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RD Sharma SolutionsClass 7Chapter 8 Linear Equations In One Variable

RD Sharma Solutions For Class 7 Maths Chapter - 8 Linear Equations in One Variable

RD Sharma Solutions For Class 7 Maths Chapter 8 Linear Equation in One Variable is the best study material for those students who are finding difficulties in solving Math problems. Students can download the pdf of RD Sharma Solutions Maths Chapter 8 Linear Equations in One Variable from the available links. These exercises problems are solved by our subject experts in Maths to clear exam fears in students.

Chapter 8, Linear Equations in One variable, contains four exercises. RD Sharma Solutions for Class 7 given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

Equation – a statement of equality which involves one or more literalsLinear equations – an equation in which the highest power of the variables involved is 1The solution of an equation in trial and error methodSystematic methodEquations involving additionEquations involving subtractionEquations involving multiplicationEquations involving divisionTransposition methodApplications of linear equations to practical problems

Chapter 8 Linear Equations in One…

Exercise

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Access answers to Maths RD Sharma Solutions For Class 7 Chapter 8 – Linear Equations in One variable

Exercise 8.1 Page No: 8.6

1. Verify by substitution that:

(i) x = 4 is the root of 3x – 5 = 7

(ii) x = 3 is the root of 5 + 3x = 14

(iii) x = 2 is the root of 3x – 2 = 8x – 12

(iv) x = 4 is the root of (3x/2) = 6

(v) y = 2 is the root of y – 3 = 2y – 5

(vi) x = 8 is the root of (1/2)x + 7 = 11

Solution:

(i) Given x = 4 is the root of 3x – 5 = 7

Now, substituting x = 4 in place of ‘x’ in the given equation, we get

= 3(4) – 5 = 7

= 12 – 5 = 7

7 = 7

Since, LHS = RHS

Hence, x = 4 is the root of 3x – 5 = 7.

(ii) Given x = 3 is the root of 5 + 3x = 14.

Now, substituting x = 3 in place of ‘x’ in the given equation, we get

= 5 + 3(3) = 14

= 5 + 9 = 14

14 = 14

Since, LHS = RHS

Hence, x = 3 is the root of 5 + 3x = 14.

(iii) Given x = 2 is the root of 3x – 2 = 8x – 12.

Now, substituting x = 2 in place of ‘x’ in the given equation, we get

= 3(2) – 2 = 8(2) – 12

= 6 – 2 = 16 – 12

4 = 4

Since, LHS = RHS

Hence, x = 2 is the root of 3x – 2 = 8x – 12.

(iv) Given x = 4 is the root of 3x/2 = 6.

Now, substituting x = 4 in place of ‘x’ in the given equation, we get

= (3 × 4)/2 = 6

= (12/2) = 6

6 = 6

Since, LHS = RHS

Hence, x = 4 is the root of (3x/2) = 6.

(v) Given y = 2 is the root of y – 3 = 2y – 5.

Now, substituting y = 2 in place of ‘y’ in the given equation, we get

= 2 – 3 = 2(2) – 5

= -1 = 4 – 5

-1 = -1

Since, LHS = RHS

Hence, y = 2 is the root of y – 3 = 2y – 5.

(vi) Given x = 8 is the root of (1/2)x + 7 = 11.

Now, substituting x = 8 in place of ‘x’ in the given equation, we get

= (1/2) (8) + 7 =11

= 4 + 7 = 11

= 11 = 11

Since, LHS = RHS

Hence, x = 8 is the root of 12x + 7 = 11.

2. Solve each of the following equations by trial – and – error method:

(i) x + 3 =12

(ii) x -7 = 10

(iii) 4x = 28

(iv) (x/2) + 7 = 11

(v) 2x + 4 = 3x

(vi) (x/4) = 12

(vii) (15/x) = 3

(vii) (x/18) = 20

Solution:

(i) Given x + 3 =12

Here LHS = x +3 and RHS = 12

xLHSRHSIs LHS = RHS11 + 3 = 412No22 + 3 = 512No33 + 3 = 612No44 + 3 = 712No55 + 3 = 812No66 + 3 = 912No77 + 3 = 1012No88 + 3 = 1112No99 + 3 = 1212Yes

Therefore, if x = 9, LHS = RHS.

Hence, x = 9 is the solution to this equation.

(ii) Given x -7 = 10

Here LHS = x -7 and RHS = 10

xLHSRHSIs LHS = RHS99 – 7 = 210No1010 -7 = 310No1111 – 7 = 410No1212 – 7 = 510No1319 – 7 = 610No1414 – 7 = 710No1515 – 7 = 810No1616 – 7 = 910No1717 – 7 = 1010Yes

Therefore if x = 17, LHS = RHS

Hence, x = 17 is the solution to this equation.

(iii) Given 4x = 28

Here LHS = 4x and RHS = 28

xLHSRHSIs LHS = RHS14 × 1 = 428No24 × 2 = 828No34 × 3 = 1228No44 × 4 = 1628No54 × 5 = 2028No64 × 6 = 2428No74 × 7 = 2828Yes

Therefore if x = 7, LHS = RHS

Hence, x = 7 is the solution to this equation.

(iv) Given (x/2) + 7 = 11

Here LHS = (x/2) + 7 and RHS = 11

Since RHS is a natural number, (x/2) must also be a natural number, so we must substitute values of x that are multiples of 2.

xLHSRHSIs LHS = RHS2(2/2) + 7 = 1 + 7 =811No4(4/2) + 7 = 2 + 7 = 911No6(6/2) + 7 = 3 + 7 = 1011No8(8/2) + 7 = 4 + 7 = 1111Yes

Therefore if x = 8, LHS = RHS

Hence, x = 8 is the solutions to this equation.

(v) Given 2x + 4 = 3x

Here LHS = 2x + 4 and RHS = 3x

xLHSRHSIs LHS = RHS12 (1) + 4 = 2 + 4 = 63 (1) = 3No22 (2) + 4 = 4 + 4 = 83 (2) = 6No32 (3) + 4 = 6 + 4 = 103 (3) = 9No42 (4) + 4 = 8 + 4 = 123 (4) = 12Yes

Therefore if x = 4, LHS = RHS

Hence, x = 4 is the solutions to this equation.

(vi) Given (x/4) = 12

Here LHS = (x/4) and RHS = 12

Since RHS is a natural number, x/4 must also be a natural number, so we must substitute values of x that are multiples of 4.

xLHSRHSIs LHS = RHS16(16/4) = 412No20(20/4) = 512No24(24/4) = 612No28(28/4) = 712No32(32/4) = 812No36(36/4) = 912No40(40/4) = 1012No44(44/4) = 1112No48(48/4) = 1212Yes

Therefore if x = 48, LHS = RHS

Hence, x = 48 is the solutions to this equation.

(vii) Given (15/x) = 3

Here LHS = (15/x) and RHS = 3

Since RHS is a natural number, 15x must also be a natural number, so we must substitute values of x that are factors of 15.

xLHSRHSIs LHS = RHS1(15/1) = 153No3(15/3) = 53No5(15/5) = 33Yes

Therefore if x = 5, LHS = RHS

Hence, x = 5 is the solutions to this equation.

(viii) Given (x/18) = 20

Here LHS = (x/18) and RHS = 20

Since RHS is a natural number, (x/18) must also be a natural number, so we must substitute values of x that are multiples of 18.

xLHSRHSIs LHS = RHS324(324/18) = 1820No342(342/18) = 1920No360(360/18) = 2020Yes

Therefore if x = 360, LHS = RHS

Hence, x = 360 is the solutions to this equation.

Exercise 8.2 Page No: 8.12

Solve each of the following equations and check your answers:

1. x – 3 = 5

Solution:

Given x – 3 = 5

Adding 3 to both sides we get,

x – 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS, we get

LHS = x – 3 and RHS = 5

LHS = 8 – 3 = 5 and RHS = 5

LHS = RHS

Hence, verified.

2. x + 9 = 13

Solution:

Given x + 9 = 13

Subtracting 9 from both sides i.e. LHS and RHS, we get

x + 9 – 9 = 13 – 9

x = 4

Verification:

Substituting x = 4 on LHS, we get

LHS = 4 + 9 = 13 = RHS

LHS = RHS

Hence, verified.

3. x – (3/5) = (7/5)

Solution:

Given x – (3/5) = (7/5)

Add (3/5) to both sides, we get

x – (3/5) + (3/5) = (7/5) + (3/5)

x = (7/5) + (3/5)

x = (10/5)

x = 2

Verification:

Substitute x = 2 in LHS of given equation, then we get

2 – (3/5) = (7/5)

(10 – 3)/5 = (7/5)

(7/5) = (7/5)

LHS = RHS

Hence, verified

4. 3x = 0

Solution:

Given 3x = 0

On dividing both sides by 3 we get,

(3x/3) = (0/3)

x = 0

Verification:

Substituting x = 0 in LHS we get

3 (0) = 0

And RHS = 0

Therefore LHS = RHS

Hence, verified.

5. (x/2) = 0

Solution:

Given x/2 = 0

Multiplying both sides by 2, we get

(x/2) × 2 = 0 × 2

x = 0

Verification:

Substituting x = 0 in LHS, we get

LHS = 0/2 = 0 and RHS = 0

LHS = 0 and RHS = 0

Therefore LHS = RHS

Hence, verified.

6. x – (1/3) = (2/3)

Solution:

Given x – (1/3) = (2/3)

Adding (1/3) to both sides, we get

x – (1/3) + (1/3) = (2/3) + (1/3)

x = (2 + 1)/3

x = (3/3)

x =1

Verification:

Substituting x = 1 in LHS, we get

1 – (1/3) = (2/3)

(3 – 1)/3 = (2/3)

(2/3) = (2/3)

Therefore LHS = RHS

Hence, verified.

7. x + (1/2) = (7/2)

Solution:

Given x + (1/2) = (7/2)

Subtracting (1/2) from both sides, we get

x + (1/2) – (1/2) = (7/2) – (1/2)

x = (7 – 1)/2

x = (6/2)

x = 3

Verification:

Substituting x = 3 in LHS we get

3 + (1/2) = (7/2)

(6 + 1)/2 = (7/2)

(7/2) = (7/2)

Therefore LHS = RHS

Hence, verified.

8. 10 – y = 6

Solution:

Given 10 – y = 6

Subtracting 10 from both sides, we get

10 – y – 10 = 6 – 10

-y = -4

Multiplying both sides by -1, we get

-y × -1 = – 4 × – 1

y = 4

Verification:

Substituting y = 4 in LHS, we get

10 – y = 10 – 4 = 6 and RHS = 6

Therefore LHS = RHS

Hence, verified.

9. 7 + 4y = -5

Solution:

Given 7 + 4y = -5

Subtracting 7 from both sides, we get

7 + 4y – 7 = -5 -7

4y = -12

Dividing both sides by 4, we get

y = -12/ 4

y = -3

Verification:

Substituting y = -3 in LHS, we get

7 + 4y = 7 + 4(-3) = 7 – 12 = -5, and RHS = -5

Therefore LHS = RHS

Hence, verified.

10. (4/5) – x = (3/5)

Solution:

Given (4/5) – x = (3/5)

Subtracting (4/5) from both sides, we get

(4/5) – x – (4/5) = (3/5) – (4/5)

– x = (3 -4)/5

– x = (-1/5)

x = (1/5)

Verification:

Substituting x = (1/5) in LHS we get

(4/5) – (1/5) = (3/5)

(4 -1)/5 = (3/5)

(3/5) = (3/5)

Therefore LHS =RHS

Hence, verified.

11. 2y – (1/2) = (-1/3)

Solution:

Given 2y – (1/2) = (-1/3)

Adding (1/2) from both the sides, we get

2y – (1/2) + (1/2) = (-1/3) + (1/2)

2y = (-1/3) + (1/2)

2y = (-2 + 3)/6 [LCM of 3 and 2 is 6]

2y = (1/6)

Now divide both the side by 2, we get

y = (1/12)

Verification:

Substituting y = (1/12) in LHS we get

2 (1/12) – (1/2) = (-1/3)

(1/6) – (1/2) = (-1/3)

(2 – 6)/12 = (-1/3) [LCM of 6 and 2 is 12]

(-4/12) = (-1/3)

(-1/3) = (-1/3)

Therefore LHS = RHS

Hence, verified.

12. 14 = (7x/10) – 8

Solution:

Given 14 = (7x/10) – 8

Adding 8 to both sides we get,

14 + 8 = (7x/10) – 8 + 8

22 = (7x/10)

Multiply both sides by 10 we get,

220 = 7x

x = (220/7)

Verification:

Substituting x = (220/7) in RHS we get,

14 = (7/10) × (220/7) – 8

14 = 22 -8

14 = 14

Therefore LHS = RHS.

Hence, verified.

13. 3 (x + 2) = 15

Solution:

Given 3 (x + 2) = 15

Dividing both sides by 3 we get,

3 (x + 2)/3 = (15/3)

(x + 2) = 5

Now subtracting 2 by both sides, we get

x + 2 -2 = 5 -2

x = 3

Verification:

Substituting x =3 in LHS we get,

3 (3 + 2) = 15

3 (5) = 15

15 = 15

Therefore LHS = RHS

Hence, verified.

14. (x/4) = (7/8)

Solution:

Given (x/4) = (7/8)

Multiply both sides by 4 we get,

(x/4) × 4 = (7/8) × 4

x = (7/2)

Verification:

Substituting x = (7/2) in LHS we get,

(7/2)/4 = (7/8)

(7/8) = (7/8)

Therefore LHS = RHS

Hence, verified.

15. (1/3) – 2x = 0

Solution:

Given (1/3) – 2x = 0

Subtract (1/3) from both sides we get,

(1/3) – 2x – (1/3) = 0 – (1/3)

– 2x = – (1/3)

2x = (1/3)

Divide both side by 2 we get,

2x/2 = (1/3)/2

x = (1/6)

Verification:

Substituting x = (1/6) in LHS we get,

(1/3) – 2 (1/6) = 0

(1/3) – (1/3) = 0

0 = 0

Therefore LHS = RHS

Hence, verified.

16. 3 (x + 6) = 24

Solution:

Given 3 (x + 6) = 24

Divide both the sides by 3 we get,

3 (x + 6)/3 = (24/3)

(x + 6) = 8

Now subtract 6 from both sides we get,

x + 6 – 6 = 8 – 6

x = 2

Verification:

Substituting x = 2 in LHS we get,

3 (2 + 6) = 24

3 (8) =24

24 = 24

Therefore LHS =RHS

Hence, verified.

17. 3 (x + 2) – 2 (x – 1) = 7

Solution:

Given 3 (x + 2) – 2 (x – 1) = 7

On simplifying the brackets, we get

3 × x + 3 × 2 – 2 × x + 2 × 1 = 7

3x + 6 – 2x + 2 = 7

3x – 2x + 6 + 2 = 7

x + 8 = 7

Subtracting 8 from both sides, we get

x + 8 – 8 = 7 – 8

x = -1

Verification:

Substituting x = -1 in LHS, we get

3 (x + 2) -2(x -1) = 7

3 (-1 + 2) -2(-1-1) = 7

(3×1) – (2×-2) = 7

3 + 4 = 7

Therefore LHS = RHS

Hence, verified.

18. 8 (2x – 5) – 6(3x – 7) = 1

Solution:

Given 8 (2x – 5) – 6(3x – 7) = 1

On simplifying the brackets, we get

(8 × 2x) – (8 × 5) – (6 × 3x) + (-6) × (-7) = 1

16x – 40 – 18x + 42 = 1

16x – 18x + 42 – 40 = 1

-2x + 2 = 1

Subtracting 2 from both sides, we get

-2x+ 2 – 2 = 1 -2

-2x = -1

Multiplying both sides by -1, we get

-2x × (-1) = -1× (-1)

2x = 1

Dividing both sides by 2, we get

2x/2 = (1/2)

x = (1/2)

Verification:

Substituting x = (1/2) in LHS we get,

(8 × (2 × (1/2)–5) – (6 × (3 × (1/2)-7) = 1

8(1 – 5) – 6(3/2 – 7) = 1

8× (-4) – (6 × 3/2) + (6 × 7) = 1

– 32 – 9 + 42 = 1

– 41 + 42 = 1

1 = 1

Therefore LHS = RHS

Hence, verified.

19. 6 (1 – 4x) + 7 (2 + 5x) = 53

Solution:

Given 6 (1 – 4x) + 7 (2 + 5x) = 53

On simplifying the brackets, we get

(6 ×1) – (6 × 4x) + (7 × 2) + (7 × 5x) = 53

6 – 24x + 14 + 35x = 53

6 + 14 + 35x – 24x = 53

20 + 11x = 53

Subtracting 20 from both sides, we get 20 + 11x – 20 = 53 – 20

11x = 33

Dividing both sides by 11, we get

11x/11 = 33/11

x = 3

Verification:

Substituting x = 3 in LHS, we get

6(1 – 4 × 3) + 7(2 + 5 × 3) = 53

6(1 – 12) + 7(2 + 15) = 53

6(-11) + 7(17) = 53

– 66 + 119 = 53

53 = 53

Therefore LHS = RHS

Hence, verified.

20. 5 (2 – 3x) -17 (2x -5) = 16

Solution:

Given 5 (2 – 3x) -17 (2x – 5) = 16

On expanding the brackets, we get

(5 × 2) – (5 × 3x) – (17 × 2x) + (17 × 5) = 16

10 – 15x – 34x + 85 = 16

10 + 85 – 34x – 15x = 16

95 – 49x = 16

Subtracting 95 from both sides, we get

– 49x + 95 – 95 = 16 – 95

– 49x = -79

Dividing both sides by – 49, we get

– 49x/ -49 = -79/-49

x = 79/49

Verification:

Substituting x = (79/49) in LHS we get,

5 (2 – 3 × (79/49) – 17 (2 × (79/49) – 5) = 16

(5 × 2) – (5 × 3 × (79/49)) – (17 × 2 × (79/49)) + (17 × 5) = 16

10 – (1185/49) – (2686/49) + 85 = 16

(490 – 1185 – 2686 + 4165)/49 = 16

784/49 = 16

16 = 16

Therefore LHS = RHS

Hence, verified.

21. (x – 3)/5 -2 = -1

Solution:

Given ((x – 3)/5) -2 = -1

Adding 2 to both sides we get,

((x -3)/5) – 2 + 2 = -1 + 2

(x -3)/5 = 1

Multiply both sides by 5 we get

(x – 3)/ 5 × 5 = 1 × 5

x – 3 = 5

Now add 3 to both sides we get,

x – 3 + 3 = 5 + 3

x = 8

Verification:

Substituting x = 8 in LHS we get,

((8 – 3)/5) -2 = -1

(5/5) – 2 = -1

1 -2 = -1

-1 = -1

Therefore LHS = RHS

Hence, v