**DAV Class 8 Maths Solution Chapter 9 Linear Equation in one variable**

DAV School Books Class 8 Maths Solution Chapter 9 Linear Equation in one variable all Question Answer.

DAV Class 8 9th chapter Linear Equation in one variable full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.

DAV School Books Class 8 Maths Solution Chapter 9 Linear Equation in one variable

**Worksheet 1 **

**1) Solve the following equations and check your answer:**

**i) (p+7) ÷ (p – 6) = 1÷3**

**ii) (x – 2) ÷ (6x+1) = 1**

**iii) 3x ÷ (5x – 5) = – 1**

**iv) (2x – 1) ÷ 5x = – (1÷6)**

**v) (4x+1) ÷ (3x – 1) = – 2**

**vi) (4z – 3) ÷ (2z+1) = (5/7)**

**vii) [(2x÷5) + 3]** **÷ [(x÷3) – 1] = (3÷5)**

**viii) [2x – (3÷4)]** **÷ [3x+ (4÷5)] = (1÷5)**

**ix) [(3x÷4) + 1]** **÷ [x + (1÷5)] = (7÷4)**

**x) (3k+5)** **÷ (4k – 3) = (4÷9)**

**xi) (0.5x – 4)** **÷ (2.4x+6) = (- 5÷3)**

**xii) [(x÷3) – (2÷5)]** **÷ [(3÷4) – (2x)] = (16÷15)**

**xiii) [(1 – 2x)+(1+2x)]** **÷ [(4x + 1) + (x – 3)] = (1÷2)**

**xiv) [x ^{2} – (x+2)(x+3)]**

**÷ (7x+1) = (2÷3)**

**Ans:**

i) **(p+7) ÷ (p – 6) = 1÷3**

Cross multiplying, we get

3p+ 21 = p – 6

Or, 3p – p = – 27

Or, p = – 27/2

Hence p = – 27/2

Check:

For p = –27/2

L.H.S

= (p+7) ÷ (p – 6)

= (–27/2 + 7) ÷ (–27/2 – 6)

= 1/3

= R.H.S

Hence p = – 27/2 is the solution

ii) **(x – 2) ÷ (6x+1) = 1**

Cross multiplying, we get

x – 2 = 6x + 1

Or, 5x = – 3

Or, x = – 3/5

Hence, x = – 3/5

Check:

For x = – 3/5

L.H.S

= (x – 2) ÷ (6x+1)

= [- 3/5 – 2] ÷ [6(- 3/5) + 1]

= [-13/5] ÷ [-13/5]

= 1

= R.H.S

Hence x = – 3/5 is the solution.

iii) **3x ÷ (5x – 5) = – 1**

Cross multiplying we get

3x = 5 – 5x

Or, 8x = 5

Or, x = 5/8

Hence, x = 5/8

Check:

For x = 5/8

L.H.S

= 3x ÷ (5x – 5)

= 3(5/8) ÷ [5 (5/8) – 5]

= (15/8) ÷ [(25/8) – 5]

= 15 ÷ (-15)

= -1

= R.H.S

Hence, x = 5/8 is the solution

iv) **(2x – 1) ÷ 5x = – (1÷6)**

Cross multiplying, we get

6 (2x – 1) = – 5x

Or, 12x – 6 = – 5x

Or, 17x = 6

Or, x = 6/17

Hence, x = 6/17

Check:

For x = 6/17

L.H.S

= (2x – 1) ÷ 5x

= [2(6/17) – 1] ÷ 5(6/17)

= (12/17 – 1) ÷ (30/17)

= (–5/17) ÷ (30/17)

= – (1/6)

= R.H.S

Hence x = 6/17 is the solution.

v)** (4x+1) ÷ (3x – 1) = – 2**

Cross multiplying, we get

4x+1 = –6x+2

Or, 10x = 1

Or, x = (1/10)

Hence, x = 1/10

Check:

For x = 1/10

L.H.S

= (4x+1) ÷ (3x – 1)

= [4(1/10) + 1] ÷ [3(1/10) – 1]

= (14/10) ÷ (–7/10)

= – 2

= R.H.S

Hence, x = 1/10 is the solution.

vi) **(4z – 3) ÷ (2z+1) = (5 ÷ 7)**

Cross multiplying, we get

7 (4z–3) = 5 (2z+1)

Or, 28z – 21=10z+5

Or, 18z = 26

Or, z = 13/9

Hence, z = 13/9

Check:

Z = 13/9

L.H.S

= (4z – 3) ÷ (2z+1)

= [4(13/9) – 3] ÷ [2(13/9) + 1]

= (25/9) ÷ (35/9)

= 5/7

= R.H.S

Hence, z = 13/9 is the solution.

vii) **[(2x÷5) + 3] ÷ [(x÷3) – 1] = (3÷5)**

Cross multiplying, we get

5 [(2x÷5) + 3] = 3 [(x÷3) – 1]

Or, 2x + 15 = x – 3

Or, x = – 18

Hence, x = –18

Check:

For, x = – 18

L.H.S

= [(2x÷5) + 3] ÷ [(x÷3) – 1]

= [(-36÷5) + 3] ÷ [(-18÷3) – 1]

= (-21/5) ÷ (-7)

= (3/5)

= R.H.S

Hence x = -18 is the solution.

viii) **[2x – (3÷4)] ÷ [3x+ (4÷5)] = (1÷5)**

Cross multiplying, we get

5 [2x – (3÷4)] = [3x+ (4÷5)]

Or, 10x – (15/4) = 3x + (4/5)

Or, 7x= (15/4) + (4/5)

Or, 7x = (75+16)/20

Or, 7x = 91/20

Or, x = 13/20

Hence, x = 13/20

Check:

For, x = 13/20

L.H.S

= [2x – (3÷4)] ÷ [3x+ (4÷5)]

= [2(13/20) – (3/4)] ÷ [3(13/20) + (4/5)]

= (11/20) ÷ (11/4)

= (1/5)

= R.H.S

Hence, x = 13/20 is the solution.

ix) **[(3x/4) + 1] ÷ [x + (1/5)] = (7÷4)**

Cross multiplying, we get

4 [(3x/4) + 1] = 7 [x + (1/5)]

Or, 3x + 4 = 7x + 7/5

Or, 4x = 4 – (7/5)

Or, 4x = 13/5

Or, x = 13/20

Hence, x = 13/20

Check:

For x = 13/20

L.H.S

= [(3x/4) + 1] ÷ [x + (1/5)]

= [{3× (13/20)/4} + 1] ÷ [(13/20) + (1/5)]

= (119/80) ÷ (17/20)

= (7/4)

= R.H.S

Hence, x = 13/20 is the solution.

x) **(3k+5) ÷ (4k – 3) = (4÷9)**

Cross multiplying, we get

9 (3k+5) = 4 (4k – 3)

Or, 27k + 45 = 16k – 12

Or, 11k = – 57

Or, k = –57/11

Hence k = –57/11

Check:

For k = – 57/11

L.H.S

= (3k+5) ÷ (4k – 3)

= [3(–57/11) +5] ÷ [4(–57/11) – 3]

= (-116/11) ÷ (-261/11)

= (4/9)

= R.H.S

Hence, k = – 57/11 is the solution

xi) **(0.5x – 4) ÷ (2.4x+6) = (–5÷3)**

Cross multiplying, we get

3(0.5x – 4) = – 5 (2.4x+6)

Or, 1.5x – 12 = – 12x – 30

Or, 13.5x = –18

Or, x = – 18/13.5

Hence x = – 18/13.5

Check:

For x = – 18/13.5

L.H.S

= (0.5x – 4) ÷ (2.4x+6)

= [0.5(– 18/13.5) – 4] ÷ [2.4(– 18/13.5) +6]

= (-14/3) ÷ (14/5)

= -5/3

= R.H.S

Hence, x = – 18/13.5 is the solution.

xii) **[(x/3) – (2/5)] ÷ [(3/4) – (2x)] = (16 / 15)**

Cross multiplying, we get

15 [(x/3) – (2/5)] = 16 [(3/4) – (2x)]

Or, 5x – 6 = 12 – 32x

Or, 37x = 18

Or, x = 18/37

Hence, x= 18/37

Check:

For x = 18/37

L.H.S

= [(x/3) – (2/5)] ÷ [(3/4) – (2x)]

= [{(18/37)/3} – (2/5)] ÷ [(3/4) – {2× (18/37)}]

= (-44/185) ÷ (-33/148)

= 16/15

= R.H.S

Hence, x = 18/37 is the solution.

xiii) **[(1 – 2x) + (1+2x)] ÷ [(4x + 1) + (x – 3)] = (1÷2)**

Cross multiplying, we get

2 [(1 – 2x) + (1+2x)] = [(4x + 1) + (x – 3)]

or, 2×2 = 5x – 2

or, 5x = 6

or, x = 6/5

Hence, x = (6/5)

Check:

For, x = (6/5)

L.H.S

= [(1 – 2x) + (1+2x)] ÷ [(4x + 1) + (x – 3)]

= [{1 – 2× (6/5)} + {1+2× (6/5)}] ÷ [{4× (6/5) + 1} + {(6/5) – 3}]

= 2÷4 = 1/2

= R.H.S

Hence x = (6/5) is the solution.

xiv) [x^{2} – (x+2) × (x+3)] ÷ (7x+1) = (2÷3)

Cross multiplying, we get

3 [x^{2} – (x+2) × (x+3)] = 2 (7x+1)

Or, 3x^{2} – 3(x^{2}+3x +2x +6) = 14x + 2

Or, 3x^{2} – 3x^{2} – 15x – 18 = 14x + 2

Or, – 15x – 14x = 2 + 18

Or, –29x = 20

Or, x = –20/29

Hence, x = –20/29

Check:

For x = – 20/29

L.H.S

= [x^{2} – (x+2) × (x+3)] ÷ (7x+1)

= [(-20/29)^{2} – {(-20/29)+2} × {(-20/29)+3}] ÷ [7×(-20/29)+1]

= (-74/29) ÷ (-111/29) = (2/3)

= R.H.S

Hence, x = -20/29 is the solution

**2) Find the positive value of x for which the given equation is satisfied.**

**i) (3 – x ^{2})**

**÷ (8+x**–

^{2}) =**3/4**

**ii) (y ^{2}+6)**

**÷ (8y**

^{2}+3) = (1/5)**iii) (x ^{2}**–

**9)**

**÷ (5+x**

^{2}) = (5/9)**iv) (y ^{2}+4)**

**÷ (3y**

^{2}+7) = (1/2)**Ans:**

i) **(3 – x ^{2}) ÷ (8+x^{2}) = –3/4**

Put x^{2} = y in the equation, Then it becomes

(3 – y) ÷ (8+y) = -3/4

Cross multiplying, we get

4(3 – y) = -3 (8+y)

Or, 12 – 4y = -24 – 3y

Or, y = 36

∵ y = x^{2}

∴ x^{2}= 36

Or, x = 6

Check for x = 6

L.H.S = (3 – x^{2}) ÷ (8+x^{2}) = (3 – 36) ÷ (8+36) = (-33)/ (44) = -3/4 = R.H.S

Hence x = 6 is the solution.

ii) **(y ^{2}+6) ÷ (8y^{2}+3) = (1/5)**

Put y^{2} = x in the equation, Then it becomes

(x+6) ÷ (8x+3) = (1/5)

Cross multiplying, we get

5(x+6) = (8x + 3)

Or, 5x + 30 = 8x+3

Or, 3x = 27

Or, x = 9

∵ y^{2}=x

∴ y^{2}=9

Or, y = 3

Check: for y = 3

L.H.S = (y^{2}+6) ÷ (8y^{2}+3) = (9+6) ÷ [(8×9) +3] = 15/75 = 1/5 = R.H.S

Hence y = 3 is the solution

iii) **(x ^{2}–9) ÷ (5+x^{2}) = (5/9)**

Put x^{2} = y in the equation, Then it becomes

(y–9) ÷ (5+y) = (5/9)

Cross multiplying, we get

9(y–9) = 5(5+y)

Or, 9y – 81 = 25 + 5y

Or, 4y = 106

Or, y = 106/4

∵ y = x^{2}

∴ x^{2}= 106/4

Or, x = √106/2

Check: For x = √106/2

L.H.S = (x^{2}–9) ÷ (5+x^{2}) = [(106/4)–9] ÷ [5+(106/4)] = 5/9 = R.H.S

Hence x = √106/2 is the solution

iv) **(y ^{2}+4) ÷ (3y^{2}+7) = (1/2)**

Put y^{2} = x in the equation, Then it becomes

(x+4) ÷ (3x+7) = (1/2)

Cross multiplying, we get

2(x+4) = (3x+7)

Or, 2x + 8 = 3x + 7

Or, x = 1

∵ y^{2} = x

∴ y^{2}= 1

Or, y = 1

Check for y = 1

L.H.S = (y^{2}+4) ÷ (3y^{2}+7) = (1+4) ÷ (10) = 1/2 = R.H.S

Hence, y = 1 is the solution.

**Worksheet 2**

**1) The present ages of A and B are in the ration 7:5. Ten years later their ages will be in the ratio 9:7. Find their present ages.**

**Ans:**

Let us assume that the ages of A and B are 7k and 5k respectively.

According to the problem ten years later the ratio of their ages will be 9:7

∴ (7k+10)/ (5k+10) = 9/7

Cross multiplying we get

7(7k+10) = 9(5k+10)

Or, 49k + 70 = 45k + 90

Or, 4k = 20

Or, k = 5

Hence the present ages of A and B are 35 and 25 respectively.

**2) Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find the present ages.**

**Ans:**

Let us assume that the present age of the father is x and the present age of the son is y.

According to the problem two years ago father was three times older than his son.

∴ (x – 2) = 3 (y – 2)

Or x – 2 = 3y – 6

Or, x – 3y = – 4

Or, x = 3y – 4………. (Equation 1)

According to the problem two years hence twice the age of the father will be equal to five times that of his son.

∴ 2 (x+2) = 5 (y+5)

Or, 2x + 4 = 5y + 25

Or, 2x – 5y = 21…..… (Equation 2)

Putting the value of x from equation 1 in equation 2 we get

2 (3y – 4) – 5y = 21

Or, 6y – 8 – 5y = 21

Or, y = 29

Now putting the value of y in equation 1 we get

x = 3×29 – 4

Or, x = 83

Hence the present age of the father is 83 and the present age of the son is 29.

**3) The sum of the digits of a 2-digit number is 8. The number obtained by interchanging the digits exceeds the given number by 18. Find the given numbers.**

**Ans:**

Let us assume that ones place digit of the given number is x.

The sum of the two digits is 8. So the tens place digit will be (8 – x)

∴ The original number is 10 (8 – x) + x = 80 – 9x

On interchanging the digits the ones place digit becomes tens place digit and the tens place digit becomes ones place digit.

∴ New number = 10x + (8 – x) = 9x + 8

According to the problem new number exceed the given number by 18.

9x+8 – 80 + 9x = 18

Or, 18x – 72 = 18

Or, 18x = 90

Or, x = 5

∴ The original number is 80 – 9x = 80 – 9×5 = 35

**4) The ones digit of a 2-digit number is twice the tens digit. When the number formed by reversing the digits is added to the original number, the sum is 99. Find the original number.**

**Ans:**

Let the digit in the tens place of the given number is x

According to the problem the ones digit is twice the tens digit

So the ones digit of the number is 2x.

∴ The given number is 10x+2x = 12x

When a new number is formed by interchanging the digits we get = 10×2x + x = 21x.

According to the problem the sum of these two numbers is 99

∴ 12x + 21x = 99

Or, 33x = 99

Or, x = 3

∴ The original number is 12x = 12×3 = 36

**5) The sum of the digits of a 2-digit number is 11. The number obtained by interchanging the digits exceeds the original number by 27. Find the number.**

**Ans:**

Let us assume that ones place digit of the given number is x.

The sum of the two digits is 11. So the tens place digit will be (11 – x)

∴ The original number is 10 (11 – x) + x = 110 – 9x

On interchanging the digits the ones place digit becomes tens place digit and the tens place digit becomes ones place digit.

∴ New number = 10x + (11 – x) = 9x + 11

According to the problem new number exceed the given number by 27.

9x+11–110 + 9x = 27

Or, 18x – 99 = 27

Or, 18x = 126

Or, x = 7

∴ The original number is 110 – 9x = 110 – 9×7 = 47.

**6) The sum of three consecutive multiples of 7 is 777. Find these multiples.**

**Ans:**

Let the first multiple of 8 be 8x.

The other two consecutive multiples would be 8(x+1) and 8(x+2).

According to the problem the sum of these three numbers will be equal to 888

∴ 8x + 8(x+1) + 8(x+2) = 888

Or, 24x + 24 = 888

Or, 24x = 888 – 24

Or, 24x = 864

Or, x = 36

The multiples of 8 are 8x = 288, 8 (x+1) = 296 and 8 (x+2) = 304.

**7) The sum of three consecutive multiples of 9 is 999. Find the multiples.**

**Ans:**

Let the first multiple of 9 be 9x.

The other two consecutive multiples would be 9(x+1) and 9(x+2).

According to the problem the sum of these three numbers will be equal to 999

∴ 9x + 9(x+1) + 9(x+2) = 999

Or, 27x + 27 = 999

Or, 27x = 999 – 27

Or, 27x = 972

Or, x = 36

The multiples of 9 are 9x = 324, 9 (x+1) = 333 and 9 (x+2) = 342.

**8) The denominator of a rational number is greater than its numerator by 7. If 3 is subtracted from the numerator and 2 is added to its denominator, the new number becomes (1/5). Find the rational number.**

**Ans:**

Let the numerator of the fractional number given in question is x. So, the denominator is (x+7)

The given fractional number is x/ (x+7).

According to the question if 3 is subtracted form numerator and 2 is added to its denominator we get 1/5

∴ (x – 3)/ (x+7+2) = 1/5

Cross multiplying we get

5 (x – 3) = x + 9

Or, 5x – 15 = x + 9

Or, 4x = 24

Or, x = 6

The given number is x/ (x+7) = 6/ 13

**9) The numerator and denominator of a rational number are in the ratio 3:4. If the denominator is increased by 3, the ratio becomes 3:5. Find the rational number.**

**Ans:**

Let the numerator and denominator of a rational number is 3k and 4k respectively.

When the denominator is increased by 3 the ratio becomes 3:5

∴ 3k / (4k+3) = 3/5

Or, 15k = 12k + 9

Or, 3k = 9

Or, k = 3

∴ The given rational number is 9/12

**10) A motor boat goes downstream and covers a distance in 4 hours while it covers the same distance upstream in 5 hours. If the speed of the stream is 3km/hr, find the speed of the motor boat in still water.**

**Ans:**

Let the speed of the motor boat in still water is x km/hr

Downstream speed of the motor boat = (x+3) km/hr

Upstream speed of the motor boat = (x – 3) km/hr

Distance covered in downstream in 4 hours = 4 (x+3) km = (4x + 12) km

Distance covered in upstream in 5 hours = 5 (x – 3) km = 5x – 15 km

According to the problem the distance covered by the boat in upstream and downstream is same.

∴ 4x + 12 = 5x – 15

Or, x = 29

Hence, Speed of the motorboat in still water is 29 km/hr.

**11) A steamer, going downstream in a river covers the distance between 2 towns in 15 hours. Coming back upstream, it covers the distance in 20 hours. The speed of the water is 3km/hr. Find the distance between two towns.**

**Ans:**

Let the distance between two towns is D km.

The speed of the steamer in still water is x km/hr

Upstream speed = speed of the steamer – speed of water = (x – 3) km/hr

Downstream speed = speed of the steamer + speed of water = (x+3) km/hr

Time taken to travel the distance in upstream is 20 hours

Now, Distance/ Speed = Time

∴ D/(x – 3) = 20

Or, D = 20x – 60

Or, x = (D+60) /20 ……….. (Equation 1)

Time taken to travel the distance in downstream is 15 hours.

Now, Distance/ Speed = Time

D/(x+3) = 15

Or, D = 15x + 45

Or, x = (D – 45) / 15 ……… (Equation 2)

Equating equation 1 and 2 we get

(D+60) /20 = (D – 45) / 15

Or, 15 (D+60) = 20(D – 45)

Or, 15D + 900 = 20D – 900

Or, 5D = 1800

Or, D = 360

The Distance between two towns is 360 km.

**12) The distance between 2 towns is 300 km. Two cars start simultaneously from these towns and move towards each other. The speed of one car is more than the other by 7km/hr. If the distance between the cars after 2 hours is 34 km, find the speed of the cars.**

**Ans:**

** **

Let the speed of the slower car is x km/hr

The speed of the faster car is 7 km/hr more than that of the slower car.

∴ The speed of the faster car is (x+7) km/hr

We know that the Distance = Speed × Time

Distance travelled by the slower car in 2 hours = 2x km

Distance travelled by the faster car in 2 hours = 2 (x+7)

After 2 hours the distance between them is 34 km

So the total distance travelled by both the cars is (300 – 34) km = 266 km

∴ 2x + 2 (x+7) = 266

Or, 4x + 14 = 266

Or, 4x = 252

Or, x = 63

The speed of the slower car is 63 km/hr

The speed of the faster car is 70 km/hr.

**13) The length of a rectangle is greater than the breadth by 18 cm. If both length and breadth are increased by 6 cm, then the area was increased by 168 cm ^{2}. Find the length and breadth of the rectangle.**

**Ans:**

Let the length of the original rectangle is x cm.

According to the problem length of the rectangle is greater than the breadth by 18 cm

So the breadth of the original rectangle is (x – 18) cm

Area of the given rectangle is x(x – 18) cm^{2}

Increased length of the modified rectangle = (x+6) cm

Increased breadth of the modified rectangle = (x – 18+6) cm = (x – 12) cm

Area of the modified rectangle = (x+6) × (x – 12) cm^{2}

According to the problem area was increased by 168 cm^{2}

∴ (x+6) × (x – 12) – x(x – 18) = 168

Or, x^{2} – 12x+6x – 72 – x^{2}+18x = 168

Or, 12x = 240

Or, x = 20

The length of the rectangle is 20 cm and the breadth of the rectangle is 2 cm.

**14) The length of a rectangle is greater than the breadth by 3cm. If the length is increased by 9 cm and the breadth is reduced by 5 cm, the area remains the same. Find the length and breadth of the rectangle.**

**Ans:**

Let the length of the original rectangle is x cm.

According to the problem length of the rectangle is greater than the breadth by 3 cm

So the breadth of the original rectangle is (x – 3) cm

Area of the given rectangle is x(x – 3) cm^{2}

Increased length of the modified rectangle = (x+9) cm

Increased breadth of the modified rectangle = (x – 3 – 5) cm = (x – 8) cm

Area of the modified rectangle = (x+9) × (x – 8) cm^{2}

According to the problem area of the modified rectangle is equal to the original rectangle.

∴ x(x – 3) = (x+9) × (x – 8)

Or, x^{2} – 3x = x^{2} – 8x +9x – 72

Or, 4x = 72

Or, x = 18

The length of the rectangle is 18 cm and the breadth of the rectangle is 15 cm.

**15) The difference between two positive integers is 30. The ration of these integers is 2:5. Find the integers.**

**Ans:**

Let us assume that the values of these positive integers are 2k and 5k.

∴ According to the problem

5k – 2k = 30

Or, 3k = 30

Or, k = 10

∴ The values of these integers are 50 and 20.

**16) The sum of two positive integers is 105. The integers are in the ratio 2:3. Find the integers.**

**Ans:**

Let us assume that the values of these positive integers are 2k and 3k.

∴ According to the problem

2k + 3k = 105

Or, 5k = 105

Or, k = 21

∴ The values of these integers are 42 and 63.

**17) A money box contains one-rupee and two-rupee coins in the ratio 5:6. If the total value of the coins in the money box is ₹85, find the number of two rupee coins.**

**Ans:**

Let the numbers of one rupee and two rupee coins are 5k and 6k.

Total value of one rupee coins is ₹5k

Total value of two rupee coins is 2×₹6k = ₹12k

Total value of coins in the money box = ₹ (5k + 12k) = ₹17k

∴ 17k = 85

Or, k = 5

∴ The number of two rupee coins = 6k = 6×5 = 30

**18) A purse has only one-rupee and two-rupee coins in it. The number of two-rupee coins is one-third the number of one-rupee coins. If the purse has ₹115, find the number of two-rupee coins.**

**Ans:**

Let us consider the number of one rupee coins is 3k and the number of two rupee coins is (3k/3) = k

Total value of one rupee coins is ₹3k.

Total value of two rupee coins is ₹ 2k

∴ According to the question

3k + 2k = 115

or, (3k + 2k) = 115

or, 5k = 115

or, k = 23

∴ The number of two-rupee coins in the purse is 23.