The document NCERT Solutions(Part- 3)- 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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**SOLVING EQUATIONS HAVING THE VARIABLE ON BOTH SIDES**

**Example:**

**Solve: **

**Ans:** We have

Transposing to RHS, we have

Transpiring 5x to LHS, we have

or

or

Dividing both sides bywe have

= -35/7 = -5

∴ x = -5

**EXERCISE 2.3 **

**Ques: **Solve the following equations and check your results.

1. 3x = 2x + 18

2. 5t – 3 = 3t – 5

3. 5x + 9 = 5 + 3x

4. 4z + 3 = 6 + 2z

5. 2x – 1 = 14 – x

6. 8x + 4 = 3(x – 1) + 7

7.

8.

9.

10. 3m = 5m 8/5

**Ans: ****1.** 3x = 2x + 18

Transposing 2x from RHS to LHS, we have

3x – 2x = 18

or

x = 18

**2.** 5t – 3 = 3t – 5

Transposing (–3) to RHS, we have

5t = 3t – 5 + 3

or

5t = 3t – 2

Transposing 3t to LHS, we have

5t – 3t = –2

or

2t = –2

Diving both sides by 2, we have

or

t = –1

**3.** 5x + 9 = 5 + 3x

Transposing 9 to RHS, we have

5x = 5 + 3x – 9

or

5x = –4 + 3x

Transposing 3x to LHS, we have

5x – 3x = –4

or

2x = –4

Dividing both sides by 2, we have

x = -4/2 = -2

or

x = –2

**4. **4z + 3 = 6 + 2z

Transposing 3 to RHS, we have

4z = 6 – 3 + 2z

or

4z = 3 + 2z

Transposing 2z to LHS, we have

4z – 2z = 3

or

2z = 3

Dividing both sides by 2, we have

z = 3/2

**5.** 2x – 1 = 14 – x

Transposing (–1) to RHS, we have

2x = 14 – x + 1

or

2x = 15 – x

Transposing (–x) to LHS, we have

2x + x = 15

or

3x = 15

Dividing both sides by 3, we have

x = 15/3 = 5

∴ x = 5

**6.** 8x + 4 = 3(x – 1) + 7

or

8x + 4 = 3x – 3 + 7

or

8x + 4 = 3x + 4

Transposing 4 to RHS, we have

8x = 3x + 4 – 4

Transposing 3x to LHS, we have

8x – 3x = 0

or

5x = 0

∴ x = 0

**7. **

or

or

Transposing x to LHS, we have

or

or x/5 = 8

Multiplying both sides by 5, we have

x = 8 * 5 = 40

∴ x = 40

**8. **

Transposing 1 to RHS, we have

or

Transposing 7x/15 to LHS, we have

or

or 3x/15 = 2

Multiplying both sides by 15, we have

3x = 2 * 15 = 30

Dividing both sides by 3, we have

∴ x = 10

**9. **

Transposing 5/3 to RHS, we have

Transposing (–y) to LHS, we have

or 3y = 7

Dividing both sides by 3, we have

3y = 3 = 7/3

∴ y = 7/3

**10.** 3m = 5m

Transposing 5m to LHS, we have

3m – 5m =

or –2m =

Dividing both sides by (–2), we have

∴ m = 4/5

**Some More Application**

**Example: **The digits of a two digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, then we get 99. Find the original number.

**Ans:** Let the digit at unit place be ‘x’.

∴ The digit at the tens place = (x + 5)

∴ Original number = 10(x + 5) + x

With interchange of digits, the new number = 10x + (x + 5)

Now, According to the condition, we have

[original number] + [New number] = 99

or [10(x + 5) + x] + [10x + (x + 5)] = 99

or [10x + 50 + x] + [10x + x + 5] = 99

or 11x + 50 + 11x + 5 = 99

or 22x + 55 = 99

Transposing 55 to RHS, we have

22x = 99 – 55 = 44

Dividing both sides by 22, we have

x = 44 ÷ 22 = 2

∴ x = 2

i.e. Unit place digit = 2

∴Tens place digit = 2 + 5 = 7

Thus the original number = 72.

**Note:** The statement of the above example is valid for both 72 and 27 and both are correct answers.

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