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

The linear equations in one variable are of the form $ax+b=0$ where $a \neq 0$ and $a,b$ are real numbers and $x$ is the variable. Also, these equations have degree 1 ie. the maximum power of $x$ is 1. These linear equations in one variable have applications in our day-to-day life, for example, if you want to purchase a pen and you have been told that the cost of 20 pens is 5 less than 125. In this case equation will be like $20x+5=125$ where $x$ is the cost of 1 pen. Linear equation in one variable in Class 8 mainly covers solving linear equations in one variable with variable on both sides and solving linear equations in the form of fractions. This chapter is an important part of the NCERT Syllabus for Class 8 Maths that lays a foundation for algebra. In this article, we provide comprehensive step-by-step solutions to all the exercise problems in the chapter.

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This Story also Contains

1. Linear Equations in One Variable Class 8 Questions And Answers PDF Free Download
2. Linear Equations in One Variable Class 8 Solutions - Important Formulae
3. Linear Equations in One Variable Class 8 NCERT Solutions (Exercise)
4. Linear Equations In One Variable Class 8 NCERT Solutions - Topics
5. NCERT Solutions for Class 7 Maths Chapter 2 Linear Equations in One Variable - Points to Remember
6. NCERT Solutions for Class 8 Maths - Chapter Wise
7. NCERT Solutions for Class 8 -Subject Wise

These NCERT solutions are highly reliable as they are developed by the subject matter experts of Careers360. These solutions are of great help for the exam preparation as the students can practice the exercise problems and check the answers to evaluate their performance, and can focus on their weaker areas.

Linear Equations in One Variable Class 8 Questions And Answers PDF Free Download

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Linear Equations in One Variable Class 8 Solutions - Important Formulae

Linear Equations in One Variable:

The linear equation in one variable is an equation that involves only one variable and is in the form $ax+b=0$ where $a \neq 0 $, $x$ is a variable and $a,b$ are real numbers.

Linear equations with variables on both sides:

• Transpose both the number and the variable to one side, maintaining the sign of the number.

• Solve (add/subtract) the equation on both sides to simplify and find the variable's value.

Linear equations that are reducible to a simpler form:

• If the equation is of the form $\frac{x+a}{c}=\frac{x+b}{d}$, cross-multiply the numerator and denominator to simplify it to a linear form, such as $(x + a) \times d = c \times (x + b)$

• Solve this linear equation with variables on both sides to find the variable's value.

Linear Equations in One Variable Class 8 NCERT Solutions (Exercise)

Q1 Solve the following equations and check your results 3x = 2x + 18

Answer: We have

3x = 2x + 18

Subtracting 2x from both sides

3x - 2x = 2x - 2x + 18

or x = 18

Check:- Put x = 18 in both LHS and the RHS.

LHS = 3x = 3(18) = 54

RHS = 2x + 18 = 2(18) + 18 = 36 + 18 = 54

Thus, LHS = RHS

Q2 Solve the following equation and check your results 5t – 3 = 3t – 5

Answer: We have

5t - 3 = 3t - 5

Transposing 3t to the LHS and -3 to the RHS, we get:

5t - 3t = -5 + 3

or 2t = -2

Dividing both sides by 2 :

t = -1

Check:- Put t = -1 in the LHS we have,

5t - 3 = 5(-1) - 3 = -5 -3 = -8

Similarly, put t = -1 in the RHS:

3t - 5 = 3(-1) - 5 = -3 - 5 = -8

Hence, LHS = RHS

Q3 Solve the following equation and check your result 5x + 9 = 5 + 3x

Answer: We have

5x + 9 = 5 + 3x

Transposing 3x to the LHS and 9 to the RHS, we get:

5x - 3x = 5 - 9

or 2x = -4

Dividing both sides by 2 :

x = -2

Check:- Put x = - 2 in both LHS and RHS

LHS :- 5x + 9 = 5(-2) + 9 = -10 + 9 = -1

RHS :- 5 + 3x = 5 + 3(-2) = 5 - 6 = -1

Hence, LHS = RHS

Q4 Solve the following equation and check your result. 4z + 3 = 6 + 2z

Answer: We have

4z + 3 = 6 +2z

Transposing 2z to the LHS and 3 to the RHS, we get:

4z - 2z = 6 - 3

or 2z = 3

Dividing both sides by 2,

$z = \frac{3}{2}$

Check:- Put $z = \frac{3}{2}$ in LHS as well as RHS, we have

LHS :- 4z + 3

$= 4\times \frac{3}{2} + 3 = 9$

RHS :- 6 + 2z

$= 6 + 2\times \frac{3}{2} = 6 + 3 = 9$

Thus, LHS = RHS

Q5 Solve the following equation and check your result 2x – 1 = 14 – x

Answer: We have

2x - 1 = 14 - x

Transposing -x to the LHS and -1 to the RHS

2x + x = 14 + 1

or 3x = 15

Dividing both sides by 3, we get

x = 5

Check:- Put x = 5 in both the LHS and the RHS

LHS :- 2x - 1 = 2(5) - 1 = 10 - 1 = 9

RHS :- 14 - x = 14 - 5 = 9

Hence, LHS = RHS

Q6 Solve the following equation and check your result 8x + 4 = 3 (x – 1) + 7

Answer: We have

8x + 4 = 3 (x - 1) + 7

or 8x + 4 = 3x - 3 + 7 = 3x + 4

Transposing 3x to the LHS and 4 to the RHS, we get

8x - 3x = 4 - 4 = 0

5x = 0

Dividing both sides by 5:

x = 0

Check:- Putting x = 0 in both LHS and RHS:

LHS :- 8x + 4 = 8(0) + 4 = 4

RHS :- 3(x-1) + 7 = 3(-1) + 7 = -3 + 7 = 4

Hence, LHS = RHS

Q7 Solve the following equations and check your results.

$x =\frac{4}{5}\times \left ( x+10 \right )$

Answer: We have

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

$= \frac{4}{5}x + 8$

Transposing $\frac{4}{5}x$ to the left-hand side:

$x - \frac{4}{5}x = 8$

or $\frac{5x-4x}{5} = 8$

or $\frac{x}{5} = 8$

Multiplying both sides by 5

x = 40

Check:- Put x = 40 in both the LHS and the RHS.

LHS :- x = 40

RHS :- $\frac{4}{5}\left ( x + 10 \right ) = \frac{4}{5}\left ( 40 + 10 \right )$

$= \frac{4}{5}\left ( 50 \right ) = 40$

Thus, LHS = RHS

Q8 Solve the following equation and check your result.

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

Answer: We have

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

Transposing $\frac{7x}{15}$ to the LHS and 1 to the RHS:

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

or $\frac{10x - 7x}{15} = 2$

or $\frac{3x}{15} = 2$

or x = 10

Check:- Put x = 10 in both the LHS and the RHS

LHS :

$\frac{2x}{3} + 1 = \frac{2}{3}\times 10 + 1 = \frac{23}{3}$

RHS :

$\frac{7x}{15} + 3 = \frac{7}{15}\times 10 + 3 = \frac{14}{3} + 3 = \frac{23}{3}$

Thus, LHS = RHS

Q9 Solve the following equations and check your results.

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

Answer: We have

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

Transposing -y to the LHS and $\frac{5}{3}$ to the RHS, we get

$2y + y = \frac{26}{3}- \frac{5}{3} = \frac{21}{3} = 7$

or 3y = 7

Dividing both sides by 3:

$y = \frac{7}{3}$ i

Check:- Put $y = \frac{7}{3}$ in both the LHS and the RHS:

LHS :

$2y + \frac{5}{3} = 2\times \frac{7}{3} + \frac{5}{3} = \frac{19}{3}$

RHS :

$\frac{26}{3} - \frac{7}{3} = \frac{19}{3}$

Hence, LHS = RHS

Q10 Solve the following equation and check your result.

$3m = 5m - \frac{8}{5}$

Answer: We have

$3m = 5m - \frac{8}{5}$

Transposing 5m to the LHS

$3m -5m = - \frac{8}{5}$

or $-2m = - \frac{8}{5}$

Dividing both sides by -2, we get

$m = \frac{4}{5}$

Check:- Put $m = \frac{4}{5}$ in the LHS and the RHS:

LHS:-

$3m = 3\times \left ( \frac{4}{5} \right ) = \frac{12}{5}$

RHS:-

$5m - \frac{8}{5} = 5 \times \frac{4}{5} - \frac{8}{5} = \frac{12}{5}$

Hence, LHS = RHS

Q1 Solve the following linear equations.

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

Answer: The detailed solution to the above question is here.

We have

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

or $\frac{x}{2} - \frac{x}{3} = \frac{1}{4} + \frac{1}{5}$

or $\frac{3x - 2x}{6} = \frac{4 +5}{20}$

or $\frac{x}{6} = \frac{9}{20}$

Multiplying both sides by 6:

$x = \frac{9}{20}\times 6 = \frac{27}{10}$

Thus $x = \frac{27}{10}$

Q2 Solve the following linear equations.

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

Answer: We have

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

Multiplying both sides by 12.

(We multiplied both sides by 12 because it is the lowest common factor.)

6n - 9n + 10n = 252

or 7n = 252

or n = 36

Q3 Solve the following linear equations.

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

Answer: We have

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

Multiplying both sides by 6, we get

6x + 42 - 16x = 17 - 15x

or -10x +42 = 17 - 15x

Transposing -15x to the LHS and 42 to the RHS :

-10x + 15x = 17 - 42

or 5x = -25

or x = -5

Q4 Solve the following linear equations.

$\frac{x-5}{3} = \frac{x-3}{5}$

Answer: We have

$\frac{x-5}{3} = \frac{x-3}{5}$

Multiply both sides by 15. (because the LCM of the denominator is 15 )

5x - 25 = 3x - 9

Transposing 3x to the LHS and -25 to the RHS.

5x - 3x = -9 + 25

or 2x = 16

x = 8

Q5 Solve the following linear equations.

$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$

Answer: We have

$\frac{3t-2}{4} -\frac{2t+3}{3} = \frac{2}{3} - t$

Multiplying both sides by 12, we get

9t - 6 - ( 8t + 12 ) = 8 - 12t

or 9t - 6 - 8t - 12 = 8 - 12t

or t - 18 = 8 - 12t

Transposing -18 to the RHS and -12t to the LHS:

12t + t = 8 + 18 = 26

or 13t = 26

Thus, t = 2

Q6 Solve the following linear equation.

$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$

Answer: We have

$m - \frac{m-1}{2} = 1 - \frac{m-2}{3}$

Multiplying both sides by 6 (as it is the LCM of the denominator)

6m - 3(m - 1) = 6 - 2(m - 2)

or 6m - 3m + 3 = 6 - 2m + 4

or 3m + 3 = 10 - 2m

Transposing -2m to the LHS and 3 to the RHS:

3m + 2m = 10 - 3

or 5m = 7

$m = \frac{7}{5}$

Q7 Simplify and solve the following linear equation. 3(t – 3) = 5(2t + 1)

Answer: Let us open the brackets :

LHS : 3(t - 3) = 3t - 9

RHS : 5(2t + 1) = 10t + 5

The equation is 3t - 9 = 10t + 5

or -9 = 10t - 3t + 5

or -9 - 5 = 7t

or -14 = 7t

Therefore t = -2

Q8 Simplify and solve the following linear equation. 15(y – 4) –2(y – 9) + 5(y + 6) = 0

Answer: Let us open the brackets

Equation becomes: 15y - 60 - 2y + 18 + 5y + 30 = 0

or 18y - 12 = 0

or 18y = 12

$y = \frac{12}{18} = \frac{2}{3}$

Q9 Simplify and solve the following linear equation. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Answer: Let us open the brackets.

LHS: 3(5z - 7) - 2(9z - 11)

= 15z - 21 - 18z + 22

= -3z + 1

RHS: 4(8z - 13) - 17

= 32z - 52 - 17

= 32z - 69

So the equation becomes : -3z + 1 = 32z - 69

or 69 + 1 = 32z + 3z

or 70 = 35z

z = 2

Q10 Simplify and solve the following linear equations. 0.25(4f – 3) = 0.05(10f – 9)

Answer: We have

0.25(4f – 3) = 0.05(10f – 9)

Let us open the brackets:

f - 0.75 = 0.5f - 0.45

or f - 0.5f = - 0.45 + 0.75

or 0.5f = 0.30

Dividing both sides by 0.5, we get

f = 0.6

Linear Equations In One Variable Class 8 NCERT Solutions - Topics

• Solving Equations having a variable on both sides
• Reducing Equations to Simpler Form

NCERT Solutions for Class 7 Maths Chapter 2 Linear Equations in One Variable - Points to Remember

Linear equations with variables on both sides:

• Transpose both the number and the variable to one side, maintaining the sign of the number.

• Solve (add/subtract) the equation on both sides to simplify and find the variable's value.

Linear equations that are reducible to a simpler form:

• If the equation is of the form $\frac{x+a}{c}=\frac{x+b}{d}$, cross-multiply the numerator and denominator to simplify it to a linear form, such as $(x + a) \times d = c \times (x + b)$

• Solve this linear equation with variables on both sides to find the variable's value.

NCERT Solutions for Class 8 Maths - Chapter Wise

NCERT Solutions for Class 8 -Subject Wise

The NCERT Subject-wise Solutions for Class 8 is an essential study resource that helps students get trustworthy solutions for each subject to make the exam preparation easier. Students can refer to those solutions using the links below.

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

Students can also refer to the NCERT Books and the NCERT Syllabus using the links below:

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8