The document RD Sharma Solutions - Ex - 8.1, 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.

All you need of Class 7 at this link: Class 7

**Question 1:**

**Verify by substitution that: (i) x = 4 is the root of 3x âˆ’ 5 = 7 (ii) x = 3 is the root of 5 + 3x = 14 (iii) x = 2 is the root of 3x âˆ’ 2 = 8x âˆ’ 12 (iv) x = 4 is the root of (v) y = 2 is the root of y âˆ’ 3 = 2y âˆ’ 5 (vi) x = 8 is the root of **

(i) *x* = 4 is the root of 3*x* âˆ’ 5 = 7.

Now, substituting x = 4 in place of 'x' in the given equation 3x âˆ’ 5 = 7,

3(4) âˆ’ 5 = 7

12 âˆ’ 5 = 7

7 = 7

LHS = RHS

Hence, *x* = 4 is the root of 3*x* âˆ’ 5 = 7.

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

Now, substituting x = 3 in place of 'x' in the given equation 5 + 3*x* = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

LHS = RHS

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

(iii) *x* = 2 is the root of 3*x* âˆ’ 2 = 8*x* âˆ’ 12.

Now, substituting x = 2 in place of 'x' in the given equation 3*x* âˆ’ 2 = 8*x* âˆ’ 12,

3(2) âˆ’ 2 = 8(2) âˆ’ 12

6 âˆ’ 2 = 16 âˆ’ 12

4 = 4

LHS = RHS

Hence, *x* = 2 is the root of 3*x* âˆ’ 2 = 8*x* âˆ’ 12.

(iv) x = 4 is the root of

Now, substituting x = 4 in place of 'x' in the given equation

LHS = RHS

Hence, x = 4 is the root of

(v) *y* = 2 is the root of *y* âˆ’ 3 = 2*y* âˆ’ 5.

Now, substituting y = 2 in place of 'y' in the given equation *y* âˆ’ 3 = 2*y* âˆ’ 5,

2 âˆ’ 3 = 2(2) âˆ’ 5

âˆ’1 = 4 âˆ’ 5

âˆ’1 = âˆ’1

LHS = RHS

Hence, *y *= 2 is the root of *y* âˆ’ 3 = 2*y* âˆ’ 5.

(vi) *x* = 8 is the root of

Now, substituting x = 8 in place of 'x' in the given equation

4 + 7 = 11

11 = 11

LHS = RHS

Hence, *x* = 8 is the root of

**Solve each of the following equations by trial-and-error method:**

(i) *x* + 3 = 12

(ii) *x* âˆ’ 7 = 10

(iii) 4*x* = 28

(i) *x* + 3 = 12

Here, LHS = x + 3 and RHS = 12.

Therefore, if x = 9, LHS = RHS.

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

(ii) *x* âˆ’ 7 = 10

Here, LHS = x âˆ’7 and RHS =10.

Therefore, if x = 17, LHS = RHS.

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

(iii) 4*x* = 28

Here, LHS = 4x and RHS = 28.

Therefore, if x = 7, LHS = RHS.

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

Here, LHS = 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.

Therefore, if x = 8, LHS = RHS.

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

(v) 2*x* + 4 = 3*x** *Here, LHS = 2*x* + 4 and RHS = 3x.

Therefore, if x = 4, LHS = RHS.

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

(vi) 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.

Therefore, if x = 48, LHS = RHS.

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

(vii) 15/x = 3

Here, LHS =15/x and RHS = 3.

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

Therefore, if x = 5, LHS = RHS.

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

(viii) 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.

Therefore, if x = 360, LHS = RHS.

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