Chapter 8: Linear Equations in One Variable Exercise – 8.1

Question: 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) x = 4 is the root of 3x - 5 = 7.

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

3(4) – 5 = 7

12 – 5 = 7

7 = 7

Since, LHS = RHS

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

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

Now, substituting x = 3 in place of ‘x’ in the given equation 5 + 3x = 14,

5 + 3(3) = 14

5 + 9 = 14

14 = 14

Since, LHS = RHS

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

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

Now, substituting x = 2 in place of ‘x’ in the given equation 3x –2 = 8x – 12,

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

Now, substituting x = 4 in place of ‘x’ in the given equation 3x/2 = 6,

(3× 4)/2 = 6

12/2 = 6

6 = 6

Since, LHS = RHS

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

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

Now, substituting y = 2 in place of ‘y’ in the given equation y – 3 = 2y – 5,

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). x = 8 is the root of 12x + 7 = 11.

Now, substituting x = 8 in place of ‘x’ in the given equation 12x + 7 = 11,

12(8) + 7 =11

4 + 7 = 11

11 = 11

Since, LHS = RHS

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

Question: 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

(viii) x/18 = 20

Solution:

(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 – 7 = 10

Here, LHS = x – 7 and RHS = 10.

Therefore, if x = 17, LHS = RHS.

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

(iii) 4x = 28

Here, LHS = 4x and RHS = 28.

Therefore, if x = 7, LHS = RHS

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

(iv) 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.

Therefore, if x = 8, LHS = RHS.

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

(v) 2x + 4 = 3x

Here, LHS = 2x + 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) 15x = 3

Here, LHS = 15x 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.

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.