In this page we have **class 8 maths chapter 2 exercise 2.1 solutions ** .This exercise has questions about simple linear equation where we have linear expression on Present on both the left and right side of the equation. Hope you like them and do not forget to like , social share
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(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$

- NCERT Solutions for Linear equations Class 8 maths Chapter 2 Exercise 2.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

Download class 8 maths chapter 2 exercise 2.1 solutions as pdf - This chapter 2 has total 2 Exercise 2.1,2.2. Exercise 2.3,2.4,2.5.2.6 has been deleted as per latest syllabus. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

**Notes****Worksheets****Ncert Solutions**- class 8 maths chapter 2 exercise 2.1 solutions
- class 8 maths chapter 2 exercise 2.2 solutions
- NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.3
- NCERT Solutions for Linear equations Class 8 maths Chapter 2 Exercise 2.4
- class 8 maths chapter 2 exercise 2.5 solutions
- class 8 maths chapter 2 exercise 2.6 solutions

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