RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.1

Solve the following equations:
Question 1.
x - 2 = 7
Answer:
Given    x - 2 = 7
or x - 2 + 2 = 7 + 2
(Adding 2 to both sides)
or x = 9

Verification: Putting x = 9 in given equation
LHS = 9 - 2 = 7 = RHS
Hence, x = 9 is a solution of given equation.

Question 2.
y + 3 = 10
Answer:
Given, y + 3= 10
or y + 3 - 3 = 10 - 3
(Subtracting 3 from both sides)
or y = 7

Verification:
Putting y = 7 in given equation
LHS = 7 + 3 = 10 = RHS
Hence, y = 7 is a solution of given equation.

Question 3.
6 = z + 2
Answer:
Given, 6 = z    + 2
or 6 - 2 = z + 2 - 2
(Subtracting 2 from both sides)
or 4 =    z
Hence, z = 4

Verification:
Putting z = 4 in the given equation
RHS =4 + 2 = 6 LHS
Hence z = 4 is a solution of given equation.

Question 4.
\(\frac{3}{7}\) + x = \(\frac{17}{7}\)
Answer:
Given \(\frac{3}{7}\)+ x = \(\frac{17}{7}\)
or \(\frac{3}{7}-\frac{3}{7}\) + x = \(\frac{17}{7}-\frac{3}{7}\)
(Subtracting \(\frac{3}{7}\) from both sides)
or x = \(\frac{14}{7}\)
or x = 2

Verification:
Putting x = 2 in the given equation
LHS = \(\frac{3}{7}\) + 2
= \(\frac{3+14}{7}=\frac{17}{7}\) = RHS
Hence x = 2 is a solution of given equation.

Question 5.
6x = 12
Answer:
Given 6x = 12
or \(\frac{6x}{6}=\frac{12}{6}\)
(Dividing both sides by 6)
or x = 2

Verification:
Putting x = 2 in the given equation
LHS = 6 × 2 = 12 = RHS
Hence x = 2 is a solution of given equation. 

Question 6.
\(\frac{t}{5}\) = 10
Answer:
Given \(\frac{t}{5}\) = 10
or \(\frac{t}{5}\) × 5 = 10 × 5
(Multiplying both sides by 5)
or t = 50

Verification:
Putting t = 50 in the given equation
LHS = \(\frac{50}{5}\) = 10
Hence t = 50 is a solution of given equation. 

Question 7.
\(\frac{2x}{3}\) = 18
Answer:
Given \(\frac{2x}{3}\) = 18
or \(\frac{2}{3}\) × x × \(\frac{3}{2}\) = 18 × \(\frac{3}{2}\)
(Multiplying both sides by \(\frac{3}{2}\))
or x = 27

Verification:
Putting x = 27 in the given equation
LHS = \(\frac{2 \times 27}{3}\) = 2 × 9 = 18
= RHS
Hence x = 27 is a solution of given equation.

Question 8.
1.6 = \(\frac{y}{1.5}\)
Answer:
Given 1.6 = \(\frac{y}{1.5}\)
or 1.6 × 1.5 = \(\frac{y}{1.5}\) × 1.5
(Multiplying both sides by 1.5)
or 2.4 = y
Hence y = 2.4

Verification:
Putting y = 2.4 in the given equation
RHS = \(\frac{2.4}{1.5}\) = 1.6 = LHS
Hence y = 2.4 is a solution of given equation.

Question 9.
7x - 9=16
Answer:
Given, 7x - 9= 16
or 7x - 9 + 9= 16 + 9
(Adding 9 to both sides)
or 7x= 25
or \(\frac{7 x}{7}=\frac{25}{7}\)
(Dividing both sides by 7)
or x= \(\frac{25}{7}\)

Verification:
Putting x = \(\frac{25}{7}\) given equation
LHS = 7 × \(\frac{25}{7}\) - 9
= 25 - 9 = 16 = RHS
Hence, x = \(\frac{25}{7}\) is a solution of given equation.

Question 10.
14y - 8 = 13
Answer:
Given, 14y - 8 = 13
or 14y - 8 + 8 = 13 + 8
(Adding 8 to both sides) 
or 14y = 21
or \(\frac{14 y}{14}=\frac{21}{14}\)
(Dividing both sides by 14)
or y = \(\frac{3}{2}\)

Verification:
Putting y = \(\frac{3}{2}\) in given equation
LHS = 14 × \(\frac{3}{2}\) -8
= 7 × 3 - 8
= 21 - 8
= 13 = RHS
Hence x = \(\frac{3}{2}\) is a solution of given equation.

Question 11.
17 + 6p = 9
Answer:
Given 17 + 6p = 9
or 17 - 17 + 6p = 9 - 17
(Subtracting 17 on both sides)
or 6p = -8
or \(\frac{6 p}{6}=\frac{-8}{6}\)
(Dividing both sides by 6)
or p = \(\frac{-4}{3}\)

Verification:
Putting p = \(\frac{-4}{3}\) in given equation
LHS = 17 + 6 × \(\left(\frac{-4}{3}\right)\)
= 17 - \(\frac{24}{3}\)
= 17 - 8 = 9 = RHS
Hence p = \(\frac{-4}{3}\) is a solution of given equation.

Question 12.
\(\frac{x}{3}\) + 1 = \(\frac{7}{15}\)
Answer:
Given \(\frac{x}{3}\) + 1 = \(\frac{7}{15}\)
or \(\frac{x}{3}\) + 1 - 1 = \(\frac{7}{15}\) - 1
(Subtracting 1 from both sides)
or \(\frac{x}{3}=\frac{7-15}{15}=\frac{-8}{15}\)
or \(\frac{x}{3}\) × 3 = \(\frac{-8}{15}\) × 3
(Multiplying both sides by 3)
or x = \(\frac{-8}{15}\)

Verification:
Putting x = \(\frac{-8}{15}\) in given equation
LHS = \(\frac{1}{3} \times \frac{-8}{5}\) + 1
= \(\frac{-8}{15}\) + 1
= \(\frac{-8+15}{15}\)
= \(\frac{7}{15}\) = RHS
Hence x = \(\frac{-8}{15}\) is a solution of given equation.