You can Download KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Questions and Answers helps you to revise the complete syllabus.
KSEEB Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1
Solve the following equations.
Question 1.
x – 2 = 7
Solution:
We have x – 2 = 7
x – 2 = 7
x = 7 + 2
x = 9
Check: L.H.S = 9 – 2 = 7 = R.H.S
Question 2.
y + 3 = 10
Solution:
y + 3 = 10
y + 3 – 3 = 10 – 3 [Subtracting 3 from both sides]
y = 7
Check: Substituting y = 7 in the given equation
L.H.S = 7 + 3 = 10 = R.H.S
Question 3.
6 = z + 2
Solution:
6 = z + 2
6 – 2 = z + 2 – 2 [Subtracting 2 from bcth sides]
4 = z
z = 4
Check: Substituting the value of z = 4 in R.H.S.
R.H.S = z + 2 = 4 + 2 = 6 = L.H.S
Question 4.
\(\frac{3}{7}+x=\frac{17}{7}\)
Solution:
\(\frac{3}{7}+x=\frac{17}{7}\)
\(\frac{3}{7}-\frac{3}{7}+x=\frac{17}{7}-\frac{3}{7}\) [Subtracting \(\frac{3}{7}\) from both sides]
x = \(\frac{17-3}{7}\)
x = \(\frac{14}{7}\)
x = 2
Check: Substituting the value of x = 2 in the given equation
L.H.S = \(\frac{3}{7}+2=\frac{3+2 \times 7}{7}\)
= \(\frac{3+14}{7}\)
= \(\frac{17}{7}\)
= R.H.S
Question 5.
6x = 12
Solution:
6x = 12
\(\frac{6 x}{6}=\frac{12}{6}\) [Dividing both sides by 6]
x = 2
Check: Substituting the value of x = 2 in the given equation
L.H.S = 6 × 2 = 12 = R.H.S
Question 6.
\(\frac{t}{5}\) = 10
Solution:
\(\frac{t}{5}\) = 10
\(\frac{t}{5}\) × 5 = 10 × 5 [Multiplying both sides by 5]
t = 50
Check: Substituting t = 50 in the given equation, we get
LHS = \(\frac{50}{5}\) = 10 = RHS
Question 7.
\(\frac{2 x}{3}\) = 18
Solution:
\(\frac{2 x}{3}\) = 18
\(\frac{2 x}{3}\) × 3 = 18 × 3 [Multiplying both sides by 3]
2x = 54
\(\frac{2 x}{2}=\frac{54}{2}\) [Dividing both sides by 2]
x = 27
Check: Substituting x = 2 in the given equation, we get
L.H.S = \(\frac{2}{3}\) × 27
= 2 × 9
= 18
= L.H.S
Question 8.
1.6 = \(\frac{y}{1.5}\)
Solution:
1.6 = \(\frac{y}{1.5}\)
1.6 × 1.5 = \(\frac{y}{1.5}\) × 1.5 [Multipljdng both sides by 1.5]
2.4 = y
y = 2.4
Check: Substituting y = 2.4 in the given equation, we get
R.H.S = \(\frac{y}{1.5}=\frac{2.4}{1.5}=\frac{24}{15}\) = 1.6 = RHS
Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
7x – 9 + 9 = 16 + 9 [Adding 9 on both sides]
7x = 25
\(\frac{7 x}{7}=\frac{25}{7}\) [Dividing both sides by 7]
x = \(\frac{25}{7}\)
Check: Substituting x = \(\frac{25}{7}\) in the given equation, we get
L.H.S = 7 × \(\frac{25}{7}\) – 9
= 25 – 9
= 16
= R.H.S
Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
14y – 8 + 8 = 13 + 8 [Adding 8 on both sides]
14y = 21
\(\frac{14 y}{14}=\frac{21}{14}\) [Dividing both sides by 14]
y = \(\frac{21}{14}\) = \(\frac{3}{2}\)
Check: Substituting y = \(\frac{3}{2}\) in the given equation, we get
L.H.S = 14 × \(\frac{3}{2}\) – 8
= 7 × 3 – 8
= 21 – 8
= 13
= R.H.S
Question 11.
17 + 6p = 9
Solution:
17 + 6p = 9
17 – 17 + 6p = 9 – 17 [Subtracting 17 on both sides]
6p = -8
\(\frac{6 p}{6}=\frac{-8}{6}\) [Dividing both sides by 6]
p = \(\frac{-4}{3}\)
Check: Substituting p = \(\frac{-4}{3}\) in the given equation, we get
L.H.S = 17 + 6 × \(\frac{-4}{3}\)
= 17 + 2 × -4
= 17 – 8
= 9
= R.H.S
Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:
\(\frac{x}{3}+1=\frac{7}{15}\)
\(\frac{x}{3}+1-1=\frac{7}{15}-1\) [Subtracting 1 on both sides]
\(\frac{x}{3}=\frac{7-15}{15}\)
\(\frac{x}{3}=\frac{-8}{15}\)
\(\frac{x}{3} \times 3=\frac{-8}{15} \times 3\) [Multiplying by 3 on both sides]
x = \(\frac{-8}{5}\)
Check : Substituting the value of x = \(\frac{-8}{5}\) in the given equation, we get
L.H.S = \(\frac{-8}{5 \times 3}+1\)
= \(\frac{-8}{15}+1\)
= \(\frac{-8+15}{15}\)
= \(\frac{7}{15}\)
= R.H.S