\] \[\text{RHS}=\frac{17}{6}\,-\frac{5}{2}\,(-5)\] \[=\frac{17}{6}+\frac{25}{2}\] \[=\frac{17+75}{6}\] \[=\frac{92}{6}=\frac{92\div 2}{6\div 2}\] \[=\frac{46}{3}\] Therefore, LHS = RHS | as desired 4. \[\frac{x-5}{3}=\frac{x-3}{5}\] We have \[\frac{x-5}{3}=\frac{x-3}{5}\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[\frac{x}{3}\,-\frac{5}{3}\,=\frac{x}{5}-\frac{x}{5}\] \[\Rightarrow \] \[\frac{x}{3}-\frac{x}{5}\,=\frac{5}{3}\,-\frac{3}{5}\] | Transposing \[\frac{x}{5}\] to LHS and \[\frac{-5}{3}\] to RHS \[\Rightarrow \] \[\frac{5x-3x}{15}\,=\frac{25-9}{15}\] \[\Rightarrow \] \[\frac{2x}{15}=\frac{16}{15}\] \[\Rightarrow \] \[x=\frac{16}{15}\times \frac{15}{2}=8\] |Multiplying both sides by \[\frac{15}{2}\] This is the required solutions. Check: LHS \[=\frac{8-5}{3}\,=\frac{3}{3}=1\] RHS \[=\frac{8-3}{5}=\frac{5}{5}=1\] Therefore, LHS = RHS |as desired 5. \[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\] We have \[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[\frac{3}{4}t-\frac{2}{4}\,-\frac{2}{3}t-\frac{3}{3}=\frac{2}{3}-t\] \[\Rightarrow \] \[\frac{3}{4}t-\frac{1}{2}-\frac{2}{3}\,t-1=\frac{2}{3}-t\] \[\Rightarrow \] \[\frac{3}{4}\,t-\frac{2}{3}t+t=\frac{2}{3}+\frac{1}{2}+1\] | Transposing \[-t\] to LHS and \[-\frac{1}{2}\] and \[-1\] RHS \[\Rightarrow \] \[\frac{9t-8t+12t}{12}=\frac{4+3+6}{6}\] \[\Rightarrow \] \[\frac{13t}{12}=\frac{13}{6}\] \[\Rightarrow \] \[t=\frac{13}{6}\times \frac{12}{13}\,=2\] | Multiplying both sides by \[\frac{12}{13}\] This is the required solution. Check: LHS \[=\frac{3}{4}\times 2-\frac{2}{4}\,-\frac{2}{3}\,\times 2-\frac{3}{3}\] \[=\frac{3}{2}-\frac{1}{2}-\frac{4}{3}-1\] \[=\frac{9-3-8-6}{6}\] \[=-\frac{8}{6}\,=-\frac{8\div 2}{6\div 2}\] \[=-\frac{4}{3}\] RHS \[=\frac{2}{3}-2\] \[=\frac{2-6}{3}=-\frac{4}{3}\] Therefore, LHS = RHS |as desired 6. \[m-\frac{m-1}{2}=1-\frac{m-2}{3}\] We have \[m-\frac{m-1}{2}=1-\frac{m-2}{3}\] It is a linear equation since it involves linear expressions only. \[\Rightarrow \] \[m-\frac{m}{2}\,+\frac{1}{2}=1-\frac{m}{3}\,+\frac{2}{3}\] \[\Rightarrow \] \[m-\frac{m}{2}+\frac{m}{3}\,=1+\frac{2}{3}\,-\frac{1}{2}\] |Transposing \[-\frac{m}{3}\] to LHS and \[\frac{1}{2}\] to RHS \[\Rightarrow \] \[\frac{6m-3m+2m}{6}=\frac{6+4-3}{6}\] \[\Rightarrow \] \[\frac{5m}{6}=\frac{7}{6}\] \[\Rightarrow \] \[m=\frac{7}{6}\times \frac{6}{5}=\frac{7}{5}\] | Multiplying both sides by \[\frac{6}{5}\] This is the required solution. Check: \[=\frac{7}{5}-\frac{1}{2}\times \frac{7}{5}+\frac{1}{2}\] \[=\frac{7}{5}-\frac{7}{10}+\frac{1}{2}\] \[=\frac{14-7+5}{10}=\frac{12}{10}\,=\frac{6}{5}\] RHS \[=1-\frac{1}{3}\times \,\frac{7}{5}\times \frac{2}{3}\] \[=1-\frac{7}{15}\,+\frac{2}{3}\] \[=\frac{15-7+10}{15}\] \[=\frac{18}{15}=\frac{6}{5}\] Therefore, LHS = RHS |as desired
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