8th Maths : Chapter 3 : Algebra: Linear Equation in One Variable : Exercise 3.6 : Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.6**

** **

**1. Fill in the blanks**

**(i) The value
of x in the equation x + 5 = 12 is --------------------**

**[Answer: 7]**

**Solution:**

Given, *x* + 5 = 12

*x* = 12 − 5 = 7 (by transposition method)

Value of *x* is 7

**(ii) The
value of y in the equation y − 9 = ( −5) +**

**[Answer: 11]**

**Solution:**

Given, *y* − 9 = (−5) + 7

*y* − 9 = 7 − 5 (re−arranging)

*y* − 9 = 2

∴ *y* = 2 + 9 = 11 (by transposition
method)

**(iii) The
value of m in the equation 8m = 56 is--------------------**

**[Answer: 7]**

**Solution:**

Given, 8*m* = 56

Divided by 8 on both sides

[ 8 × *m* ] / 8 = 56
/ 8

∴ *m* = 7

**(iv) The
value of p in the equation 2p/3 = 10 is --------------------**

**[Answer: 15]**

**Solution:**

Given, 2*p* / 3 = 10

Multiplying by 3 on both sides,

[2*p* / 3] × 3 = 10 × 3

Dividing by 2 on both sides

2*p* / 2 = 30 / 2

∴ *p* = 15

**(v) The linear
equation in one variable has -------------------- solution.**

**[Answer: one]**

** **

**2. Say True or False.**

(i) The shifting
of a number from one side of an equation to other is called transposition. **[Answer: True]**

(ii) Linear
equation in one variable has only one variable with power 2. **[Answer: False]**

[** ****correct statement : **Linear equation in one variable has only one variable with
power one ]

** **

**3. Match the following :**

**(a) x/2 = 10 : (i) x = 4 **

**(b) 20= 6 x – 4 : (ii) x = 1**

**(c) 2 x – 5 = 3 – x : (iii) x =
20**

**(d) 7 x – 4 – 8x = 20 : (iv) x = 8/3 **

**(e) 4/11 − x = −7/11 : (v) x = –24**

(A) (i),
(ii), (iv), (iii), (v)

(B) (iii),
(iv), (i), (ii), (v)

(C) (iii),
(i), (iv), (v), (ii)

(D) (iii),
(i), (v), (iv), (ii)

**[Answer: (C) (iii), (i), (iv), (v), (ii)]**

**Solution:**

**a.**** x/2 = 10**, multiplying by 2
on both sides, we get

[*x* / 2] × 2 = 10 × 2 ⇒ *x ***= 20**

**b. ****20 = 6 x – 4** by
transposition ⇒ 20 + 4 = 6

6*x* = 24 dividing by 6 on both sides,

6*x* / 6 = 24 / 6 ⇒ *x*** = 4**

**c.**** 2 x − 5 = 3 − x**

By transposing the variable ‘*x*’, we get

2*x* − 5 + *x* = 3

by transposing − 5 to other side,

2*x* + *x* = 3 + 5

∴ 3*x* = 8 , 3*x */ 3 = 8 / 3

∴ *x* = 8 / 3

**d.**** 7 x − 4 − 8x = 20**

by transposing − 4 to other side,

7*x* − 8*x* = 20 + 4

− *x* = 24

∴ *x* = − 24

**e. ****4/11 – x = −7 / 11**

Transposing 4/11 to other side,

−*x* = [−7/11] [–4/11] = [−7 – 4] / 11 = −11/11 = −1

∴ − *x* = − 1 ⇒ *x *= 1

**4. Find x and p **

**(i) 2 x/3 − 4 = 10/3 (ii) y + 1/6 − 3y = 2/3 (iii)
1/3 – x/3 = 7x/12 + 5/4**

**(i) ****2 x/3 – 4 = 10/3**

**Solution:**

Transposing − 4 to other side, it becomes + 4

∴ 2*x* / 3 = 10/3 + 4 Taking LCM & adding,

2*x*/3 = 10/3 + 4/1 = [10 + 12] / 3 = 22 / 3

2*x*/3 = 22/3 Multiplying by 3 on
both sides

[2*x*/3] × 3 = [22/3] × 3 ⇒ 2*x* = 22, dividing by 2 on both sides,

We get 2*x* / 2 = 22 / 2

∴ *x* = 11

**(ii) **

**Solution:**

Transposing 1/6 to other side,

*y* – 3*y* = 2/3 – 1/6

Taking LCM,

−2*y* = 2/3 – 1/6 = [2 × 2 – 1] / 6 = 3/6 = 1/2

∴ −2*y* = 1/2 ⇒ 2*y* = −1/2, dividing by 2 or
both sides.

2*y */ 2 = –1/2 × 1/2
⇒ *y* = −1/4

**(iii) **

Transposing −*x*/3 to other side, it becomes + *x*/3

∴ 1/3 = 7*x*/12 + 5/4 + *x*/3

Transposing 5/4 to other side,
it becomes −5/4

1/3 – 5/4 = 7*x*/12 + *x*/3

Multiply by 12 throughout [we look at the denominators 3, 4, 12,
3 and take the LCM, which is 12]

[ (1/3) × 12 ] – [ (5/4) × 12 ] = [ 7*x*/12 × 12 ] + [ *x*/3
× 12 ]

4 − 15 = 7*x* + *x* × 4

−11 = 7*x* + 4*x*

11*x* = − 11

*x* = − 1

** **

**5. Find x (i) –3(4x + 9) = 21 (ii)
20 – 2 ( 5 – p) = 8 (iii) (7x – 5) –
4(2 + 5x) = 10(2 – x)**

**(i) −3(4 x + 9) = 21**

**Solution:**

− 3(4*x* + 9) = 21

Expanding the bracket,

−3 × 4*x* + (−3) × 9 = 21

∴ −12*x* + (−27) = 21

−12*x* − 27 = 21

Transposing − 27 to other side, it becomes +27

− 12*x* = 21 + 27 = 48

∴ −12*x* = 48 ⇒
12*x* = − 48

Dividing by 12 on both sides

12*x* / 12 = − 48 / 12 ⇒
*x* = − 4

**(ii) 20 − 2 ( 5 − p) = 8 **

**Solution:**

20 − 2(5 − *p*) = 8

Expanding the bracket,

20 − 2 × 5 − 2 × (−*p*) = 8

20 − 10 + 2*p* = 8 ( −2 × –*p* =
2*p*)

10 + 2*p *= 8 transposing 10 to other side,

2*p* = 8 – 10 = − 2

∴ 2*p* = − 2 ∴ *p* = − 1

**(iii)** **(7 x − 5) − 4(2 + 5x)
= 10(2 − x)**

**Solution:**

(7*x* − 5) − 4(2 + 5*x*) = 10(2 − *x*)** **

Expanding the brackets,

7*x* – 5 − 4 × 2 − 4 × 5*x* = 10 × 2 + 10 × (− *x*)

7*x* − 5 − 8 − 20*x* = 20 − 10*x*

7*x* – 13 − 20*x* = 20 − 10*x*

Transposing 10*x* & − 13, we get

7*x* − 13 − 20*x* + 10*x* = 20

7*x* − 20*x* + 10*x* = 20 +13, Simplifying,

−3*x* = 33

∴ 3*x* = −33

*x* = −33 / 3 = − 11

*x* = −11

** **

**6. Find x and m **

**(i) (3 x−2)/4 − (x−3)/5 = −1 (ii) (m + 9) /
(3m + 15) = 5/3**

**(i)**** [(3 x – 2) / 4] – [(x − 3) / 5] = −1**

**Solution:**

Taking LCM on LHS,

[ (3*x* − 2) × 5 − (*x* − 3) × 4 ] / 20 = −1

Expanding brackets,

[ 3*x* × 5 – 2 × 5 – *x* × 4 − (−3) × 4 ] / 20 = −1

[ 15*x* − 10 − 4*x* − (−12) ] / 20 = − 1

[ 15*x* − 10 − 4*x +*12 ] / 20 = −1 ⇒ [ 11*x* + 2 ] / 20 = −l

Multiplying both sides by 20

[(11*x* + 2) / 20] × 20 = −l × 20

∴ 11*x* + 2 = − 20

∴ 11*x * = − 20 – 2 = −
22

*x*
= −22/11 = − 2** **

∴ **Answer** : *x* = − 2

**(ii) [**** m + 9 ] / [ 3m + 15 ] = 5/3**

Cross multiplying, we get

[*m* + 9 ] / [3*m* + 15] cross over [5 / 3]

(*m* + 9) × 3 = 5 × (3*m* + 15)

*m* × 3 + 9 × 3 = 5 × 3*m* + 5 × 15

3*m* + 27 = 15*m *+ 75 Transporting 3m
& 75, we get

27 − 75 = 15*m* − 3*m*

−48 = 12*m*

12*m* / 12 = − 48 /
12

⇒* **m * = − 4

**Answer:**

**Exercise 3.6 **

**1. (i) x = 7 (ii) y = 11 (iii) m = 7 (iv) p = 15 (v) One **

**2. (i) True (ii) False
**

**3. (c) (iii),(i), (iv),
(v), (ii) **

**4. (i) x = 11 (ii) y = 1/-4 (iii) x = −1 **

**5. (i) x = −4 (ii) p = −1 (iii) x = −11**

**6. (i) x = −2 (ii) m = −4**

** **

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8th Maths : Chapter 3 : Algebra : Exercise 3.6 (Linear Equation in One Variable) | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths

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