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

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

Edited By Ramraj Saini | Updated on Feb 29, 2024 05:57 PM IST

Linear Equations in One Variable Class 8 Questions And Answers provided here. These NCERT Solutions are prepared by subject matter expert of Careers360 considering latest syllabus and pattern of CBSE 2023-24. this chapter created a strong foundation for algebra unit. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable is comprehensively covering each problem related to this particular topic.

In this chapter of NCERT Syllabus for Class 8 Maths, there will be an expression consisting of variables as well as numbers and you have to find the value of that variable. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable are covering solutions to word problems using the linear equations of one variable. This chapter has a total of 6 exercises consisting of a total of 65 questions in all. NCERT solutions for Class 8 Maths chapter 2 Linear Equations in One Variable has the solution to all 65 questions to boost your overall preparation level.

## Linear Equations in One Variable Class 8 Solutions - Important Formulae

Linear equations with a linear expression on one side:

• Transpose the number to the side with all the numbers, maintaining the sign.

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

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 with a number in the denominator and variables on both sides:

• Find the LCM (Least Common Multiple) of the denominators on both sides.

• Multiply both sides by the LCM to simplify the equation.

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

Linear equations that are reducible to the linear form:

• If the equation is of the form (x + a / x + b) = c / d, cross-multiply the numerator and denominator to simplify it to a linear form, such as (x + a) * d = c * (x + b).

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

Free download NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable for CBSE Exam.

## Linear Equations in One Variable Class 8 NCERT Solutions (Intext Questions and Exercise)

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

Answer: 1. Transposing -2 to the RHS, we get

x = 7 + 2 = 9

Answer: Transposing 3 to the RHS, we get

y = 10 - 3 = 7

y = 7

Answer: Transposing 2 to the LHS, we get

6 - 2 = z => z= 4

Thus z = 4

$\frac{3}{7} + x = \frac{17}{7}$

Answer: 4. Transposing $\frac{3}{7}$ to the RHS, we get

$x = \frac{17}{7}-\frac{3}{7} = \frac{14}{7} = 2$

Thus x =2

Answer: Dividing both sides by 6, we get

$x = \frac{12}{6} => 2$

Thus x = 2

$\frac{t}{5} = 10$

Answer: Multiplying both sides by 5, we get

$t = 10\times 5 = 50$

Thus t = 50

$\frac{2x}{3} = 18$

$\frac{2x}{3} = 18$

Multiplying both sides by 3, we get

$2x = 18\times 3 = 54$

Now, dividing both sides by 2, we get

$x = \frac{54}{2} = 27$

$1.6 = \frac{y}{1.5}$

Answer: Multiplying both sides by 1.5, we get

$1.6\times 1.5 = y = 2.4$

Thus y = 2.4

Answer: Transposing 9 to the RHS, we get

7x = 16 + 9 = 25

Now, dividing both sides by :

$x = 25\times \frac{1}{7} = \frac{25}{7}$

Answer: Transposing -8 to the RHS, we get

14y = 13 + 8 = 21

Now dividing both sides by 14:

$y = \frac{21}{14} = \frac{3}{2}$

Answer: At first, transposing 17 to the RHS:

6p = 9 - 17 = -8

Now, dividing both sides by 6, we get

$p = \frac{-8}{6} = \frac{-4}{3}$

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

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

Transposing 1 to the RHS, we get

$\frac{x}{3} = \frac{7}{15} -1 = \frac{7-15}{15} = \frac{-8}{15}$

Now multiplying both sides by 3, it becomes

$x = \frac{-8}{15}\times 3 = \frac{-8}{5}$

Class 8 maths chapter 2 question answer - Exercise: 2.2

Answer: Assume the number to be x.

Thus according to the question,

$\left ( x-\frac{1}{2} \right )\times \frac{1}{2} = \frac{1}{8}$

Multiplying both sides by 2, we get

$\left ( x-\frac{1}{2} \right ) = \frac{1}{8}\times 2 = \frac{1}{4}$

Now transposing $-\frac{1}{2}$ to the RHS, we get

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

Thus the number was x = 3/4

According to the question, the length of the pool = 2x + 2 m.

Perimeter of recatangle = 2(l + b) = 154 m .

i.e., 2(2x + 2 + x) = 154

2(3x + 2) = 154

Dividing both sides by 2, we get

3x + 2 = 77

Now transposing 2 to the RHS, we get

3x = 77 - 2 = 75

Dividing both sides by 3, we get

$x = \frac{75}{3} = 25$

Thus breadth of pool = 25 m

and lenght of the pool = 2 $\times$ 25 + 2 = 52 m

Answer: In isosceles triangles, we have 2 sides of equal length.

Given that its perimeter is $\frac{62}{15}$ cm.

Let's assume the length of the equal side is x cm.

Also,

$Perimeter = x + x + \frac{4}{3} = \frac{62}{15}$

Transposing $\frac{4}{3}$ to the RHS side it becomes,

$2x = \frac{62}{15}-\frac{4}{3} = \frac{62-20}{15} = \frac{42}{15} = \frac{14}{5}$

Now dividing both sides by 2, we get

$x = \frac{14}{5}\times \frac{1}{2} = \frac{7}{5}$

Hence, the length of the equal sides of the isosceles triangle is $\frac{7}{5}$ cm.

Answer: Assume one number to be x. Then the other number will be x + 15.

Now it is given that sum of the two numbers is 95.

Thus quation becomes x + x + 15 = 95

or 2x +15 = 95

Transposing 15 to the RHS, it becomes

2x = 95 - 15 = 80

Dividng both sides by 2, we get

$x = \frac{80}{2} = 40$

Thus two numbers are 40 and 55

Answer: Given that numbers are in the ratio of 5:3, so we can assume numbers to be 5x and 3x.

Also, the difference between these numbers is 18, so the equation becomes

5x - 3x = 18

2x = 18

Dividing both sides by 2, we get

x = 9

Hence the two numbers are 5 $\times$ 9 = 45 and 3 $\times$ 9 = 27

Answer: Let the consecutive integers be x, x+1, x+2.

Sum of these integers is given to be 51.

Thus equation becomes : x + x + 1 + x + 2 = 51

3x + 3 = 51

Transposing 3 to the RHS,

3x = 48

Now dividing both sides by 3, we get

$x = \frac{48}{3} = 16$

Hence the consecutive integers are 16, 17, 18.

Answer: Let x be the multiple of 8.

Then the three consecutive integers (multiple of 8) are x, x+8, x+16.

Given their sum is 888,

thus the equation becomes: x + x + 8 + x + 16 = 888

or 3x + 24 = 888

Transposing 24 to the RHS

3x = 888 - 24 = 864

Now dividing both sides by 3, we get

$x = \frac{864}{3} = 288$

Thus the required consecutive integers are 288, 296, 304.

Answer: Let the consecutive integers be x, x+1, x+2.

Now according to the question equation becomes,

2x + 3(x+1) + 4(x+2) = 74

or 2x + 3x + 3 + 4x + 8 = 74

or 9x + 11 = 74

Transposing 11 to the RHS we get,

9x = 74 - 11 = 63

Divinding both sides by 9.

$x = \frac{63}{9} = 7$

Therefore required consecutive integers are 7, 8, 9.

Answer: Let the current age of Rahul and Haroon be 5x and 7x respectively(Since there age ratio is given as 5:7).

Four years later their ages become 5x + 4 and 7x + 4 respectively.

According to the question,

5x + 4 +7x + 4 = 56

or 12x + 8 = 56

Transposing 8 to the RHS we get,

12x = 48

Dividing both sides by 12:

$x = \frac{48}{12} = 4$

Thus current age of Rahul = 5 $\times$ 4 = 20

Haroon = 7 $\times$ 4 = 28

Answer: Let us assume the number of boys and the number of girls is 7x and 5x.

According to the given data in the question,

7x = 5x + 8

Transposing 5x to the LHS we get,

7x - 5x = 8

2x = 8

Dividing both sides by 2

x = 4

Hence number of boys = 7 $\times$ 4 = 28

and number of girls = 5 $\times$ 4 = 20

So total number of students = 28 + 20 = 48.

Answer: Let the age of Baichung be x years.

Then according to question age of Baichung's father = x + 29 years

and age of Baichung's grandfather = x + 29 + 26 years

The sum of their ages is given 135 years. According to that, x + x + 29 + x + 29 + 26 = 135

3x + 84 = 135

Transposing 81 on the RHS we get,

3x = 135 - 84 = 51

Dividing both sides by 3

$x =\frac{51}{3} = 17$

Thus, the age of Baichung = 17 years

The age of Baichung's father = 17 + 29 = 46 years

The age of Baichung's grandfather = 46 + 26 = 72 years

Answer: Let us assume Ravi's present age to be x years.

According to the question,

x + 15 = 4x

Transposing x to the RHS

15 = 3x

Now dividing both sides by 3, we get

$x = \frac{15}{3} = 5$

Hence Ravi's current age is 5 years.

Answer: Let the rational number be x.

According to the question,

$x\times \frac{5}{2} + \frac{2}{3} = -\frac{7}{12}$

Transposing $\frac{2}{3}$ to the RHS:

$x\times \frac{5}{2} = -\frac{7}{12} - \frac{2}{3} = \frac{-15}{12} = \frac{-5}{4}$

Multiplication by 2 in both sides, we get

$5x = \frac{-5}{4}\times 2 = \frac{-5}{2}$

Dividing both sides by 5, we get

$x = \frac{-5}{2}\times \frac{1}{5} = \frac{-1}{2}$

Answer: Since the ratio of currency notes is 2:3:5. Therefore,

Let the number of currency notes of Rs.100, Rs.50 and Rs.10 be 2x, 3x, 5x respectively.

Hence according to the question equation becomes,

100 $\times$ 2x + 50 $\times$ 3x + 10 $\times$ 5x = 400000

or 200x + 150x + 50x = 400000

or 400x = 400000

Dividing both sides by 400 we get,

x = 1000 .

No. of Rs.100 notes = 2 $\times$ 1000 = 2000 notes

No. of Rs.50 notes = 3 $\times$ 1000 = 3000 notes

No. of Rs.10 notes = 5 $\times$ 1000 = 5000 notes

Answer: Let the no. of Rs.1 coin be y.

And the no. of Rs.5 coin be x.

Thus according to question no. of Rs.2 coin will be 3x.

Also, the total no. of coins = 160

This implies : y + x + 3x = 160

or y + 4x = 160

Transposing 4x to the RHS

y = 160 - 4x = No. of Rs.1 coin.

Now, it is given that total amount is Rs.300.

i.e., 1(160-4x) + 2(3x) + 5(x) = 300

or 160 - 4x + 6x + 5x = 300

or 160 + 7x = 300

Transposing 160 to the RHS :

7x = 300 - 160 = 140

Dividing both sides by 7:

$x = \frac{140}{7} = 20$

x = 20

Thus number of Rs.1 coin = 160 - 4x = 160 - 80 = 80

number of Rs.2 coin = 3x = 60

number of Rs.5 coin = x = 20

Answer: Let the no. of winners be x.

Since total no. of participants is 63, thus no. of participants who does not win = 63 - x

According to the question, equation becomes:

x(100) + (63-x)(25) = 3000

or 100x + 1575 - 25x = 3000

or 75x + 1575 = 3000

Transposing 1575 to the RHS :

75x = 3000 - 1575

or 75x = 1425

Dividing both sides by 75, we get

$x = \frac{1425}{75} = 19$

x = 19

Therefore no. of winners = 19

Class 8 maths chapter 2 ncert solutions - Exercise: 2.3

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 8 linear equations in one variable NCERT solutions - Exercise: 2.4

Answer: Let us assume that the number Amina thought was x.

As per the question,

$\left ( x-\frac{5}{2} \right )\times8 = 3x$

or $8x - 20 = 3x$

Transposing 3x to the LHS and -20 to the RHS, we get

8x - 3x = 20

or 5x = 20

Dividing both sides by 5.

x = 4

Therefore the number Amina thought was 4.

Answer: Let the No.1 = x

Then the second no. No.2 = 5x.

Now the question states that if we add 21 to both numbers one number becomes twice of the other.

i.e., 2(x + 21) = 5x + 21

or 2x + 42 = 5x + 21

Transposing 2x to the RHS and 21 to the LHS:

21 = 5x - 2x = 3x

or x = 7

Thus the two numbers are 7 and 35 . (Since No.2 = 5x)

Answer: Given that sum of the two digits is 9.

Let us assume the digit of units place be x.

Then the digit of tens place will be 9-x.

Thus the two digit number is 10(9-x) + x

Now if we reverse the digits, the number becomes 10x + (9-x).

As per the question,

10x + (9-x) = 10(9-x) + x + 27

or 9x + 9 = 90 - 10x + x + 27

or 9x + 9 = 117 - 9x

Transposing -9x to the LHS and 9 to the RHS:

9x + 9x = 117 - 9

or 18x = 108

x = 6

Thus two digit number is 36.

Answer: Let us assume that the units place of a two digit number is x.

Then as per the question, tens place of the number is 3x.

Thus the two digit number is 10(3x) + x.

If we interchange the digits the number becomes : 10x + 3x

According to the question,

10(3x) + x + 10x + 3x = 88

or 30x + 14x = 88

or 44x = 88

Dividing both sides by 44, we get

x = 2

Hence the two digit number is 10(6) + 2 = 62

Note that the two-digit number can also be found if we reverse the digits. So one more possible number is 26.

Answer: Let us assume that the Shobo's present age be x.

So, Shobo's mother's present age = 6x.

Shobo's age five years from now will be = x + 5.

It is given that Shobo’s age five years from now will be one-third of his mother’s present age.

So,

$x + 5 = \left ( \frac{1}{3} \right )\times6x = 2x$

or x + 5 = 2x

Transposing x to the RHS,

5 = 2x - x = x

Hence Shobo's present age is x = 5 years

and Shobo's mother's present age is 6x = 30 years

Answer: Let the length and breadth of the plot be 11x and 4x respectively.

(Since they are in the ratio11:4)

We know that the fencing will be done on the boundary of the plot, so we need to calculate its perimeter.

Perimeter of rectangle = 2(lenght + breadth)

Total cost to fence the plot = Rs.75000 = Rs.100(Perimeter of plot)

Thus the equation becomes,

100 [2(11x + 4x)] = 75000

or 200(15x) = 75000

or 3000x = 75000

Dividing both sides by 3000:

x = 25

Hence length of plot is 11x = 275 m

and breadth of the plot is 4x = 100m

Answer: Let the total shirt material that Hasan bought be x metre.

and the total trouser material is y metre.

As per the question,

$\frac{x}{3} = \frac{y}{2}$

Thus trouser material $y = \frac{2}{3}x$

Now we will make another equation by using cost of the materials.

Profit for shirt is 12% i.e., selling price of shirt is

$=50 +\frac{12}{100}\times 50 = 50 + 6 = 56$

and selling price for trouser is

$=90 +\frac{10}{100}\times 90 = 90 + 9 = 99$

Hence equation becomes:

$x\left ( 56 \right ) + \frac{2}{3}x\left ( 99 \right ) = 36600$

or 56x + 66x = 36600

or 122x = 36600

Dividing both sides by 122:

x = 300

Thus trouser material that Hasan bought was = 200 meter

$\left ( \frac{2}{3}x = \frac{2}{3}\times300 = 200 \right )$

Answer: Let us assume the number of deer in the herd to be x.

Half of a herd of deer are grazing in the field i.e., $\frac{x}{2}$ .

Now half are remaining.

Three-fourths of the remaining are playing nearby, implies :

$\frac{3}{4}\times \frac{x}{2} = \frac{3x}{8}$

The remaining number of deer will be :

$\frac{x}{2} - \frac{3x}{8} = \frac{4x - 3x}{8} = \frac{x}{8}$

Given that rest is 9, so

$\frac{x}{8} = 9$

Hence the number of deer in the herd = 72

Answer: Let the age of granddaughter is x years.

So the age of grandfather is x + 54.

Also, grandfather is ten times older than his granddaughter.

i.e., x + 54 = 10x

Transposing x to the RHS :

54 = 10x - x

or 54 = 9x

or x = 6

Therefore age of granddaughter = 6 years

and the age of grandfather = 6 + 54 = 60 years

Answer: Let the age of Aman's son is x years.

So the age of Aman will be 3x years.

Ten years ago, the age of Aman's son = x - 10 and Aman's age = 3x - 10.

So according to question,

5(x - 10) = 3x - 10

or 5x - 50 = 3x - 10

Transposing 3x to the LHS and -50 to the RHS:

5x - 3x = -10 + 50 = 40

or 2x = 40

or x = 20

So Aman's age = 3x = 60 years and

Aman's son's age = x = 20 years.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Excercise: 2.5

Answer: The detailed solution of the above-written 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 $x = \frac{27}{10}$

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

Multiplying both sides 12.

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

6n - 9n + 10n = 252

or 7n = 252

or n = 36

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

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

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

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

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

Multiplying both sides by 15 . (because the LCM of denomenator 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

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

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

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

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

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

Multiplying both sides by 6 (as it is 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

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

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

Answer: Let us open the brackets

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

or 18y - 12 = 0

or 18y = 12

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

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

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

NCERT linear equations in one variable class 8 questions and answers - Exercise: 2.6

$\frac{8x-3}{3x} = 2$

Answer: We observe that the equation is not a linear equation since LHS is not linear.

So we multiply 3x to both the sides to make it linear.

We get, 8x - 3 = 2 $\times$ 3x

or 8x - 3 = 6x

or 8x - 6x = 3

or 2x = 3

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

Answer: We will convert the given equation into a linear equation by multiplying

(7 - 6x) to both sides.

Equation becomes: 9x = 15(7 - 6x)

or 9x = 105 - 90x

or 99x = 105

$x = \frac{105}{99} = \frac{35}{33}$

$\frac{z}{z+15} = \frac{4}{9}$

$\frac{z}{z+15} = \frac{4}{9}$

Cross-multiplication gives:

9z = 4z + 60

or 9z - 4z = 60

or 5z = 60

or z = 12

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

We have

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

By cross-multiplication we get:

5(3y + 4) = -2(2 - 6y)

or 15y + 20 = -4 + 12y

or 15y - 12y = -4 - 20

or 3y = -24

y = -8

$\frac{7y+4}{y+2} = \frac{-4}{3}$

$\frac{7y+4}{y+2} = \frac{-4}{3}$

Cross-multiplication gives,

3(7y + 4) = -4(y + 2)

or 21y + 12 = -4y - 8

or 21y + 4y = -8 - 12

or 25y = -20

$y = \frac{-20}{25} = \frac{-4}{5}$

Answer: Let the present ages of Hari and Harry be 5x and 7x respectively.

(Since there are in the ratio of 5:7)

Four years from age of Hari will be = 5x + 4

and Harry's age will be = 7x + 4.

According to question four years from now, the ratio of their ages will be 3:4.

So the equation becomes :

$\frac{5x + 4}{7x + 4} = \frac{3}{4}$

Cross-multiplication gives:

4(5x + 4) = 3(7x + 4)

or 20x + 16 = 21x + 12

or x = 4

Hence present age of Hari = 5x = 20 and

present age of Harry = 7x = 28

Answer: Let the numerator of the rational number be x.

Then denominator will be x + 8.

Further, according to question,

$\frac{x + 17}{x+ 8 - 1} = \frac{3}{2}$

or $\frac{x + 17}{x+ 7} = \frac{3}{2}$

Cross-multiplication gives:

2(x + 17) = 3(x + 7)

or 2x + 34 = 3x + 21

or x = 13

Hence the rational number is $\frac{13}{21}$ .

## Linear Equations In One Variable Class 8 NCERT Solutions - Topics

• Solving Equations which have Linear Expressions on one Side and Numbers on the other Side
• Some Applications
• Solving Equations having the Variable on both Sides
• Some More Applications
• Reducing Equations to Simpler Form
• Equations Reducible to the Linear Form

## NCERT Solutions for Class 8 Maths - Chapter Wise

 Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

## Key Features Of Linear Equations In One Variable Class 8 Solutions

Step-by-Step Solutions: Detailed step-by-step solutions of maths chapter 2 class 8 to each problem, ensuring clarity and understanding for students.

Variety of Problems: A diverse set of problems covering different aspects of linear equations in one variable. Practice class 8 maths ch 2 question answer to command the concepts.

Concept Clarity: A focus on explaining the fundamental concepts and principles related to linear equations in a clear and concise manner provided in this ch 2 maths class 8.

## NCERT Solutions for Class 8 -Subject Wise

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

• Go through some examples to get an idea about how to solve a question.
• Once you have identified the process, do the practice problems given in the exercises.
• For better preparation and time saving, you must use NCERT solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable.
• You can also practice previous year papers' questions to get perfection over the chapter.

Keep working hard and happy learning!

## Also Check NCERT Books and NCERT Syllabus here

1. What are the important topics of chapter Linear Equations in One Variable ?

Multiplication and division of algebraic equations, identities, factorization, and the method of solving linear equations in one variable are the important topics of this chapter.

2. Does CBSE class 8 maths is tough ?

CBSE class 8 maths is base and foundation for the upcoming classes. Most of the topics are related to previous classes. It is basic maths and very easy.

3. Does CBSE provides the solutions of NCERT for class 8 ?

No, CBSE doesn’t provide NCERT solutions for any class or subject.

4. Where can I find the complete solutions of NCERT for class 8 ?

Here you will get the detailed NCERT solutions for class 8 by clicking on the link.

5. Where can I find the complete solutions of NCERT for class 8 maths ?

Here you will get the detailed NCERT solutions for class 8 maths by clicking on the link.

6. Which is best book for CBSE class 8 maths ?

NCERT textbook is the best book for CBSE class 8 maths. You don't need to buy any supplementary book. All you need to do is a rigorous practice of NCERT problems.

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4 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Automation Test Engineer

An Automation Test Engineer job involves executing automated test scripts. He or she identifies the project’s problems and troubleshoots them. The role involves documenting the defect using management tools. He or she works with the application team in order to resolve any issues arising during the testing process.

2 Jobs Available