You can Download KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers helps you to revise the complete syllabus.
KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3
Solve the following equations and check your results:
Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
⇒ 3x – 2x = 18
⇒ x = 18
Check: LHS = 3 × 18 = 54
R.H.S = 2 × 18 + 18 = 36 + 18 = 54
54 = 54
L.H.S = R.H.S
Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
⇒ 5t – 3t = -5 + 3
⇒ 2t = -2
⇒ t = -1
Check: L.H.S = 5(-1) – 3 = -5 – 3 = -8
R.H.S = 3(-1) – 5 = -3 – 5 = -8
-8 = -8
L.H.S = R.H.S
Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
⇒ 5x – 3x = 5 – 9
⇒ 2x = -4
⇒ x = -2
Check: L.H.S = 5(-2) + 9 = -10 + 9 = -1
R.H.S = 5 + 3(-2) = 5 – 6 = -1
-1 = -1
L.H.S = R.H.S
Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
⇒ 4z – 2z = 6 – 3
⇒ 2z = 3
⇒ z = \(\frac{3}{2}\)
L.H.S = 4(\(\frac{3}{2}\)) + 3 = 6 + 3 = 9
R.H.S = 6 + 2(\(\frac{3}{2}\)) = 6 + 3 = 9
9 = 9
L.H.S = R.H.S
Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
⇒ 2x + x = 14 + 1
⇒ 3x = 15
⇒ x = \(\frac{15}{3}\) = 5
Check: L.H.S = 2(5) – 1 = 10 – 1 = 9
R.H.S = 14 – 5 = 9
9 = 9
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 – 3x = -3 + 7 – 4
⇒ 5x = -7 + 7
⇒ 5x = 0
⇒ x = 0
Check: L.H.S = 8(0) + 4 = 0 + 4 = 4
R.H.S = 3(0 – 1) + 7 = -3 + 7 = 4
4 = 4
L.H.S = R.H.S
Question 7.
x = \(\frac{4}{5}\) (x + 10)
Solution:
x = \(\frac{4}{5}\) (x + 10)
⇒ 5x = 4(x + 10)
⇒ 5x = 4x + 40
⇒ 5x – 4x = 40
⇒ x = 40
Check: LHS = 5 × 40 = 200
RHS = 4(40 + 10) = 4(50) = 200
200 = 200
L.H.S = R.H.S
Question 8.
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Solution:
Check:
Question 9.
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:
\(y=\frac{21}{3 \times 3}=\frac{7}{3}\)
Check:
Question 10.
3m = 5m – \(\frac{8}{5}\)
Solution:
Check:
