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## NCERT Class 8 Maths Textbook Chapter 2 With Answer Book PDF Free Download

### Chapter 2: Linear Equations In One Variable

#### 2.3 Some Applications

We begin with a simple example. The sum of the two numbers is 74. One of the numbers is 10 more than the other. What are the numbers?

We have a puzzle here. We do not know either of the two numbers, and we have to find them. We are given two conditions.

(i) One of the numbers is 10 more than the other.

(ii) Their sum is 74.

We already know from Class VII how to proceed. If the smaller number is taken to be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that the sum of these two numbers x and x + 10 is 74.

This means that x + (x + 10) = 74.

or 2x + 10 = 74

Transposing 10 to RHS, 2x = 74 – 10

or 2x = 64

Dividing both sides by 2, x = 32. This is one number.

The other number is x + 10 = 32 + 10 = 42

The desired numbers are 32 and 42. (Their sum is indeed 74 as given and also one

the number is 10 more than the other.)

We shall now consider several examples to show how useful this method is

#### 2.4 Solving Equations having the Variable on Both Sides

An equation is the equality of the values of two expressions. In the equation 2x – 3 = 7, the two expressions are 2x – 3 and 7. In most examples that we have come across so far, the RHS is just a number.

But this need not always be so; both sides could have expressions with variables.

For example, the equation 2x – 3 = x + 2 has expressions with a variable on both sides; the expression on the LHS is (2x – 3) and the expression on the RHS is (x + 2).

We now discuss how to solve such equations which have expressions with the variable on both sides.

Author | NCERT |

Language | English |

No. of Pages | 16 |

PDF Size | 1567 KB |

Category | Mathematics |

Source/ Credits | ncert.nic.in |

### NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable pdf

**1. x – 2 = 7**

**Solution:**

x – 2 = 7

x=7+2

x=9

**2. y + 3 = 10**

**Solution:**

y + 3 = 10

y = 10 –3

y = 7

**3. 6 = z + 2**

**Solution:**

6 = z + 2

z + 2 = 6

z = 6-2

z=4

**4. 3/7 + x = 17/7**

**Solution:**

3/7 + x = 17/7

x = 17/7 – 3/7

x = 14/7

x = 2

**5. 6x = 12**

**Solution:**

6x = 12

x = 12/6

x = 2

**6. t/5 = 10**

**Solution:**

t/5 = 10

t = 10 × 5

t = 50

**7. 2x/3 = 18**

**Solution:**

2x/3 = 18

2x = 18 × 3

2x = 54

x = 54/2

x = 27

**8. 1.6 = y/15**

**Solution:**

1.6 = y/1.5

y/1.5 = 1.6

y = 1.6 × 1.5

y = 2.4

**9. 7x – 9 = 16**

**Solution:**

7x – 9 = 16

7x = 16+9

7x = 25

x = 25/7

**10. 14y – 8 = 13**

**Solution:**

14y – 8 = 13

14y = 13+8

14y = 21

y = 21/14

y = 3/2

**11. 17 + 6p = 9**

**Solution:**

17 + 6p = 9

6p = 9 – 17

6p = -8

p = -8/6

p = -4/3

**12. x/3 + 1 = 7/15**

**Solution:**

x/3 + 1 = 7/15

x/3 = 7/15 – 1

x/3 = (7 -15)/15

x/3 = -8/15

x = -8/15 × 3

x = -8/5

NCERT Class 8 Maths Textbook Chapter 2 With Solution Book PDF Free Download