Ex 2.1 NCERT Solutions: Linear Equations in One Variable

The linear equations in one variable is an equation which is expressed in the form of ax+b = 0 or ax+b= c, where  x is a variable, a is the coefficient of x, b and c are constants

• For example, 2x+3=8 is a linear equation having a single variable in it. Therefore, this equation has only one solution, which is x = 5/2. 

• Whereas if we speak about linear equations in two variables, it has two solutions.

Let's have a look at NCERT Solutions of Linear Equations in One Variable:

Solve the following equations

Q1. x - 2 = 7 
Ans:  Transposing -2 to RHS (change its sign) gives:
⇒ x = 7 + 2
⇒ x = 9
∴ x = 9

Note: By transposing a term from one side to another side, we mean changing its sign and carrying it to the other side. 

Q2. y + 3 = 10 
Ans: Transposing +3 to RHS (change its sign):
⇒ y = 10 - 3
⇒ y = 7
∴ y = 7

Q3. 6 = z + 2
Ans: Transpose +2 to LHS (change its sign):
⇒ 6 - 2 = z
⇒ 4 = z
∴ z = 4

Q4. (3/7) + x = (17/7) 
Ans: Transposing 3/7 to RHS (change its sign):

⇒ x = 14/7

⇒ x = 2

Q5. 6x = 12

Ans: Divide both sides by 6 to isolate x:
⇒ x = 12 ÷ 6

⇒ x = 2

Q6. (t/5) = 10 

Ans: Multiply both sides by 5 to get t alone:

∴ t = 50

Q7. (2x/3) = 18
Ans: Multiply both sides by 3 to remove denominator:

⇒ 2x = 18 × 3 = 54

Now divide both sides by 2:

⇒ x = 54 ÷ 2

⇒ x = 27

Q8. 1.6 = (y/1.5) 
Ans: Multiply both sides by 1.5 to isolate y:

⇒ 

⇒ y = 2.4

Q9. 7x - 9 = 16

Ans: Transpose -9 to RHS (change its sign):
7x = 16 + 9 = 25
Divide both sides by 7:

⇒ x = 25/7

Q10. 14y - 8 = 13
Ans: Transpose -8 to RHS (change its sign):
14y = 13 + 8 = 21
Now divide both sides by 14:

⇒

⇒ y = 3/2

Q11. 17 + 6p = 9

Ans: Transpose +17 to RHS (change its sign):
6p = 9 - 17 = -8
Now divide both sides by 6:

⇒

∴  

Q12. (x/3) + 1 = (7/15)

Ans: Transpose +1 to RHS (change its sign):

 

Multiplying both sides by 3, we have:

⇒ x = -8/5