NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.1 - Linear Equations in One Variable

1. Solve and Verify the Equation \[\text{3x=2x+18}\].

Ans: We have an equation \[\text{3x=2x+18}\]

To solve the equation, we will shift \[\text{2x}\] to left hand side 

\[\text{3x-2x=18}\]

\[\text{x=18}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{3x=3 }\!\!\times\!\!\text{ 18}\]

\[\text{54}\]

R.H.S

\[\text{2x+18=2 }\!\!\times\!\!\text{ 18+18}\]

\[\text{54}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

2. Solve and Verify the Equation \[\text{5t-3=3t-5}\].

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

To solve the equation, we will shift \[\text{3t}\] to left hand side and \[\text{3}\] to the right hand side, now,

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

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

Dividing the equation by \[\text{2}\]

\[\text{t=-1}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{5t-3=}\left( \text{5 }\!\!\times\!\!\text{ -1} \right)\text{-3}\]

\[\text{-5-3=-8}\]

R.H.S

\[\text{3t-5=}\left( \text{3 }\!\!\times\!\!\text{ -1} \right)\text{-5}\]

\[\text{-3-5=-8}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

3. Solve and Verify the Equation \[\text{5x+9=5+3x}\].

Ans: We have an equation \[\text{5x+9=5+3x}\]

To solve the equation, we will shift \[\text{3x}\] to left hand side and \[\text{9}\] to the right hand side, now,

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

\[\text{2x=-4}\]

Dividing the equation by \[\text{2}\]

\[\text{x=-2}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{5x+9=}\left( \text{5 }\!\!\times\!\!\text{ -2} \right)\text{+9}\]

\[\text{-10+9=-1}\]

R.H.S

\[\text{5+3x=5+}\left( \text{3 }\!\!\times\!\!\text{ -2} \right)\]

\[\text{5-6=-1}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

4. Solve and Verify the Equation \[\text{4z+3=6+2z}\].

Ans: We have an equation \[\text{4z+3=6+2z}\]

To solve the equation, we will shift \[\text{2z}\] to left hand side and \[\text{3}\] to right hand side, now,

\[\text{4z-2z=6-3}\]

\[\text{2z=3}\]

Dividing the equation by \[\text{2}\]

\[\text{z=}\frac{\text{3}}{\text{2}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{4z+3=}\left( \text{4 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)\text{+3}\]

\[\text{6+3=9}\]

R.H.S

\[\text{6+2z=6+}\left( \text{2 }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{2}} \right)\]

\[\text{6+3=9}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

5. Solve and Verify the Equation \[\text{2x-1=14-x}\].

Ans: We have an equation \[\text{2x-1=14-x}\]

To solve the equation, we will shift \[\text{x}\] to left hand side and \[\text{1}\] to the right hand side, now,

\[\text{2x+x=14+1}\]

\[\text{3x=15}\]

Dividing the equation by \[\text{3}\]

\[\text{x=5}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{2x-1=}\left( \text{2 }\!\!\times\!\!\text{ 5} \right)\text{-1}\]

\[\text{10-1=9}\]

R.H.S

\[\text{14-x=14-5}\]

\[\text{14-5=9}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

6. Solve and Verify the Equation \[\text{8x+4=3}\left( \text{x-1} \right)\text{+7}\].

Ans: We have an equation \[\text{8x+4=3}\left( \text{x-1} \right)\text{+7}\]

\[\text{8x+4=3x-3+7}\]

\[\text{8x+4=3x+4}\]

To solve the equation, we will shift \[\text{3x}\] to left hand side and \[4\] to the right hand side, now,

\[\text{8x-3x=4-4}\]

\[\text{5x=0}\]

\[\text{x=0}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{8x+4=}\left( \text{8 }\!\!\times\!\!\text{ 0} \right)\text{+4}\]

\[\text{0+4=4}\]

R.H.S

\[\text{3}\left( \text{x-1} \right)\text{+7=3}\left( \text{0-1} \right)\text{+7}\]

\[\text{-3+7=4}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

7. Solve and Verify the Equation \[\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\].

Ans: We have an equation \[\text{x=}\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\]

\[\text{x=}\frac{\text{4x+40}}{\text{5}}\]

Multiplying the equation by \[\text{5}\]

\[\text{5x=4x+40}\]

To solve the equation, we will shift \[\text{4x}\] to left hand side, now,

\[\text{5x-4x=40}\]

\[\text{x=40}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{x=40}\]

R.H.S

\[\frac{\text{4}}{\text{5}}\left( \text{x+10} \right)\text{=}\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)\]

\[\frac{\text{4}}{\text{5}}\left( \text{40+10} \right)\text{=}\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}\]

\[\frac{\text{4 }\!\!\times\!\!\text{ 50}}{\text{5}}\text{=40}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

8. Solve and Verify the Equation \[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}\].

Ans: We have an equation \[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\text{7x}}{\text{15}}\text{+3}\]

To solve the equation, we will shift \[\frac{\text{7x}}{\text{15}}\] to left hand side and \[\text{1}\] to right hand side, now,

\[\frac{\text{2x}}{\text{3}}\text{-}\frac{\text{7x}}{\text{15}}\text{=3-1}\]

\[\frac{\text{10x-7x}}{\text{15}}\text{=2}\]

Multiplying \[15\] to the equation, we get

\[\text{3x=30}\]

\[\text{x=10}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\frac{\text{2x}}{\text{3}}\text{+1=}\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}\]

\[\frac{\left( \text{2 }\!\!\times\!\!\text{ 10} \right)\text{+3}}{\text{3}}\text{=}\frac{\text{23}}{\text{3}}\]

R.H.S

\[\frac{\text{7x}}{\text{15}}\text{+3=}\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}\]

\[\frac{\left( \text{7 }\!\!\times\!\!\text{ 10} \right)\text{+45}}{\text{15}}\text{=}\frac{\text{115}}{\text{15}}\]

\[\frac{\text{115}}{\text{15}}\text{=}\frac{\text{23}}{\text{3}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

9. Solve and Verify the Equation \[\text{2y+}\frac{\text{5}}{\text{3}}\text{+}\frac{\text{26}}{\text{3}}\text{-y}\].

Ans: We have an equation \[\text{2y+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{26}}{\text{3}}\text{-y}\]

To solve the equation, we will shift \[y\] to left hand side and \[\frac{\text{5}}{\text{3}}\] to right hand side, now,

\[\text{2y+y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{5}}{\text{3}}\]

\[\text{3y=}\frac{\text{21}}{\text{3}}\]

\[\text{3y=7}\]

\[\text{y=}\frac{\text{7}}{\text{3}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{2y+}\frac{\text{5}}{\text{3}}\text{=2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}\]

\[\text{2 }\!\!\times\!\!\text{ }\frac{\text{7}}{\text{3}}\text{+}\frac{\text{5}}{\text{3}}\text{=}\frac{\text{14+5}}{\text{3}}\]

\[\frac{\text{14+5}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}\]

R.H.S

\[\frac{\text{26}}{\text{3}}\text{-y=}\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}\]

\[\frac{\text{26}}{\text{3}}\text{-}\frac{\text{7}}{\text{3}}\text{=}\frac{\text{19}}{\text{3}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

10. Solve and Verify the Equation \[\text{3m=5m-}\frac{\text{8}}{\text{5}}\].

Ans: We have an equation \[\text{3m=5m-}\frac{\text{8}}{\text{5}}\]

To solve the equation, we will shift \[\text{5m}\] to left hand side, now,

\[\text{3m-5m=-}\frac{\text{8}}{\text{5}}\]

\[\text{-2m=-}\frac{\text{8}}{\text{5}}\]

Dividing the equation by \[\text{-2}\], we get,

\[\text{m=}\frac{4}{\text{5}}\]

Now to verify the result, solve L.H.S and R.H.S

L.H.S

\[\text{3m=3 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{5}}\]

\[\frac{\text{3 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{=}\frac{\text{12}}{\text{5}}\]

R.H.S

\[\text{5m-}\frac{\text{8}}{\text{5}}\text{=}\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}\]

\[\frac{\text{5 }\!\!\times\!\!\text{ 4}}{\text{5}}\text{-}\frac{\text{8}}{\text{5}}=\frac{12}{\text{5}}\]

L.H.S\[\text{=}\]R.H.S

Hence Proved

Conclusion

Ex 2.1 Class 8  Maths Chapter 2 is crucial for mastering linear equations in one variable. It focuses on building a strong foundation by teaching students how to solve equations through simple steps. This exercise emphasizes understanding the process of isolating the variable and verifying solutions. Students should concentrate on practicing various types of problems to enhance their problem-solving skills. It's important to pay attention to the methods used and ensure accuracy in each step. Consistent practice will help in gaining confidence and proficiency in handling linear equations.

Class 8 Maths Chapter 2: Exercises Breakdown

CBSE Class 8 Maths Chapter 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.

Important Related Links for CBSE Class 8 Maths