Class 7 Linear Equations In One Variable Exercise 8.1

\begin{array}{l} \text {Q1. Verify by substitution that:} \\ \text {(i) } \quad x=4 \text { is the root of } 3 x-5=7 \\ \text {(ii) } \quad x=3 \text { is the root of } 5+3 x=14 \\ \text {(iii) } \quad x=2 \text { is the root of } 3 x-2=8 x-12 \\ \text {(iv) } \quad x=4 \text { is the root of } \frac{3 x}{2}=6 \\ \text {(v) } \quad y=2 \text { is the root of } y-3=2 y-5 \\ \text {(vi) } \quad x=8 \text { is the root of } \frac{1}{2} x+7=11 \end{array}\begin{array}{l} \text {Sol. } \\\text {(i) Given that } x=4 \text { is the root of } 3 x-5=7 \\\text {Putting } x=4 \text { on LHS, we have } \\ \text {LHS} =3 \times 4 -5=7 = \text {RHS} \\ \text {Hence, x=4 is root of } 3 x-5=7 \\ \\\text {(ii) Given that } x=3 \text { is the root of } 5+3x=14 \\ \text {Putting } x=3 \text { on LHS, we have } \\ LHS =5 + 3 \times 3 =15 = RHS \\ \text {Hence, x=3 is root of } 5+3x=14 \\ \\\text {(iii) Given that } x=2 \text { is the root of } 3 x-2=8 x-12 \\ \text {Putting } x=2 \text { on LHS, we have } \\ LHS = 3 \times 2 -2 =6-2 =4 \\ RHS = 8 \times 2 -12 =16-12 =4 \\ \Rightarrow \quad LHS = RHS \\ \text {Hence, x=2 is root of } 3 x-2=8 x-12 \\ \\\text {(iv) Given that } x=4 \text { is the root of } \frac{3 x}{2}=6 \\\text {Putting } x=4 \text { on LHS, we have } \\ LHS = \frac{3 \times 4}{2}=6 = RHS \\ \text {Hence, x=4 is root of } \frac{3 x}{2}=6 \\ \\\text {(v) Given that } y=2 \text { is the root of } y-3=2 y-5 \\ \text {Putting } y=2 \text { on LHS, we have } \\ LHS = 2 -3 =-1 \\ RHS = 2 \times 2 -5 =-1 \\ \Rightarrow \quad LHS = RHS \\ \text {Hence, y=2 is root of } y-3=2 y-5 \\ \\\text {(vi) Given that } x=8 \text { is the root of } \frac{1}{2} x+7=11 \\ \text {Putting } x=8 \text { on LHS, we have } \\ LHS = \frac{1}{2} \times 8 +7 =4+7 =11 = RHS \\\text {Hence, x=8 is root of } \frac{1}{2} x+7=11 \\ \\\end{array}\begin{array}{l} \text {Q2. Solve each of the following equations by trial-and-error method: } \\ \text {(i) } \quad x+3=12 \quad \text {(ii) } \quad x-7=10 \quad \text {(iii) } \quad 4 x=28 \quad \text {(iv) } \quad \frac{x}{2}+7=11 \\ \text {(v) } \quad 2 x+4=3 x \quad \text {(vi) } \quad \frac{x}{4}=12 \quad \text {(vii) } \quad \frac{15}{x}=3 \quad \text {(viii) } \quad \frac{x}{18}=20 \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) We have LHS } = x+3 \text { and RHS }= 12 \\ \end{array} \begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 1 & 1+3=4 & 12 & No \\ \hline 2 & 2+3=5 & 12 & No \\ \hline 3 & 3+3=6 & 12 & No \\ \hline 4 & 4+3=7 & 12 & No \\ \hline 5 & 5+3=8 & 12 & No \\ \hline 6 & 6+3=9 & 12 & No \\ \hline 7 & 7+3=10 & 12 & No \\ \hline 8 & 8+3=11 & 12 & No \\ \hline 9 & 9+3=12 & 12 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 9 \\ \text {Hence, }x = 9 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(ii) We have LHS } = x-7 \text { and RHS }= 10 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 9 & 9-7=2 & 10 & No \\ \hline 10 & 10-7=3 & 10 & No \\ \hline 11 & 11-7=4 & 10 & No \\ \hline 12 & 12-7=5 & 10 & No \\ \hline 13 & 19-7=6 & 10 & No \\ \hline 14 & 14-7=7 & 10 & No \\ \hline 15 & 15-7=8 & 10 & No \\ \hline 16 & 16-7=9 & 10 & No \\ \hline 17 & 17-7=10 & 10 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 17 \\ \text {Hence, }x = 17 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(iii) We have LHS } = 4x \text { and RHS }= 28 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 1 & 4 \times 1=4 & 28 & No \\ \hline 2 & 4 \times 2=8 & 28 & No \\ \hline 3 & 4 \times 3=12 & 28 & No \\ \hline 4 & 4 \times 4=16 & 28 & No \\ \hline 5 & 4 \times 5=20 & 28 & No \\ \hline 6 & 4 \times 6=24 & 28 & No \\ \hline 7 & 4 \times 7=28 & 28 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 7 \\ \text {Hence, }x = 7 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(iv) We have LHS } = \frac{x}{2}+7 \text { and RHS }= 11 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 2 & \frac {2}{2}+7=1+7=8 & 11 & No \\ \hline 4 & \frac {4}{2}+7=2+7=9 & 11 & No \\ \hline 6 & \frac {6}{2}+7=3+7=10 & 11 & No \\ \hline 8 & \frac {8}{2}+7=4+7=11 & 11 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 8 \\ \text {Hence, }x = 8 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(v) We have LHS } = 2 x+4 \text { and RHS }= 3x \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 1 & 2(1)+4=2+4=6 & 3(1)=3 & No \\ \hline 2 & 2(2)+4=4+4=8 & 3(2)=6 & No \\ \hline 3 & 2(3)+4=6+4=10 & 3(3)=9 & No \\ \hline 4 & 2(4)+4=8+4=12 & 3(4)=12 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 4 \\ \text {Hence, }x = 4 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(vi) We have LHS } = \frac{x}{4} \text { and RHS }= 12 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\\hline 16 & \frac {16}{4}=4 & 12 & No \\ \hline 20 & \frac {20}{4}=5 & 12 & No \\ \hline 24 & \frac {24}{4}=6 & 12 & No \\ \hline 28 & \frac {28}{4}=7 & 12 & No \\ \hline 32 & \frac {32}{4}=8 & 12 & No \\ \hline 36 & \frac {36}{4}=9 & 12 & No \\ \hline 40 & \frac {40}{4}=10 & 12 & No \\ \hline 44 & \frac {44}{4}=11 & 12 & No \\ \hline 48 & \frac {48}{4}=12 & 12 & Yes \\\hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 48 \\ \text {Hence, }x = 48 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(vii) We have LHS } = \frac{15}{x} \text { and RHS }= 3 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\\hline 1 & \frac {15}{1}=15 & 3 & No \\ \hline 3 & \frac {15}{3}=5 & 3 & No \\ \hline 5 & \frac {15}{5}=3 & 3 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 5 \\ \text {Hence, }x = 5 \text { is the solution of given equation.}\end{array}\begin{array}{l} \text {(viii) We have LHS } = \frac{x}{18} \text { and RHS }= 20 \\ \end{array}\begin{array}{|c|c|c|c|} \hline x & \text {LHS} & \text {RHS} & \text {Is LHS = RHS} \\ \hline 324 & \frac {324}{18}=18 & 20 & No \\ \hline 342 & \frac {342}{18}=19 & 20 & No \\ \hline 360 & \frac {360}{18}=20 & 20 & Yes \\ \hline \end{array}\begin{array}{l} \text {Clearly, LHS = RHS for } x = 360 \\ \text {Hence, }x = 360 \text { is the solution of given equation.}\end{array}