NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable - Exercise 2.2

Exercise 2.2

1. Find the solution of the linear equation\[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

Ans: We have an equation \[\frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}}\text{=}\frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\]

The L.C.M of denominators \[\text{2,3,4}\] and \[\text{5}\] is \[\text{60}\]

Multiplying both the sides by \[\text{60}\],

\[\Rightarrow \text{60}\left( \frac{\text{x}}{\text{2}}\text{-}\frac{\text{1}}{\text{5}} \right)\text{=60}\left( \frac{\text{x}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}} \right)\]

\[\Rightarrow \text{60}\left( \frac{\text{5x-2}}{\text{10}} \right)\text{=60}\left( \frac{\text{4x+3}}{\text{12}} \right)\]

\[\Rightarrow \text{6}\left( \text{5x-2} \right)\text{=5}\left( \text{4x+3} \right)\]

\[\Rightarrow \text{30x-12=20x+15}\]

Shifting \[\text{20x}\] to left hand side and \[\text{12}\] on right hand side

\[\Rightarrow \text{30x-20x=15+12}\]

\[\Rightarrow \text{10x=27}\]

\[\Rightarrow \text{x=}\frac{\text{27}}{\text{10}}\]

2. Find the solution of the linear equation\[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

Ans: We have an equation \[\frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}}\text{=21}\]

The L.C.M of denominators \[\text{2,4}\] and \[\text{6}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{n}}{\text{2}}\text{-}\frac{\text{3n}}{\text{4}}\text{+}\frac{\text{5n}}{\text{6}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{6n-9n+10n}}{\text{12}} \right)\text{=12}\left( \text{21} \right)\]

\[\Rightarrow \text{7n=252}\]

Dividing the equation by \[\text{7}\]

\[\Rightarrow \text{n=36}\]

3. Find the solution of the linear equation\[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

Ans: We have an equation \[\text{x+7-}\frac{\text{8x}}{\text{3}}\text{=}\frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}}\]

The L.C.M of denominators \[\text{2,3}\] and \[\text{6}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \text{x+7-}\frac{\text{8x}}{\text{3}} \right)\text{=6}\left( \frac{\text{17}}{\text{6}}\text{-}\frac{\text{5x}}{\text{2}} \right)\]

\[\Rightarrow \text{6}\left( \frac{\text{6x+42-16x}}{\text{6}} \right)\text{=6}\left( \frac{\text{17-15x}}{\text{6}} \right)\]

\[\Rightarrow \text{6x+42-16x=17-15x}\]

Shifting \[\text{15x}\] to left hand side and \[\text{42}\] to right hand side

\[\Rightarrow \text{-10x+15x=17-42}\]

\[\Rightarrow \text{5x=-25}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{x=-5}\]

4. Find the solution of the linear equation\[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

Ans: We have an equation \[\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{x-3}}{\text{5}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{5}\] is \[\text{15}\]

Multiplying both the sides by \[\text{15}\],

\[\Rightarrow \text{15}\left( \frac{\text{x-5}}{\text{3}} \right)\text{=15}\left( \frac{\text{x-3}}{\text{5}} \right)\]

\[\Rightarrow \text{5}\left( \text{x-5} \right)\text{=3}\left( \text{x-3} \right)\]

\[\Rightarrow \text{5x-25=3x-9}\]

Shifting \[\text{3x}\] to left hand side and \[\text{25}\] to right hand side

\[\Rightarrow \text{5x-3x=-9+25}\]

\[\Rightarrow \text{2x=16}\]

Dividing both the sides by \[\text{2}\]

\[\Rightarrow \text{x=8}\]

5. Find the solution of the linear equation\[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

Ans: We have an equation \[\frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\text{-t}\]

The L.C.M of denominators \[\text{3}\] and \[\text{4}\] is \[\text{12}\]

Multiplying both the sides by \[\text{12}\],

\[\Rightarrow \text{12}\left( \frac{\text{3t-2}}{\text{4}}\text{-}\frac{\text{2t+3}}{\text{3}} \right)\text{=12}\left( \frac{\text{2}}{\text{3}}\text{-t} \right)\]

\[\Rightarrow \text{12}\left( \frac{\text{9t-6-8t-12}}{\text{12}} \right)\text{=12}\left( \frac{\text{8-12t}}{\text{12}} \right)\]

\[\Rightarrow \text{t-18=8-12t}\]

Shifting \[\text{12t}\] to left hand side and \[\text{18}\] to right hand side

\[\Rightarrow \text{t+12t=8+18}\]

\[\Rightarrow \text{13t=26}\]

Dividing both the sides by \[\text{13}\]

\[\Rightarrow \text{t=2}\]

6. Find the solution of the linear equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

Ans: We have an equation \[\text{m-}\frac{\text{m-1}}{\text{2}}\text{=1-}\frac{\text{m-2}}{\text{3}}\]

The L.C.M of denominators \[\text{3}\] and \[\text{2}\] is \[\text{6}\]

Multiplying both the sides by \[\text{6}\],

\[\Rightarrow \text{6}\left( \frac{\text{2m-m+1}}{\text{2}} \right)\text{=6}\left( \frac{\text{3-m+2}}{\text{3}} \right)\]

\[\Rightarrow \text{3}\left( \text{m+1} \right)\text{=3}\left( \text{-m+5} \right)\]

\[\Rightarrow \text{3m+3=10-2m}\]

Shifting \[\text{2m}\] to left hand side and \[\text{3}\] to right hand side

\[\Rightarrow \text{3m+2m=10-3}\]

\[\Rightarrow \text{5m=7}\]

Dividing both the sides by \[\text{5}\]

\[\Rightarrow \text{m=}\frac{\text{7}}{\text{5}}\]

7. Find the solution of the linear equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

Ans: We have an equation \[\text{3}\left( \text{t-3} \right)\text{=5}\left( \text{2t+1} \right)\]

\[\Rightarrow \text{3t-9=10t+5}\]

Shifting \[\text{5}\] to left hand side and \[\text{3t}\] to right hand side

\[\Rightarrow \text{-9-5=10t-3t}\]

\[\Rightarrow \text{-14=7t}\]

Dividing both the sides by \[\text{7}\]

\[\Rightarrow \text{-2=t}\]

8. Find the solution of the linear equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

Ans: We have an equation \[\text{15}\left( \text{y-4} \right)\text{-2}\left( \text{y-9} \right)\text{+5}\left( \text{y+6} \right)\text{=0}\]

\[\Rightarrow \text{15y-60-2y+18+5y+30=0}\]

\[\Rightarrow \text{18y-12=0}\]

Shifting \[\text{12}\] to right hand side

\[\Rightarrow \text{18y=12}\]

Dividing both the sides by \[\text{18}\]

\[\Rightarrow \text{y=}\frac{\text{12}}{\text{18}}\]

\[\Rightarrow \text{y=}\frac{\text{2}}{\text{3}}\]

9. Find the solution of the linear equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

Ans: We have an equation \[\text{3}\left( \text{5z-7} \right)\text{-2}\left( \text{9z-11} \right)\text{=4}\left( \text{8z-13} \right)\text{-17}\]

\[\Rightarrow \text{15z-21-18z+22=32z-52-17}\]

\[\Rightarrow \text{-3z+1=32z-69}\]

Shifting \[\text{69}\] to left hand side and \[\text{3z}\] to right hand side

\[\Rightarrow \text{1+69=32z+3z}\]

\[\Rightarrow \text{70=35z}\]

Dividing both the sides by \[\text{35}\]

\[\Rightarrow \text{z=2}\]

10. Find the solution of the linear equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

Ans: We have an equation \[\text{0}\text{.25}\left( \text{4f-3} \right)\text{=0}\text{.05}\left( \text{10f-9} \right)\]

\[\Rightarrow \text{f-0}\text{.75=0}\text{.5f-0}\text{.45}\]

Shifting \[\text{0}\text{.5f}\] to left hand side and \[\text{0}\text{.75}\] to right hand side

\[\Rightarrow \text{0}\text{.5f=-0}\text{.45+0}\text{.75}\]

\[\Rightarrow \text{0}\text{.5f=0}\text{.3}\]

Dividing both the sides by \[\text{0}\text{.5}\]

\[\Rightarrow \text{f=0}\text{.6}\]

Conclusion

Class 8 Maths Chapter 2 Exercise 2.2 solutions, "Linear Equations in One Variable," focuses on solving linear equations involving one variable. Exercise 2.2 Class 8 introduces complex problems, requiring careful manipulation. It covers isolating the variable, simplifying the equation, and verifying solutions. Exercises include fractions, integers, and word problems, solidifying understanding of different forms of linear equations. Practice is crucial for developing problem-solving skills and preparing for exams.

Class 8 Maths Chapter 2: Exercises Breakdown

CBSE Class 8 Maths Chapter 2 Exercise 2.2 Other Study Materials

Chapter-Specific NCERT Solutions for Class 8 Maths

Given below are the chapter-wise NCERT Solutions for Class 8 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

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