\begin{array}{l}
\text {QSolve each of the following equations. Also, verify the result in each case. }
\end{array}\begin{array}{l}
\text {Q1. } \quad 6 x+5=2 x+17
\end{array}\begin{array}{l}
\text {Sol. } \\
6 x+5=2 x+17 \\
\text {Transposing 2x to LHS and 5 to RHS, we get } \\
6 x-2 x=17-5 \\
\Rightarrow \quad 4x=12 \\
\text {Dividing both sides by 4, we get } \\
\frac{4 x}{4}=\frac{12}{4} \\
\Rightarrow \quad x=3 \\\text {Verification : } \\\text {Putting } x=3 \text { in the given equation, we get } \\
6 \times 3+5=2 \times 3+17 \\
\Rightarrow \quad 18+5=6+17 \\
\Rightarrow \quad 23=23 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}\end{array}\begin{array}{l}
\text {Q2. } \quad 2(5 x-3)-3(2 x-1)=9
\end{array}\begin{array}{l}
\text {Sol. } \\
2(5 x-3)-3(2 x-1)=9 \\
\text { On expanding the brackets, we get } \\
2 \times 5 x-2 \times 3-3 \times 2 x+3 \times 1=9 \\
\Rightarrow \quad 10 x-6-6 x+3=9 \\
\Rightarrow \quad 4 x-3=9 \\
\text {Adding 3 to both sides, we get } \\
4 x-3+3=9+3 \\
\Rightarrow \quad 4 x=12 \\
\text {Dividing both sides by 4, we get } \\
\frac{4 x}{4}=\frac{12}{4} \\\Rightarrow \quad x=3 \\
\text {Verification : } \\
\text {Putting } x=3 \text { in LHS, we get } \\
LHS=2(5 \times 3-3)-3(2 \times 3-1) \\
=2 \times 12-3 \times 5 \\
=24-15 \\
=9 = RHS \\
\text {Hence, verified.}\end{array}\begin{array}{l}
\text {Q3. } \quad \frac{x}{2}=\frac{x}{3}+1
\end{array}\begin{array}{l}
\text {Sol. } \\\frac{x}{2}=\frac{x}{3}+1 \\
\text {Transposing } \frac{x}{3} \text { to LHS , we get } \\
\frac{x}{2}-\frac{x}{3}=1 \\
\Rightarrow \quad \frac{3 x-2 x}{6}=1 \\
\Rightarrow \quad \frac{x}{6}=1 \\
\text {Multiplying both sides by 6, we get } \\
\frac{x}{6} \times 6=1 \times 6 \\
\Rightarrow \quad x=6 \\
\text {Verification: } \\
\text {Putting } x=6 \text { in the given equation, we get } \\
\frac{6}{2}=\frac{6}{3}+1 \\
\Rightarrow \quad 3=2+1 \\
\Rightarrow \quad 3=3 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}\end{array}\begin{array}{l}
\text {Q4. } \quad \frac{x}{2}+\frac{3}{2}=\frac{2 x}{5}-1
\end{array}\begin{array}{l}
\text {Sol. } \\
\frac{x}{2}+\frac{3}{2}=\frac{2 x}{5}-1 \\
\text {Transposing } \frac{2 x}{5} \text { to LHS and } \frac{3}{2} \text { to RHS, we get } \\
\Rightarrow \quad \frac{x}{2}-\frac{2 x}{5}=-1-\frac{3}{2} \\
\Rightarrow \quad \frac{5 x-4 x}{10}=\frac{-2-3}{2} \\
\Rightarrow \quad \frac{x}{10}=\frac{-5}{2} \\
\text {Multiplying both sides by 10, we get } \\
\frac{x}{10} \times 10=\frac{-5}{2} \times 10 \\
\Rightarrow \quad x=-25 \\\text {Verification : } \\
\text {Putting } x=-25 \text { in the given equation, we get } \\
\frac{-25}{2}+\frac{3}{2}=\frac{2 \times(-25)}{5}-1 \\
\Rightarrow \quad \frac{-25+3}{2}=-10-1 \\
\Rightarrow \quad \frac{-22}{2}=-11 \\
\Rightarrow \quad -11=-11 \\\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q5. } \quad \frac{3}{4}(x-1)=x-3
\end{array}\begin{array}{l}
\text {Sol. } \\\frac{3}{4}(x-1)=x-3 \\
\text {On expanding the brackets on both sides, we get } \\
\frac{3}{4} x-\frac{3}{4}=x-3 \\\text {Transposing } \frac {3}{4}x \text { to RHS and 3 to LHS, we get } \\
3-\frac{3}{4}=x-\frac{3}{4} x \\
\Rightarrow \quad \frac{12-3}{4}=\frac{4 x-3 x}{4} \\
\Rightarrow \quad \frac{9}{4}=\frac{x}{4} \\
\text {Multiplying both sides by 4, we get } \\
\frac{9}{4} \times 4=\frac{x}{4} \times 4 \\
\Rightarrow \quad x=9 \\
\text {Verification : } \\
\text {Putting } x=9 \text { in the given equation, we get } \\
\frac{3}{4}(9-1)=9-3 \\
\Rightarrow \quad \frac{3}{4} \times 8=6 \\
\Rightarrow \quad 6=6 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q6. } \quad 3(x-3)=5(2 x+1)
\end{array}\begin{array}{l}
\text {Sol. } \\3(x-3)=5(2 x+1) \\
\text {On expanding the brackets on both sides, we get } \\
3 x-9=10 x+5 \\
\text {Transposing } 10x \text { to LHS and 9 to RHS, we get } \\
3 x-10 x=9+5 \\
\Rightarrow \quad -7 x=14 \\
\text {Dividing both sides by 7, we get } \\
\frac{-7 x}{7}=\frac{14}{7} \\
\Rightarrow \quad x=-2 \\
\text {Verification : } \\
\text {Putting } x=-2 \text { in the given equation, we get } \\
3(-2-3)=5\{2(-2)+1\} \\
\Rightarrow \quad 3(-5)=5(-3) \\
\Rightarrow \quad -15=-15 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q7. } \quad 3 x-2(2 x-5)=2(x+3)-8
\end{array}\begin{array}{l}
\text {Sol. } \\
3 x-2(2 x-5)=2(x+3)-8 \\\text {On expanding the brackets on both sides, we get } \\
3 x-4 x+10=2 x+6-8 \\
\Rightarrow \quad -x+10=2 x-2 \\
\text {Transposing } x \text { to RHS and 2 to LHS, we get } \\
10+2=2 x+x \\
\Rightarrow \quad 3 x=12 \\
\text {Dividing both sides by 3, we get } \\
\frac{3 x}{3}=\frac{12}{3} \\
\Rightarrow \quad x=4 \\
\text {Verification: } \\
\text {Putting } x=4 \text { in the given equation, we get } \\
3(4)-2\{2(4)-5\}=2(4+3)-8 \\
\Rightarrow \quad 12-2(8-5)=14-8 \\
\Rightarrow \quad 12-6=6 \\
\Rightarrow \quad 6=6 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}\end{array}\begin{array}{l}
\text {Q8. } \quad x-\frac{x}{4}-\frac{1}{2}=3+\frac{x}{4}
\end{array}\begin{array}{l}
\text {Sol. } \\x-\frac{x}{4}-\frac{1}{2}=3+\frac{x}{4} \\
\text {Transposing } \frac{x}{4} \text { to LHS and } -\frac{1}{2} \text { to RHS, we get } \\
x-\frac{x}{4}-\frac{x}{4}=3+\frac{1}{2} \\
\Rightarrow \quad \frac{4 x-x-x}{4}=\frac{6+1}{2} \\
\Rightarrow \quad \frac{2 x}{4}=\frac{7}{2} \\
\text {Multiplying both sides by 4, we get } \\
\frac{2 x}{4} \times 4=\frac{7}{2} \times 4 \\
\Rightarrow \quad 2 x=14 \\
\text {Dividing both sides by 2, we get } \\
\frac{2 x}{2}=\frac{14}{2} \\
\Rightarrow \quad x=7 \\ \\\text {Verification : } \\
\text {Putting } x=7 \text { in the given equation, we get } \\
7-\frac{7}{4}-\frac{1}{2}=3+\frac{7}{4} \\
\Rightarrow \quad \frac{28-7-2}{4}=\frac{12+7}{4} \\
\Rightarrow \quad \frac{19}{4}=\frac{19}{4} \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q9. } \quad \frac{6 x-2}{9}+\frac{3 x+5}{18}=\frac{1}{3}
\end{array}\begin{array}{l}
\text {Sol. } \\\frac{6 x-2}{9}+\frac{3 x+5}{18}=\frac{1}{3} \\
\Rightarrow \quad \frac{(6 x -2) \times 2 + (3 x+5)}{18}=\frac{1}{3} \\
\Rightarrow \quad \frac{12 x-4+3 x+5}{18}=\frac{1}{3} \\
\Rightarrow \quad \frac{15 x+1}{18}=\frac{1}{3} \\
\text {Multiplying both sides by 18, we get } \\
\frac{15 x+1}{18} \times(18)=\frac{1}{3} \times(18) \\
\Rightarrow \quad 15 x+1=6 \\
\text {Transposing 1 to RHS, we get } \\
15 x=6-1 \\
\Rightarrow \quad 15 x=5 \\
\text {Dividing both sides by 15, we get } \\
\frac{15 x}{15}=\frac{5}{15} \\\Rightarrow \quad x= \frac {1}{3} \\
\text {Verification : } \\
\text {Putting } x=\frac {1}{3} \text { in the given equation, we get } \\
\frac{6(\frac{1}{3})-2}{9}+\frac{3(\frac{1}{3})+5}{18}=\frac{1}{3} \\
\Rightarrow \quad \frac{2-2}{9}+\frac{1+5}{18}=\frac{1}{3} \\
\Rightarrow \quad 0+ \frac {6}{18}=\frac {1}{3} \\
\Rightarrow \quad \frac{1}{3}=\frac{1}{3} \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q10. } \quad m-\frac{m-1}{2}=1-\frac{m-2}{3}
\end{array}\begin{array}{l}
\text {Sol. } \\
m-\frac{m-1}{2}=1-\frac{m-2}{3} \\
\Rightarrow \quad \frac{2 m- (m-1)}{2}=\frac{3-(m-2)}{3} \\
\Rightarrow \quad \frac{2m -m+1}{2}=\frac{3-m+2}{3} \\
\Rightarrow \quad \frac{m+1}{2}=\frac{5-m}{3} \\
\Rightarrow \quad \frac{m}{2}+\frac{1}{2}=\frac{5}{3}-\frac{m}{3} \\
\text {Transposing } \frac {m}{3} \text { to LHS and } \frac {1}{2} \text { to RHS, we get } \\
\frac{m}{2}+\frac{m}{3}=\frac{5}{3}-\frac{1}{2} \\
\Rightarrow \quad \frac{3 m+2 m}{6}=\frac{10-3}{6} \\
\Rightarrow \quad \frac{5 m}{6}=\frac{7}{6} \\
\text {Multiplying both sides by 6, we get } \\
\frac{5 m}{6} \times 6=\frac{7}{6} \times 6 \\
\Rightarrow \quad 5m=7 \\
\text {Dividing both sides by 5, we get } \\
\frac{5 m}{5}=\frac{7}{5} \\
\Rightarrow \quad \mathrm{m}=\frac{7}{5} \\
\text {Verification: } \\
\text {Putting } m=\frac{7}{5} \text { in the given equation, we get } \\
\frac{7}{5}-\frac{\frac{7}{5}-1}{2}=1-\frac{\frac{7}{5}-2}{3} \\
\Rightarrow \quad \frac{7}{5}-\frac{\frac{7-5}{5}}{2}=1-\frac{\frac{7-10}{5}}{3} \\
\Rightarrow \quad \frac{7}{5}-\frac{2}{5 \times 2}=1-\frac{-3}{5 \times 3} \\
\Rightarrow \quad \frac{7}{5}-\frac{1}{5}=1+\frac{1}{5} \\
\Rightarrow \quad \frac{6}{5}=\frac{6}{5} \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q11. } \quad \frac{(5 x-1)}{3}-\frac{(2 x-2)}{3}=1
\end{array}\begin{array}{l}
\text {Sol. } \\
\frac{(5 x-1)}{3}-\frac{(2 x-2)}{3}=1 \\
\Rightarrow \quad \frac{(5 x-1)-(2 x-2)}{3}=1 \\
\Rightarrow \quad \frac{5 x-1-2 x+2}{3}=1 \\
\Rightarrow \quad \frac{3 x+1}{3}=1 \\
\text {Multiplying both sides by 3, we get } \\
(\frac{3 x+1}{3}) \times 3=1 \times 3 \\
\Rightarrow \quad 3 x+1=3 \\
\text {Subtracting 1 from both sides, we get } \\
3 x+1-1=3-1 \\
\Rightarrow \quad 3 x=2 \\
\text {Dividing both sides by 3, we get } \\
\frac{3 x}{3}=\frac{2}{3} \\
\Rightarrow \quad x=\frac{2}{3} \\
\text {Verification: } \\
\text {Putting } x=\frac{2}{3} \text { in LHS, we get} \\
LHS=\frac{5(\frac{2}{3})-1}{3}-\frac{2(\frac{2}{3})-2}{3} \\
=\frac{\frac{10}{3}-1}{3}-\frac{\frac{4}{3}-2}{3} \\
=\frac{\frac{10-3}{3}}{3}-\frac{\frac{4-6}{3}}{3} \\
=\frac{7}{3 \times 3}-(\frac{-2}{3 \times 3}) \\
=\frac{7}{9}+\frac{2}{9} \\
=\frac{9}{9} \\
=1=RHS \\\text {Hence, verified.}
\end{array}\begin{array}{l}
\text {Q12. } \quad 0.6 x+\frac{4}{5}=0.28 x+1.16
\end{array}\begin{array}{l}
\text {Sol. } \\
0.6 x + \frac{4}{5} = 0.28 x + 1.16 \\
\text {Transposing } 0.28x \text { to LHS and } \frac {4}{5} \text { to RHS, we get } \\
0.6 x-0.28 x=1.16-\frac{4}{5} \\
\Rightarrow \quad 0.32 x=1.16-0.8 \\
\Rightarrow \quad 0.32 x=0.36 \\
\text {Dividing both sides by 0.32, we get } \\
\frac{0.32 x}{0.32}=\frac{0.36}{0.32} \\
\Rightarrow \quad x=\frac{9}{8} \\
\text {Verification: } \\
\text {Putting } x=\frac{9}{8} \text { in the given equation, we get } \\
0.6(\frac{9}{8})+\frac{4}{5}=0.28(\frac{9}{8})+1.16 \\
\Rightarrow \quad \frac{5.4}{8}+\frac{4}{5}=\frac{2.52}{8}+1.16 \\
\Rightarrow \quad 0.675+0.8=0.315+1.16 \\
\Rightarrow \quad 1.475=1.475 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}\end{array}\begin{array}{l}
\text {Q13. } \quad 0.5 x+\frac{x}{3}=0.25 x+7
\end{array}\begin{array}{l}
\text {Sol. } \\
0.5 x+\frac{x}{3}=0.25 x+7 \\
\frac{05}{10} x+\frac{x}{3}=\frac{25 x}{100}+7 \\
\frac{x}{2}+\frac{x}{3}=\frac{x}{4}+7 \\
\text {Transposing } \frac {x}{4} \text { to LHS, we get } \\
\frac{x}{2}+\frac{x}{3}-\frac{x}{4}=7 \\
\Rightarrow \quad \frac{6 x+4 x-3 x}{12}=7 \\
\Rightarrow \quad \frac{7 x}{12}=7 \\
\text {Multiplying both sides by 12, we get } \\
\frac{7 x}{12} \times 12=7 \times 12 \\
\Rightarrow \quad 7 x=84 \\
\text {Dividing both sides by 7, we get } \\
\frac{7 x}{7}=\frac{84}{7} \\
\Rightarrow \quad x=12 \\
\text {Verification: } \\
\text {Putting } x=12 \text { in the given equation, we get } \\
0.5(12)+\frac{12}{3}=0.25(12)+7 \\
\Rightarrow \quad 6+4=3+7 \\
\Rightarrow \quad 10=10 \\
\Rightarrow \quad LHS = RHS \\
\text {Hence, verified.}
\end{array}
InfiniteGrades
Learners today, leaders tomorrow