In this page we have **NCERT Solutions for Linear equations Class 8 Mathematics ** for
Exercise 2.3 . Hope you like them and do not forget to like , social share
and comment at the end of the page.

(1) Validation or check result means to verify the answer by putting the values in the linear equation

(2) Calculate the LHS(Left Hand side) by substituting the variable and similary calculate the RHS(Right hand side) by substituting the variable

(3) Both the LHS and RHS should be equal

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

$3x= 2x+ 18$

Transposing 2x to L.H.S, we obtain

$3x- 2x= 18$

x= 18

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 3x= 3 × 18 = 54

R.H.S = 2x+ 18 = 2 × 18 + 18 = 36 + 18 = 54

L.H.S. = R.H.S.

It proves that result is correct

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

$5t- 3 = 3t- 5$

Transposing 3t to L.H.S and -3 to R.H.S, we obtain

$5t- 3t= -5 - (-3)$

2t= -2

Dividing both sides by 2

t= -1

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 5t- 3 = 5 × (-1) - 3 = -8

R.H.S = 3t- 5 = 3 × (-1) - 5 = - 3 - 5 = -8

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result

$5x+ 9 = 5 + 3x$

$5x+ 9 = 5 + 3x$

Transposing 3x to L.H.S and 9 to R.H.S, we obtain

$5x- 3x= 5 - 9$

2x= -4

Dividing both sides by 2, we obtain

x= -2

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 5x+ 9 = 5 × (-2) + 9 = -1

R.H.S = 5 + 3x= 5 + 3 × (-2) = -1

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

$4z+ 3 = 6 + 2z$

$4z+ 3 = 6 + 2z$

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

z=3/2

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 4z+ 3 = 4 × (3/2)+ 3 = 6 + 3 = 9

R.H.S = 6 + 2z= 6 + 2 × (3/2)= 6 + 3 = 9

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

$2x- 1 = 14 -x$

$2x- 1 = 14 -x$

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

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 2x- 1 = 2 × (5) - 1 = 10 - 1 = 9

R.H.S = 14 -x= 14 - 5 = 9

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

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

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

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

Transposing 3x to L.H.S and 4 to R.H.S, we obtain

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

$5x= - 7 + 7$

5x=0

x=0

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = 8x+ 4 = 8 × (0) + 4 = 4

R.H.S = 3(x- 1) + 7 = 3 (0 - 1) + 7 = - 3 + 7 = 4

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

Multiplying both sides by 5, we obtain

$5x= 4(x+ 10)$

$5x= 4x+ 40$

Transposing 4x to L.H.S, we obtain

5x- 4x= 40

x= 40

Let us evaluate both the LHS and RHS for validate the answer

L.H.S =x= 40

R.H.S = =40

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

Transposing 7x/15 on LHS and 1 on RHS

Multiplying by 15 on both sides

10x-7x= 30

3x=30

Dividing by 3 on both the sides

x=10

Let us evaluate both the LHS and RHS for validate the answer

L.H.S = =23/3

R.H.S= =23/3

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result

Transposingyto L.H.S and5/3 to R.H.S, we obtain

3y=21/3

3y=7

Dividing both sides by 3, we obtain

y=7/3

Let us evaluate both the LHS and RHS for validate the answer

L.H.S. = R.H.S.

It proves that result is correct

Solve and check result:

Transposing 3m to R.H.S and 8/5 to L.H.S

8/5= 2m

Dividing both sides by 2

m=4/5

Let us evaluate both the LHS and RHS for validate the answer

L.H.S =3m=12/5

R.H.S =5m –(8/5)=12/5

L.H.S. = R.H.S.

It proves that result is correct

(2) Solve and check the result $3x - 4 = 1 - 2 x$

(3) Solve and check the result $ \frac {m}{2} -11= 2$

(4) Solve and check the result $ 11y =6y - \frac {1}{3}$

(5) Solve the check the result $ \frac {x}{5} + 8=30$

Download Linear equation Exercise 2.3 as pdf

**Notes****Worksheets****Ncert Solutions**- NCERT Solution Linear equation in One variable Exercise 2.1
- NCERT Solution Linear equation in One variable Exercise 2.2
- NCERT Solution Linear equation in One variable Exercise 2.3
- NCERT Solution Linear equation in One variable Exercise 2.4
- NCERT Solution Linear equation in One variable Exercise 2.5
- NCERT Solution Linear equation in One variable Exercise 2.6

Class 8 Maths Class 8 Science

Given below are the links of some of the reference books for class 8 Math.

- Mathematics Foundation Course for JEE/Olympiad : Class 8 This book can take students maths skills further. Only buy if child is interested in Olympiad/JEE foundation courses.
- Mathematics for Class 8 by R S Aggarwal Detailed Mathematics book to clear basics and concepts. I would say it is a must have book for class 8 student.
- Pearson Foundation Series (IIT -JEE / NEET) Physics, Chemistry, Maths & Biology for Class 8 (Main Books) | PCMB Combo : These set of books could help your child if he aims to get extra knowledge of science and maths. These would be helpful if child wants to prepare for competitive exams like JEE/NEET. Only buy if you can provide help to the child while studying.
- Reasoning Olympiad Workbook - Class 8 :- Reasoning helps sharpen the mind of child. I would recommend students practicing reasoning even though they are not appearing for Olympiad.

You can use above books for extra knowledge and practicing different questions.

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