NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3

December 4, 2021

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.

Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
Transposing 2x from RHS to L.H.S, we get
3x – 2x = 18
x = 18

Check:
Put x = 18 in L.H.S. and R.H.S. of the equation.
L.H.S. = 3x = 3 × 18 = 54
R.H.S. = 2x + 18 = 2(18) + 18 = 36 +18 = 54
∴ L.H.S. = R.H.S

Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
Transposing (-3) to R.H.S. and 3t to L.H.S., we get
5t – 3t = -5 + 3
2t = -2
Dividing both sides by 2, we get
2t/2=−2/2
t = -1

Check:
Put t = -1 in L.H.S. and RHS of the equation
L.H.S. = 5t – 3 = 5(-1) – 3 = -5 – 3 = -8
R.H.S. = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8
Hence, L.H.S. = R.H.S.

Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
Transposing 9 to R.H.S. and 3x to L.H.S.
5x – 3x = 5 – 9
2x = -4
Dividing both sides by 2, we get
2x/2=−4/2
x = -2

Check:
Put x = -2 in L.H.S. and R.H.S. of the equation
L.H.S. = 5x + 9
= 5 (-2) + 9
= -10 + 9
= -1
R.H.S. = 5 + 3x
= 5 + 3(-2)
= 5 – 6 = -1
Hence, L.H.S. = R.H.S.

Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
Transposing 3 to R.H.S. and 2z to LHS
4z – 2z = 6 – 3
2z = 3
Dividing both sides by 2, we get
2z/2=3/2
z = 3/2

Check:
Put z = 3/2 in L.H.S. and R.H.S of the equation
L.H.S. = 4z + 3
= 4(3/2) + 3
= 2(3) + 3
= 6 + 3
= 9
R.H.S = 6 + 2z
= 6 + 2(3/2)
= 6 + 3
= 9
Hence, L.H.S. = R.H.S.

Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
Transposing -1 to RHS and -x to L.H.S.
2x + x = 14 + 1
3x = 15
Dividing both sides by 3, we get
3x/3=15/3
x = 5

Check:
Put x = 5 in LHS and RHS of the equation
LHS = 2x – 1
= 2(5) – 1
= 10 – 1
= 9
R.H.S = 14 – x
= 14 – 5
= 9
Hence, L.H.S. = R.H.S.

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4
Transposing 4 to R.H.S. and 3x to LHS, we get
8x – 3x = 4 – 4
5x = 0
Dividing both sides by 5
5x/5=0/5
x = 0

Check:
Put x = 0 in L.H.S. and R.H.S. of the equation
L.H.S. = 8x + 4
= 8(0) + 4
= 4
R.H.S = 3(x – 1) + 7
= 3(0 – 1) + 7
= -3 + 7
= 4
Hence, L.H.S. = R.H.S.

Question 7.