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

• Exercise 2.1: This exercise contains 10 questions. This exercise covers how to solve equations with variables on both sides.

• Exercise 2.2: This exercise contains 10 questions. This exercise deals with how to reduce equations into simpler form.

Access NCERT Solutions for Class 8 Maths Chapter 2 – Linear Equations in One Variable

Exercise 2.1

1. Solve and check result: $3x=2x+18$

Ans: 

\[3x=2x+18\]

On Transposing \[2x\] to L.H.S, we obtain 

\[3x-2x=18\]

\[x=18\]

L.H.S \[=3x=3\times 18=54\]

R.H.S \[=2x+18=2\times 18+18=36+18=54\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct.

2. Solve and check result: $5t-3=3t-5$

Ans: 

\[5t-3=3t-5\]

On Transposing \[3t\] to L.H.S and \[-3\] to R.H.S, we obtain 

\[5t-3=-5-\left( -3 \right)\]

\[2t=-2\]

On dividing both sides by\[2\], we obtain 

\[t=-1\]

L.H.S \[=5t-3=5\times \left( -1 \right)-3=-8\]

R.H.S \[=3t-5=3\times \left( -1 \right)-5=-3-5=-8\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

3. Solve and check result: $5x+9-5+3x$

Ans: 

\[5x+9=5+3x\]

On Transposing \[3x\] to L.H.S and \[9\] to R.H.S, we obtain 

\[5x-3x=5-9\]

\[2x=-4\]

On dividing both sides by\[2\], we obtain 

\[x=-2\]

L.H.S \[=5x+9=5\times \left( -2 \right)+9=-10+9=-1\]

R.H.S \[=5+3x=5+3\times \left( -2 \right)=5-6=-1\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct.

4. Solve and check result: $4z+3=6+2z$

Ans: 

\[4z+3=6+2z\]

On Transposing \[2z\] to L.H.S and \[3\] to R.H.S, we obtain 

\[4z-2z=6-3\]

\[2z=3\]

Dividing both sides by\[2\] , we obtain 

L.H.S \[=4z+3=4\times \left( \frac{3}{2} \right)+3=6+3=9\]

R.H.S \[=6+2z=6+2\times \left( \frac{3}{2} \right)=6+3=9\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

5. Solve and check result: $2x-1=14-x$

Ans: 

\[2x-1=14-x\]

Transposing x to L.H.S and $1$ to R.H.S, we obtain 

\[2x+x=14+1\]

\[3x=15\]

Dividing both sides by \[3\], we obtain 

\[x=5\]

L.H.S \[=2x-1=2\times \left( 5 \right)-1=10-1=9\]

R.H.S \[=14-x=14-5=9\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

6. Solve and check result: $8x+4=3\left( x-1 \right)+7$

Ans: 

\[8x+4=3\left( x-1 \right)+7\]

\[8x+4=3x-3+7\]

Transposing \[3x\] to L.H.S and $4$ to R.H.S, we obtain 

\[8x-3x=-3+7-4\]

\[5x=-7+7\]

\[x=0\]

L.H.S \[=8x+4=8\times \left( 0 \right)+4=4\]

R.H.S \[=3\left( x-1 \right)+7=3\left( 0-1 \right)+7=-3+7=4\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

7. Solve and check result: $x=\frac{4}{5}\left( x+10 \right)$

Ans: 

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

Multiplying both sides by\[5\], we obtain 

\[5x=4\left( x+10 \right)\]

\[5x=4x+40\]

Transposing \[4x\] to L.H.S, we obtain 

\[5x=4x+40\]

\[x=40\]

L.H.S \[=x=40\]

R.H.S   \[=\frac{4}{5}\left( x+10 \right)=\frac{4}{5}\left( 40+10 \right)=\frac{4}{5}\times 50=40\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

8. Solve and check result: $\frac{2x}{3}+1=\frac{7x}{15}+3$

Ans: 

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

Transposing \[\frac{7x}{15}\] to L.H.S and $1$ to R.H.S, we obtain 

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

\[\frac{5\times 2x-7x}{15}=2\]

\[\frac{3x}{15}=2\]

\[\frac{x}{5}=2\]

Multiplying both sides by\[5\] , we obtain 

\[x=10\]

L.H.S \[=\frac{2x}{3}+1=\frac{2\times 10}{3}+1=\frac{2\times 10+1\times 3}{3}=\frac{23}{3}\]

R.H.S\[=\frac{7x}{15}+3=\frac{7\times 10}{15}+3=\frac{7\times 2}{3}+3=\frac{14}{3}+3=\frac{14+3\times 3}{3}=\frac{23}{3}\] 

L.H.S. = R.H.S. 

Hence, the result obtained above is correct. 

9. Solve and check result: $2y+\frac{5}{3}=\frac{26}{3}-y$

Ans: 

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

Transposing y to L.H.S and \[\frac{5}{3}\] to R.H.S, we obtain 

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

\[3y=\frac{21}{3}=7\]

Dividing both sides by$3$, we obtain 

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

L.H.S \[=2y+\frac{5}{3}=2\times \frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}\]

R.H.S = \[\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}=\frac{19}{3}\]

L.H.S. = R.H.S. Hence, the result obtained above is correct. 

10. Solve and check result: $3m=5m-\frac{8}{5}$

Ans: 

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

Transposing \[5m\] to L.H.S, we obtain 

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

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

Dividing both sides by\[-2\] , we obtain 

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

L.H.S \[=3m=3\times \frac{4}{5}=\frac{12}{5}\]

R.H.S \[5m-\frac{8}{5}=5\times \frac{4}{5}-\frac{8}{5}=\frac{12}{5}\]

L.H.S. = R.H.S. 

Hence, the result obtained above is correct.

Exercise 2.2

1. Solve the linear equation $\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

Ans: 

\[\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\]

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

Multiplying both sides by 60, we obtain 

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

\[\Rightarrow 30x-12=20x+15\] (Opening the brackets) 

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

\[\Rightarrow 10x=27\]

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

2. Solve the linear equation$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$ 

Ans: 

\[\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21\]

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

Multiplying both sides by\[12\], we obtain 

\[6n-9n+10n=252\]

\[\Rightarrow 7n=252\]

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

\[\Rightarrow n=36\]

3. Solve the linear equation $x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

Ans: 

\[x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}\]

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

Multiplying both sides by \[6\], we obtain 

\[6x+42-16x=17-15x\]

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

\[\Rightarrow 5x=-25\]

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

\[\Rightarrow x=-5\]

4. Solve the linear equation $\frac{x-5}{3}=\frac{x-3}{5}$

Ans: 

\[\frac{x-5}{3}=\frac{x-3}{5}\]

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

Multiplying both sides by\[15\], we obtain 

\[5\left( x-5 \right)=3\left( x-3 \right)\]

\[\Rightarrow 5x-25=3x-9\] (Opening the brackets) 

\[\Rightarrow 5x-3x=25-9\]

\[\Rightarrow 2x=16\]

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

\[\Rightarrow x=8\]

5. Solve the linear equation $\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t$

Ans: 

\[\frac{3t-2}{4}-\frac{2t+3}{3}=\frac{2}{3}-t\]

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

Multiplying both sides by \[12\], we obtain 

\[3\left( 3t-2 \right)-4\left( 2t+3 \right)=8-12t\]

\[\Rightarrow 9t-6-8t-12=8-12t\] (Opening the brackets) 

\[\Rightarrow 9t-8t+12t=8+6+12\]

\[\Rightarrow 13t=26\]

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

\[\Rightarrow t=2\]

6. Solve the linear equation$m-\frac{m-1}{2}=1-\frac{m-2}{3}$  

Ans: 

\[m-\frac{m-1}{2}=1-\frac{m-2}{3}\]  

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

Multiplying both sides by \[6\], we obtain 

\[6m-3\left( m-1 \right)=6-2\left( m-2 \right)\]

\[\Rightarrow 6m-3m+3=6-2m+4\] (Opening the brackets) 

\[\Rightarrow 6m-3m+2m=6+4-3\]

\[\Rightarrow 5m=7\]

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

7. Simplify and solve the linear equation $3\left( t-3 \right)=5\left( 2t+1 \right)$

Ans: 

\[3\left( t-3 \right)=5\left( 2t+1 \right)\]

\[\Rightarrow 3t-9=10t+5\] (Opening the brackets) 

\[\Rightarrow -9-5=10t-3t\]

\[\Rightarrow -14=7t\]

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

\[\Rightarrow t=-2\]

8. Simplify and solve the linear equation$15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0$  

Ans:

\[15\left( y-4 \right)-2\left( y-9 \right)+5\left( y+6 \right)=0\]  

\[\Rightarrow 15y-60-2y+18+5y+30=0\] (Opening the brackets) 

\[\Rightarrow 18y-12=0\]

\[\Rightarrow 18y=12\]

\[\Rightarrow y=\frac{12}{8}=\frac{2}{3}\]

9.Simplify and solve the linear equation $3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17$   

Ans: 

\[3\left( 5z-7 \right)-2\left( 9z-11 \right)=4\left( 8z-13 \right)-17\]  

\[\Rightarrow 15z-21-18z+22=32z-52-17\] (Opening the brackets) 

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

\[\Rightarrow -3z-32z=-69-1\]

\[\Rightarrow -35z=-70\]

\[\Rightarrow z=\frac{70}{~35}=2\]

10. Simplify and solve the linear equation $0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)$

Ans: 

\[0.25\left( 4f-3 \right)=0.05\left( 10f-9 \right)\]

$\frac{1}{4}\left( 4f-3 \right)=\frac{1}{20}\left( 10f-9 \right)$

Multiplying both sides by\[20\], we obtain 

\[5\left( 4f-3 \right)=10f-9\]

\[\Rightarrow 20f-15=10f-9\] (Opening the brackets)

\[\Rightarrow 20f-10f=-9+15\]  

\[\Rightarrow 10f=6\]

\[\Rightarrow f=\frac{3}{5}=0.6\]

Overview of Deleted Syllabus for CBSE Class 8 Maths Linear Equations In One Variable 

Class 8 Maths Chapter 2: Exercises Breakdown

Conclusion 

NCERT Maths Class 8 Solutions Vedantu's Linear Equations in One Variable provide a thorough understanding of this significant subject. Students can build a solid foundation in Linear Equations by concentrating on important ideas such as reducing equations into simpler forms, solving equations with variables on both sides. It's important to pay attention to the step-by-step solutions provided in the NCERT Solutions, as they help clarify concepts and reinforce problem-solving techniques. In previous years' exams, typically 2 to 3 questions have been asked from Ch 2 Maths Class 8. These questions often cover a variety of problem types, including basic equation solving, application-based word problems, and questions involving equations with variables on both sides. This pattern has been consistent across multiple exam papers and sources.

Other Study Material for CBSE Class 8 Maths Chapter 2

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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