The document NCERT Solutions(Part- 5)- Linear Equations in One Variable Class 8 Notes | EduRev is a part of the Class 8 Course Mathematics (Maths) Class 8.

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**REDUCING EQUATIONS TO SIMPLER FORM**

**Ques 1: **

**Ans: **

Multiplying both sides by 6, we have (∵ LCM of 3 and 6 is 6)

or

x + 3 + 6 = 2(6x – 1)

or

x + 9 = 12x – 2

9 to RHS and 12x to LHS, we have

x – 12x = –2 – 9

or

–11x = –11

or

x = -11/-11 = 1 (Dividing both sides by (–11))

∴ x = 1

Check:

∴ LHS = RHS

**Ques 2:** **Solve: 2(3x + 1) – 5x **

**Ans:** First opening the brackets,

LHS = 2(3x + 1) – 5x

= 6x + 2 – 5x

= (6 – 5)x + 2 = x + 2

Now, the equation is

Transposing 2 to RHS and –4x to LHS, we have

or

or

(Dividing both sides by 5)

∴ x = -5/2

Check: LHS = 2(3x + 1) – 5x

∴ LHS = RHS

**EXERCISE 2.5 ****Question 1:** **Solve the following linear equations.**

1.** **

2.

3.

4.

5.

6.

**Ans: 1. **

Transposing to RHS and x/3 to LHS, we have

or

or

or

(Multiplying both sides by 6)

Check:

∴ LHS = RHS

**2. **

∵ LCM of 2, 4 and 6 = 12

∴ Multiplying both sides by 12, we have

or

6n – 9n + 10n = 252

or

7n = 252

or

n = 252/7 = 36

∴ n = 36

Check:

∴ LHS = RHS

**3. **

∵ LCM of 3, 6 and 2 is 6.

∴ Multiplying both sides by 6, we have

or 6x + 42 – 16x = 17 – 15x

or (6 – 16)x + 42 = 17 – 15x

or –10x + 42 = 17 – 15x

Transposing 42 to RHS and –15x to LHS, we have

–10x + 15x = 17 – 42 or 5x = –25

or

5x = –25

or

x = -25/5 = -5 (Dividing both sides by 5)

∴ x = –5

Check:

∴ LHS = RHS

**4. **

∵ LCM of 3 and 5 is 15.

∴ Multiplying both sides by 15, we have

or

5(x – 5) = 3(x – 3)

or

5x – 25 = 3x – 9

Transposing (–25) to RHS and 3x to LHS, we have

5x – 3x = –9 + 25

or

2x = 16

or

x = 16/2 (Dividing both sides by 2)

∴ x = 8

Check:

= 3/3 = 1

∴ LHS = RHS

**5. **

∵ LCM of 4 and 3 is 12.

∴ Multiplying both sides by 12, we have

or

3(3t – 2) – 4(2t + 3) = (4 x2) – 12t

or

9t – 6 – 8t – 12 = 8 – 12t

or

(9 – 8)t – (6 + 12) = 8 – 12t

or

t – 18 = 8 – 12t

Transposing –18 to RHS and –12t to LHS, we have

t + 12t = 8 + 18

or

13t = 26

or

t = 26/13

∴ t = 2

Check:

∴ LHS = RHS

**6. **

Since, LCM of 2 and 3 is 6.

∴ Multiplying both sides by 6, we have

or

6m – 3(m – 1) = 6 – 2(m – 2)

or

6m – 3m + 3 = 6 – 2m + 4

or

(6 – 3)m + 3 = (6 + 4) – 2m

or

3m + 3 = 10 – 2m

Transposing 3 to RHS and –2m to LHS, we have

3m + 2m = 10 – 3

or 5m = 7

or

m = 7/5 (Dividing both sides by 5)

Check:

∴ LHS = RHS

**Question 2:** **Simplify and solve the following linear equations.7 . 3(t – 3) = 5(2t + 1)8. 15(y – 4) – 2(y – 9) + 5(y + 6) = 09. 3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 1710. 0.25(4f – 3) = 0.05(10f – 9)**

**Ans:** **7.** 3(t – 3) = 5(2t + 1)

or

3t – 9 = 10t + 5

Transposing (–9) to RHS and 10t to LHS, we have

3t – 10t = 5 + 9

or

–7t = 14

or

t = 14/-7 = -2 (Dividing both sides by –7)

∴ t = –2

**8.** 15(y – 4) – 2(y – 9) + 5(y + 6) = 0

Opening the brackets, we have

15y – 60 – 2y + 18 + 5y + 30 = 0

Collecting the like terms,

(15 – 2 + 5)y + (–60 + 18 + 30) = 0

or

18y + (–12) = 0

Transposing (–12) to RHS, we have

18y = 12

or

y = 12/18 = 2/3 (Dividing both sides by 18)

∴ y = 2/3

**9. **3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17

Opening the brackets, we have

15z – 21 – 18z + 22 = 32z – 52 – 17

Collecting the like terms on both sides,

(15 – 18)z + (–21 + 22) = 32z + (–52 – 17)

or

–3z + 1 = 32z – 69

Transposing 1 to RHS and 32z to LHS, we have

–3z – 32z = –69 – 1

or

–35z = –70

or z = -70/-35 = 2 (Dividing both sides by –35)

∴ z = 2

**10.** 0.25(4f – 3) = 0.05(10f – 9)

Opening the brackets, we have

0.25 * 4f – 3 * 0.25 = 0.05 *10f – 0.05 * 9

or

or

Transposing to RHS and to LHS, we have

or

or

(Dividing both sides by 1/2)

or

**EQUATIONS REDUCIBLE TO THE LINEAR FORM**

REMEMBER

When we multiply the numerator on LHS with the denominator on RHS and vice-versa and equate the products so obtained, it is known as cross-multiplication.

**Ques 1:**** Solve: **

**Ans:** We have:

By cross multiplication,

8(x – 1) = 5(2x + 3)

or

8x – 8 = 10x + 15

or

8x = 10x + 15 + 8 (Transposing (–8) to RHS)

or

8x – 10x = 23 (Transposing 10x to LHS)

or

–2x = 23

Dividing both sides by (–2), we have

Check:

= LHS

i.e. LHS = RHS

**Ques 2: The denominator of a rational number greater than its numerator by 3. If the numerator is increased by 16 and the denominator is decreased by 1, then the number obtained is 5/3. Find the rational number.****Ans: **Let the numerator = x

∴ Denominator = x + 3

∴ The number =

New numerator = (x + 16)

and New denominator = (x + 3) – 1

= (x + 2)

∴ New number =

But the new number = 5/3

= 5/3

By cross multiplication, we have

3(x + 16) = 5(x + 2)

or

3x + 48 = 5x + 10

Transposing 48 to RHS and 10 to LHS, we have

3x – 5x = 10 – 48

or

–2x = –38

Dividing both sides by –2, we have

or

x = 19 ⇒ Numerator

and x + 3 = 19 + 3 = 22 ⇒ Denominator

∴ The rational number = 19/22

Check: (i) 22 – 19 = 3, i.e. numerator is 3 less than the denominator.

(ii) New number = which is true.

**Ques 3: Present ages of Shaloo and Preeti are in the ratio 5 : 4. 5 years from now the ratio of their ages will be 6 : 5. Find their ages.****Ans: **Let the present age of Shaloo = 5x years

Let the present age of Preeti = 4x years

After 5 years, Age of Shaloo = (5x + 5) years

Age of Preeti = (4x + 5) years

According to the condition, we have

By cross multiplication, we have

5(5x + 5) = 6(4x + 5)

or

25x + 25 = 24x + 30

Transposing 25x to RHS and 24x to LHS, we have

25x – 24x = 30 – 25

or

x = 5

∴ 5x = 5 * 5 = 25

and 4x = 4 * 5 = 20

Therefore, Present age of Shaloo = 25 years

Present age of Preeti = 20 years.

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