The document NCERT Solutions(Part- 1)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by Full Circle.

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**Ques: **Solve the following equations.

1. x – 2 = 17

2. y + 3 = 10

3. 6 = z + 2

4.

5. 6x = 12

9. 7x – 9 = 16

10. 14y – 8 = 13

11. 17 + 6p = 9

**Ans: ****1.** x – 2 = 7

Transposing (–2) to RHS, we have

x = 7 + 2

∴ x = 9

**2.** y + 3 = 10

Transposing 3 to RHS, we have

y = 10 – 3

∴ y = 7

**3.** 6 = z + 2

Transposing 2 to LHS, we have

6 – 2 = z or

4 = z

∴ z = 4

Transposing 3/7 to RHS, we have

= 14/7 = 2

= 2

**5.** 6x = 12 Dividing both sides by 6, we have:

∴ x = 2

**6.** t/5 = 10

Multiplying both sides by 5, we have

∴ t = 50

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

Multiplying both sides by 3, we have

or 2x = 54

Dividing both sides by 2, we have

or

x = 54/2 = 27

∴ x = 27

Dividing both sides by 2, we hav**8.** 1.6 = y/1.5

Multiplying both sides by 1.5, we have

or

or

2.4 = y

∴ y = 2.4

**9.** 7x – 9 = 16

Transposing (–9) to RHS, we have

7x = 16 + 9 = 25

Dividing both sides by 7, we have

∴ x = 25/7

**10.** 14y – 8 = 13

Transposing –8 to RHS, we have

14y = 13 + 8 = 21

Dividing both sides by 14, we have

or

∴ y = 3/2

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

Transposing 17 to RHS, we have

6p = 9 – 17 = –8

Dividing both sides by 6, we have

or

∴

Transposing 1 to RHS, we have

or

Multiplying both sides by 3, we have

∴ x = -8/5

**SOME APPLICATIONS**

**Example 1: What should be subtracted from twice the rational number ****-5/3 to get 2/5 ?**

**Ans: **Twice of

Let the required number to be subtracted be x.

∴

Transposing -10/3 to RHS, we have

∴

Thus, the required number is

**Example 2: ****The perimeter of a rectangle is 17 cm. If its width is ****cm, then find its length.**

**Ans:**

Let the length of the rectangle = l cm

∴ Since, width of the rectangle (b) = cm

and perimeter = 17 cm

We know that perimeter of a rectangle = 2(l + b) cm

Then,

or

or

or

or

or

Dividing both sides by 2, we have

Length of the rectangle =

**Example 3:** **The sum of three consecutive multiples of 7 is 63. Find these multiples.****Ans:** Let x be a multiple of 7.

∴ The next two consecutive multiples of 7 are x + 7 and x + 7 + 7 (i.e. x + 14).

∴ Three consecutive multiples of 7 are: x, x + 7 and x + 14

According to the condition,

x + (x + 7) + (x + 14) = 63

∴ 3x + 21 = 63

Transposing 21 to RHS, we have

3x = 63 – 21 = 42

Dividing both sides by 3, we have

3x/3 = 42/14

x = 14

∴ x + 7 = 14 + 7 = 21

and x + 14 = 14 + 14 = 28

Thus, the three consecutive multiples of 7 are: 14, 21 and 28.

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