The document RD Sharma Solutions (Part - 2) - Ex - 8.2, Linear Equations in One Variable, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

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**Question 13:**

**Solve each of the following equations and check your answers: 3 ( x + 2) = 15**

3 (*x* + 2) = 15

Dividing both sides by 3, we get

Subtracting 2 from both sides, we get

â‡’ *x* + 2 âˆ’- 2 = 5 âˆ’- 2

â‡’ *x = 3*__Verification:__

Substituting x = 3 in LHS, we get

LHS = 3 (*x* + 2)= 3 (3+2) = 3Ã—Ã—5 = 15, and RHS = 15

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers:**

Multiplying both sides by 4, we get

__Verification:__

Substituting x = 7/2 in LHS, we get

LHS =

LHS = RHS

Hence, verified.

**Question 15:**

**Solve each of the following equations and check your answers:**

Subtracting 1/3 from both sides, we get

Multiplying both sides by âˆ’-1, we get

Dividing both sides by 2, we get

__Verification:__

Substituting x = 1/6 in LHS, we get

LHS =

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers: 3( x + 6) = 24**

3(*x* + 6) = 24

Dividing both sides by 3, we get

Subtracting 6 from both sides, we get

â‡’ *x* + 6 âˆ’ 6 = 8 âˆ’ 6

â‡’ *x = *2__Verification:__

Substituting x = 2 in LHS, we get

LHS = 3 (*x* + 6) = 3 (2 + 6) = 3Ã—8 = 24, and RHS = 24

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers: 3( x + 2) âˆ’ 2(x âˆ’ 1) = 7**

3(*x* + 2) âˆ’ 2(*x* âˆ’ 1) = 7

On expanding the brackets, we get

â‡’ (3Ã— x) +( 3 Ã— 2) âˆ’ (2 Ã— x) + (2 Ã— 1) = 73Ã— x + 3 Ã— 2 - 2 Ã— x + 2 Ã— 1 = 7

â‡’ 3x + 6 âˆ’ 2x + 2 = 7

â‡’ 3x âˆ’ 2x + 6 + 2 = 7

â‡’ x + 8 = 7

Subtracting 8 from both sides, we get

â‡’ *x* + 8 âˆ’ 8 = 7 âˆ’ 8

â‡’ *x = *âˆ’1__Verification:__

Substituting x = âˆ’1 in LHS, we get

LHS = 3 (*x* + 2) âˆ’2(x âˆ’1), and RHS = 7

LHS = 3 (âˆ’1 + 2) âˆ’2(âˆ’1âˆ’1) = (3Ã—1) âˆ’ (2Ã—âˆ’2) = 3 + 4 = 7, and RHS = 7

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers: 8(2 x âˆ’ 5) âˆ’ 6(3x âˆ’ 7) = 1**

8(2*x* âˆ’ 5) âˆ’ 6(3*x* âˆ’ 7) = 1

On expanding the brackets, we get

â‡’ (8Ã—2x) âˆ’ (8 Ã— 5) âˆ’ (6 Ã— 3x) + (âˆ’6)Ã—(âˆ’7) = 1

â‡’ 16x âˆ’ 40 âˆ’ 18x + 42 = 1

â‡’ 16x âˆ’ 18x + 42 âˆ’ 40 = 1

â‡’ âˆ’2x + 2 = 1

Subtracting 2 from both sides, we get

â‡’ âˆ’2x + 2 âˆ’2 = 1 âˆ’ 2

â‡’ âˆ’2x = âˆ’1

Multiplying both sides by âˆ’1, we get

â‡’ âˆ’2x Ã—(âˆ’1) = âˆ’1Ã—(âˆ’1)

â‡’ 2x = 1

Dividing both sides by 2, we get

__Verification:__

Substituting x = 1/2 in LHS, we get

LHS = RHS

Hence, verified

**Solve each of the following equations and check your answers: 6(1 âˆ’ 4 x) + 7(2 + 5x) = 53**

6(1 âˆ’ 4*x*) + 7(2 + 5*x*) = 53

On expanding the brackets, we get

â‡’ (6Ã—1) âˆ’ (6 Ã— 4x) + (7 Ã—2) + (7Ã—5x) = 53

â‡’ 6 âˆ’24x + 14 + 35x = 53

â‡’ 6 + 14 + 35x âˆ’24x = 53

â‡’ 20 + 11x = 53

Subtracting 20 from both sides, we get

â‡’ 20 + 11x âˆ’20 = 53 âˆ’- 20

â‡’ 11x = 33

â‡’ Dividing both sides by 11, we get

11x/11 = 33/11

â‡’ x = 3

Verification:

Substituting x = 3 in LHS, we get

= 6(1 âˆ’ 4Ã—Ã—3) + 7(2 + 5Ã—Ã—3)

= 6(1 âˆ’ 12) + 7(2 + 15)

= 6(âˆ’11) + 7(17)

= âˆ’66 + 119 = 53 = RHS

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers: 5(2 âˆ’ 3 x) âˆ’ 17(2x âˆ’ 5) = 16**

5(2 âˆ’ 3*x*) âˆ’ 17(2*x* âˆ’ 5) = 16

On expanding the brackets, we get

â‡’ (5Ã—Ã—2) âˆ’- (5 Ã—Ã— 3x) âˆ’ (17 Ã—Ã—2x ) + (17Ã—Ã—5) = 16

â‡’ 10 âˆ’- 15x âˆ’ 34x + 85 = 16

â‡’ 10 + 85 âˆ’ 34x âˆ’- 15x = 16

â‡’ 95 - 49x = 16

Subtracting 95 from both sides, we get

â‡’ - 49x + 95 âˆ’- 95 = 16 âˆ’- 95

â‡’ - 49x = -79

Dividing both sides by -49, we get

Verification:

Substituting x = 79/49 in LHS, we get

= RHS

So, LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers:**

Adding 2 to both sides, we get

Multiplying both sides by 5, we get

Adding 3 to both sides, we get

â‡’ x âˆ’3 + 3 = 5 + 3

â‡’ x = 8

Verification:

Substituting x = 8 in LHS, we get

= RHS

LHS = RHS

Hence, verified.

**Solve each of the following equations and check your answers: 5( x âˆ’ 2) + 3(x + 1) = 25**

5(*x* âˆ’ 2) + 3(*x* + 1) = 25

On expanding the brackets, we get

â‡’ (5 Ã—Ã— x) âˆ’- (5 Ã—Ã— 2) + (3 Ã—Ã— x) + (3Ã—Ã—1) = 25

â‡’ 5x âˆ’- 10 + 3x + 3 = 25

â‡’ 5x + 3x âˆ’- 10 + 3 = 25

â‡’ 8x âˆ’- 7 = 25

Adding 7 to both sides, we get

â‡’ 8x âˆ’- 7 + 7 = 25 +7

â‡’ 8x = 32

Dividing both sides by 8, we get

Verification:

Substituting x = 4 in LHS, we get

= 5(4 âˆ’ 2) + 3(4 + 1)

= 5(2) + 3(5)

= 10 + 15

= 25

= RHS

LHS = RHS

Hence, verified.