Class 7 Linear Equations In One Variable Exercise 8.2

\begin{array}{l} \text {Solve each of the following equations and check your answers: } \end{array}\begin{array}{l} \text {Q1. } \quad x-3=5 \end{array}\begin{array}{l} \text {Sol. } \\ x-3=5 \text {Adding 3 to both sides, we get } \\ x-3+3=5+3 \\ \Rightarrow \quad x=8 \\ \text {Verification: } \\ \text {Putting x=8 on LHS, we get } \\ LHS =x-3 = 8-3 =5 \\ RHS = 5 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q2. } \quad x+9=13 \end{array}\begin{array}{l} \text {Sol. } \\ x+9=13 \text {Subtracting 9 from both sides, we get } \\ x+9-9=13-9 \\ \Rightarrow \quad x=4 \\ \text {Verification: } \\ \text {Putting x=4 on LHS, we get } \\ LHS =x+9 = 4+9 =13 \\ RHS = 13 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q3. } \quad x-\frac{3}{5}=\frac{7}{5} \end{array}\begin{array}{l} \text {Sol. } \\x-\frac{3}{5}=\frac{7}{5} \\ \text {Adding } \frac{3}{5} \text { to both sides, we get } \\ x-\frac{3}{5}+\frac{3}{5}=\frac{7}{5}+\frac{3}{5} \\ \Rightarrow \quad x=\frac{7+3}{5} \\ \Rightarrow \quad x=\frac{10}{5} =2 \\\text {Verification: } \\ \text {Putting } x=2 \text { on LHS, we get } \\ LHS =x-\frac{3}{5} = 2-\frac{3}{5} =\frac{7}{5} \\ RHS = \frac{7}{5} \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q4. } \quad 3 x=0 \end{array}\begin{array}{l} \text {Sol. } \\ 3 x=0 \\ \text {Dividing both sides by 3, we get } \\ \frac{3 x}{3}=\frac{0}{3} \\ \Rightarrow \quad x=0 \\\text {Verification: } \\ \text {Putting } x=0 \text { on LHS, we get } \\ LHS =3x = 3 \times 0 =0 \\ RHS = 0 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q5. } \quad \frac{x}{2}=0 \end{array}\begin{array}{l} \text {Sol. } \\ \frac{x}{2}=0 \\ \text {Multiplying both sides by 2, we get } \\ \frac{x}{2} \times 2=0 \times 2 \\ \Rightarrow \quad x=0 \\\text {Verification: } \\ \text {Putting } x=0 \text { on LHS, we get } \\ LHS = \frac{x}{2} = \frac{0}{2} =0 \\ RHS = 0 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q6. } \quad x-\frac{1}{3}=\frac{2}{3} \end{array}\begin{array}{l} \text {Sol. } \\ x-\frac{1}{3}=\frac{2}{3} \\ \text {Adding } \frac{1}{3} \text { to both sides , we get } \\ x-\frac{1}{3} + \frac{1}{3}=\frac{2}{3} + \frac{1}{3} \\ \Rightarrow \quad x=\frac {2+1}{3} =1 \\\text {Verification: } \\ \text {Putting } x=1 \text { on LHS, we get } \\ LHS = x-\frac{1}{3} = 1-\frac{1}{3} =\frac{2}{3} \\ RHS = \frac{2}{3} \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q7. } \quad x+\frac{1}{2}=\frac{7}{2} \end{array}\begin{array}{l} \text {Sol. } \\ x+\frac{1}{2}=\frac{7}{2} \\ \text {Subtracting } \frac{1}{2} \text { from both sides , we get } \\ x+\frac{1}{2} – \frac{1}{2}=\frac{7}{2} – \frac{1}{2} \\ \Rightarrow \quad x=\frac {7-1}{2} =3 \\\text {Verification: } \\ \text {Putting } x=3 \text { on LHS, we get } \\ LHS = x+\frac{1}{2} = 3+\frac{1}{2} =\frac{7}{2} \\ RHS = \frac{7}{2} \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q8. } \quad 10-y=6 \end{array}\begin{array}{l} \text {Sol. } \\ 10-y=6 \\ \text {Subtracting } 10 \text { from both sides , we get } \\ 10-y -10=6 -10 \\ \Rightarrow \quad y=4 \\\text {Verification: } \\ \text {Putting } y=4 \text { on LHS, we get } \\ LHS = 10-y = 10-4 =6 \\ RHS = 6\\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q9. } \quad 7+4y=-5 \end{array}\begin{array}{l} \text {Sol. } \\ 7+4y=-5\\ \text {Subtracting } 7 \text { from both sides , we get } \\ 7+4y-7=-5-7 \\ \Rightarrow \quad 4y=-12 \\ \Rightarrow \quad y=-3 \\\text {Verification: } \\ \text {Putting } y=-3 \text { on LHS, we get } \\ LHS = 7+4y = 7+4 \times (-3) =7-12=-5 \\ RHS = -5\\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q10. } \quad \frac{4}{5}-x=\frac{3}{5} \end{array}\begin{array}{l} \text {Sol. } \\ \frac{4}{5}-x=\frac{3}{5} \\ \text {Subtracting } \frac{4}{5} \text { from both sides , we get } \\ \frac{4}{5}-x – \frac{4}{5}=\frac{3}{5}- \frac{4}{5} \\ \Rightarrow \quad -x=\frac{3-4}{5}\\ \Rightarrow \quad x=\frac{1}{5} \\\text {Verification: } \\ \text {Putting } x=\frac{1}{5} \text { on LHS, we get } \\ LHS = \frac{4}{5}-x = \frac{4}{5}-\frac{1}{5} =\frac{4-1}{5}=\frac{3}{5} \\ RHS = \frac{3}{5}\\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q11. } \quad 2y-\frac{1}{2}=-\frac{1}{3} \end{array}\begin{array}{l} \text {Sol. } \\ 2y-\frac{1}{2}=-\frac{1}{3} \\ \text {Adding } \frac{1}{2} \text { on both sides , we get } \\ 2y-\frac{1}{2} + \frac{1}{2}=-\frac{1}{3} + \frac{1}{2} \\ \Rightarrow \quad 2y=\frac{-2+3}{6}\\ \Rightarrow \quad 2y=\frac{1}{6} \\ \Rightarrow \quad y=\frac{1}{12} \\\text {Verification: } \\ \text {Putting } y=\frac{1}{12} \text { on LHS, we get } \\ LHS = 2y-\frac{1}{2} = 2 \times \frac{1}{12} – \frac{1}{2} \\ =\frac{1}{6} – \frac{1}{2} \\ =\frac{1-3}{6} \\ =\frac{-2}{6} \\ =\frac{-1}{3} \\ RHS = \frac{-1}{3} \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q12. } \quad 14=\frac{7 x}{10}-8 \end{array}\begin{array}{l} \text {Sol. } \\ 14=\frac{7 x}{10}-8 \\ \text {Adding } 8 \text { on both sides , we get } \\ 14 + 8 =\frac{7 x}{10}-8 +8\\ \Rightarrow \quad 22=\frac{7 x}{10} \\ \Rightarrow \quad x=\frac{22 \times 10}{7} \\ \Rightarrow \quad x=\frac{220}{7} \\\text {Verification: } \\ \text {Putting } x=\frac{220}{7} \text { on RHS, we get } \\ RHS = \frac{7 x}{10}-8 = \frac{7 \times \frac{220}{7}}{10}-8 \\ = \frac{220}{10}-8 \\ = 22-8 =14 \\RHS = 14 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q13. } \quad 3(x+2)=15 \end{array}\begin{array}{l} \text {Sol. } \\ 3(x+2)=15 \\ \text {Dividing both sides by 3 , we get } \\ \frac {3(x+2)}{3}=\frac {15}{3} \\ \Rightarrow \quad x+2=5 \\ \Rightarrow \quad x=3 \\\text {Verification: } \\ \text {Putting } x=3 \text { on LHS, we get } \\ LHS = 3(x+2) = 3(3+2) = 15\\RHS = 15 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q14. } \quad \frac{x}{4}=\frac{7}{8} \end{array}\begin{array}{l} \text {Sol. } \\ \frac{x}{4}=\frac{7}{8} \\ \text {Multiplying both sides by 4 , we get } \\ \frac{x}{4} \times 4=\frac{7}{8} \times 4\\ \Rightarrow \quad x =\frac{7}{2} \\\text {Verification: } \\ \text {Putting } x=\frac{7}{2} \text { on LHS, we get } \\ LHS = \frac{x}{4} = \frac{\frac{7}{2}}{4} = \frac{7}{8} \\RHS = \frac{7}{8} \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q15. } \quad \frac{1}{3}-2 x=0 \end{array}\begin{array}{l} \text {Sol. } \\ \frac{1}{3}-2 x=0 \\ \text {Subtracting } \frac{1}{3} \text { from both sides , we get } \\ \frac{1}{3}-2x- \frac{1}{3}=-\frac{1}{3} \\ \Rightarrow \quad -2x=-\frac{1}{3} \\ \text {Dividing both sides by -2 , we get } \\ \Rightarrow \quad x=\frac{1}{6} \\\text {Verification: } \\ \text {Putting } x=\frac{1}{6} \text { on LHS, we get } \\ LHS =\frac{1}{3}-2 x = \frac{1}{3}-2 \times \frac{1}{6} = 0 \\ RHS = 0 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q16. } \quad 3(x+6)=24 \end{array}\begin{array}{l} \text {Sol. } \\ 3(x+6)=24 \\ \text {Dividing both sides by 3 , we get } \\ \frac {3(x+6)}{3}=\frac {24}{3} \\ \Rightarrow \quad x+6=8 \\ \Rightarrow \quad x=2 \\\text {Verification: } \\ \text {Putting } x=2 \text { on LHS, we get } \\ LHS = 3(x+2) = 3(2+6) = 24\\RHS = 24 \\ LHS = RHS \\ \text {Hence, verified.}\end{array}\begin{array}{l} \text {Q17. } \quad 3(x+2)-2(x-1)=7 \end{array}\begin{array}{l} \text {Sol. } \\ 3(x+2)-2(x-1)=7 \\ \text {On expanding the brackets, we get } \\ (3 \times x)+(3 \times 2)-(2 \times x)+(2 \times 1)=7 \\ \Rightarrow \quad 3 x+6-2 x+2=7 \\ \Rightarrow \quad 3 x-2 x+6+2=7 \\ \Rightarrow \quad x+8=7 \\ \text {Subtracting 8 from both sides, we get } \\ x+8-8=7-8 \\ \Rightarrow \quad x = -1 \\ \text {Verification: } \\ \text {Putting } x = -1 \text { on LHS, we get } \\ LHS = 3(x+2)-2(x-1) = 3(-1+2)-2(-1-1) = 3 -2(-2) = 3+4 =7\\RHS = 7 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}\begin{array}{l} \text {Q18. } \quad 8(2x-5)-6(3x-7)=1 \end{array}\begin{array}{l} \text {Sol. } \\ 8(2 x-5)-6(3 x-7)=1 \\ \text {On expanding the brackets, we get } \\ 8 \times 2x – 8 \times 5 -6 \times 3x + 6 \times 7=1 \\ \Rightarrow \quad 16x – 40 -18x + 42=1 \\ \Rightarrow \quad -2x + 2=1 \\\text {Subtracting 2 from both sides, we get } \\ -2x + 2-2=1 -2\ \Rightarrow \quad -2x = -1 \\ \Rightarrow \quad x = \frac {1}{2} \\\text {Verification: } \\ \text {Putting } x= \frac {1}{2} \text { on LHS, we get } \\ LHS = 8(2x-5)-6(3x-7) = 8(2 \times \frac {1}{2} -5)-6(3 \times \frac {1}{2}-7) \\ = 8(1 -5)-6(\frac {3}{2}-7) \\ = 8(-4)-6(\frac {3-14}{2}) \\ = -32 – 3 \times (-11) \\ = -32 + 33 = 1 \\RHS = 1 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}\begin{array}{l} \text {Q19. } \quad 6(1-4 x)+7(2+5 x)=53 \end{array}\begin{array}{l} \text {Sol. } \\ 6(1-4 x)+7(2+5 x)=53 \\ \text {On expanding the brackets, we get } \\ (6 \times 1)-(6 \times 4 x)+(7 \times 2)+(7 \times 5 x)=53\\ \Rightarrow \quad 6-24 x+14+35 x=53 \\ \Rightarrow \quad 6+14+35 x-24 x=53 \\ \Rightarrow \quad 20+11 x=53 \\ \text {Subtracting 20 from both sides, we get } \\ 20+11 x-20=53-20 \\ \Rightarrow \quad 11 x=33 \\ \text {Dividing both sides by 11 , we get } \\ \frac{11 x}{11}=\frac{33}{11} \\ \Rightarrow \quad x=3 \\\text {Verification: } \\ \text {Putting } x= 3 \text { on LHS, we get } \\ LHS = 6(1-4 x)+7(2+5 x) = 6(1-4 \times 3)+7(2+5 \times 3) \\ =6(1-12)+7(2+15) \\ =6(-11)+7(17) \\ =-66+119 \\ =53 \\RHS = 53 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}\begin{array}{l} \text {Q20. } \quad 5(2-3 x)-17(2 x-5)=16 \end{array}\begin{array}{l} \text {Sol. } \\ 5(2-3 x)-17(2 x-5)=16 \\ \text {On expanding the brackets, we get } \\ (5 \times 2)-(5 \times 3 x)-(17 \times 2 x)+(17 \times 5)=16 \\ \Rightarrow \quad 10-15 x-34 x+85=16 \\ \Rightarrow \quad 10+85-34 x-15 x=16 \\ \Rightarrow \quad 95-49 x=16 \\ \text {Subtracting 95 from both sides, we get } \\ -49 x+95-95=16-95 \\ \Rightarrow \quad -49 x=-79 \\ \text {Dividing both sides by -49, we get } \\ \frac{-49 x}{-49}=\frac{-79}{-49} \\ \Rightarrow \quad x=\frac{79}{49} \\\text {Verification: } \\ \text {Putting } x= \frac{79}{49} \text { on LHS, we get } \\ LHS = 5(2-3 x)-17(2 x-5) = =5(2-3 \times \frac{79}{49})-17(2 \times \frac{79}{49}-5) \\ =(5 \times 2)-(5 \times 3 \times \frac{79}{49})-(17 \times 2 \times \frac{79}{49})+(17 \times 5) \\ =10-\frac{1185}{49}-\frac{2686}{49}+85 \\ =\frac{490-1185-2686+4165}{49} \\ =16 \\RHS = 16 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}\begin{array}{l} \text {Q21. } \quad \frac{x-3}{5}-2=-1 \end{array}\begin{array}{l} \text {Sol. } \\ \frac{x-3}{5}-2=-1 \\ \text {Adding 2 to both sides, we get } \\ \frac{x-3}{5}-2+2=-1+2 \\ \Rightarrow \quad \frac{x-3}{5}=1 \\ \text {Multiplying both sides by 5, we get } \\ (\frac{x-3}{5}) \times 5=1 \times 5 \\ \Rightarrow \quad x-3=5 \\ \text {Adding 3 to both sides, we get } \\ x-3+3=5+3 \Rightarrow \quad x=8 \\\text {Verification: } \\ \text {Putting } x= 8 \text { on LHS, we get } \\ LHS = \frac{x-3}{5}-2=-1 \\ =\frac{8-3}{5}-2 =\frac{5}{5}-2 = 1-2 = -1 \\RHS = -1 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}\begin{array}{l} \text {Q22. } \quad 5(x-2)+3(x+1)=25 \end{array}\begin{array}{l} \text {Sol. } \\ 5(x-2)+3(x+1)=25 \\ \text {On expanding the brackets, we get } \\ (5 \times x)-(5 \times 2)+3 \times x+3 \times 1=25 \\ \Rightarrow \quad 5 x-10+3 x+3=25 \\ \Rightarrow \quad 5 x+3 x-10+3=25 \\ \Rightarrow \quad 8 x-7=25 \\ \text {Adding 7 to both sides, we get } \\ 8 x-7+7=25+7 \\ \Rightarrow \quad 8 x=32 \\ \text {Dividing both sides by 8, we get } \\ \frac{8 x}{8}=\frac{32}{8} \\ \Rightarrow \quad x=4 \\\text {Verification: } \\ \text {Putting } x= 4 \text { on LHS, we get } \\ LHS = 5(x-2)+3(x+1) \\ =5(4-2)+3(4+1) \\ =5(2)+3(5) \\ =10+15=25 \\RHS = 25 \\ LHS = RHS \\ \text {Hence, verified.} \end{array}