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
(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.
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
(i) x + 3 = 12
Here, LHS = x + 3 and RHS = 12
| x | LHS | RHS | Is LHS = RHS |
| 1 | 1 + 3 = 4 | 12 | No |
| 2 | 2 + 3 = 5 | 12 | No |
| 3 | 3 + 3 = 6 | 12 | No |
| 4 | 4 + 3 = 7 | 12 | No |
| 5 | 5 + 3 = 8 | 12 | No |
| 6 | 6 + 3 = 9 | 12 | No |
| 7 | 7 + 3 = 10 | 12 | No |
| 8 | 8 + 3 = 11 | 12 | No |
| 9 | 9 + 3 = 12 | 12 | Yes |
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.
| x | LHS | RHS | Is LHS = RHS |
| 9 | 9 – 7 = 2 | 10 | No |
| 10 | 10 – 7 = 3 | 10 | No |
| 11 | 11 – 7 = 4 | 10 | No |
| 12 | 12 – 7 = 5 | 10 | No |
| 13 | 13 – 7 = 6 | 10 | No |
| 14 | 14 – 7 = 7 | 10 | No |
| 15 | 15 – 7 = 8 | 10 | No |
| 16 | 16 – 7 = 9 | 10 | No |
| 17 | 17 – 7 = 10 | 10 | Yes |
Therefore, if x = 17, LHS = RHS.
Hence, x = 17 is the solution to this equation.
(iii) 4x = 28
Here, LHS = 4x and RHS = 28.
| x | LHS | RHS | Is LHS = RHS |
| 1 | 4 x 1 = 4 | 28 | No |
| 2 | 4 x 2 = 8 | 28 | No |
| 3 | 4 x 3 = 12 | 28 | No |
| 4 | 4 x 4 = 16 | 28 | No |
| 5 | 4 x 5 = 20 | 28 | No |
| 6 | 4 x 6 = 24 | 28 | No |
| 7 | 4 x 7 = 28 | 28 | Yes |
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.
| x | LHS | RHS | Is LHS = RHS |
| 2 | 2/2 + 7 = 8 | 11 | No |
| 4 | 4/2 + 7 = 9 | 11 | No |
| 6 | 6/2 + 7 = 10 | 11 | No |
| 8 | 8/2 + 7 = 11 | 11 | Yes |
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.
| x | LHS | RHS | Is LHS = RHS |
| 1 | 2(1) + 4 = 6 | 3(1) = 3 | No |
| 2 | 2(2) + 4 = 8 | 3(2) = 6 | No |
| 3 | 2(3) + 4 =10 | 3(3) = 9 | No |
| 4 | 2(4) + 5 = 12 | 3(4) = 12 | Yes |
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.
| X | LHS | RHS | Is LHS = RHS |
| 16 | 16/4 = 4 | 12 | No |
| 20 | 20/4 = 5 | 12 | No |
| 24 | 24/4 = 6 | 12 | No |
| 28 | 28/4 = 7 | 12 | No |
| 32 | 32/4 = 8 | 12 | No |
| 36 | 36/4 = 9 | 12 | No |
| 40 | 40/4 = 10 | 12 | No |
| 44 | 44/4 = 11 | 12 | No |
| 48 | 48/4 = 12 | 12 | Yes |
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.
| x | LHS | RHS | Is LHS = RHS |
| 1 | 15/1= 15 | 3 | No |
| 3 | 15/3 = 5 | 3 | No |
| 5 | 15/5 = 3 | 3 | Yes |
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.
| X | LHS | RHS | Is LHS = RHS |
| 324 | 324/18 = 18 | 20 | No |
| 342 | 342/18 = 19 | 20 | No |
| 360 | 360/18 = 20 | 20 | Yes |
Therefore, if x = 360, LHS = RHS.
Hence, x = 360 is the solution to this equation.