Exercise 8.1
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
After that, LHS = RHS
Then, 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
After that, LHS = RHS
Then, 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
After that, LHS = RHS
Then, 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
After that, LHS = RHS
Then, 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
After that, LHS = RHS
Then, 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
After that, LHS = RHS
Then, 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
(viii) x/18 = 20
Solution:
(i)Given: x + 3 = 12
Here LHS = X + 3 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 | No |
So, if x = 9, LHS.
Thus, x- 9 is the solution to this equation.
(ii) Given 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 | 19-7=6 | 10 | No |
14 | 14-7=7 | 10 | No |
15 | 17-7= 8 | 10 | No |
16 | 16-7=9 | 10 | No |
17 | 17-7=10 | 10 | No |
(iii) Given 4x = 28
Here LHS = 4X and RHS = 28
X | LHS | RHS | Is LHS = RHS |
1 | 4×1= 1 | 28 | No |
2 | 4×2=8 | 28 | No |
3 | 4×3=12 | 28 | No |
4 | 4×4=16 | 28 | No |
5 | 4×5= 20 | 28 | No |
6 | 4×6=24 | 28 | No |
7 | 4×7=28 | 28 | No |
So, if = 7, LHS = RHS
Then, x = 7 is the solution to this equation.
(iv) Given x/2 + 7 = 11
Here LHS = x/2 + 7 = 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 LSH = RHS |
2 | (2/2)+7=1+7=8 | 11 | No |
4 | (4/2)+7=2+7=9 | 11 | No |
6 | (6/2) + 7 = 3 + 7 = 10 | 11 | No |
8 | (8/2) + 7 = 4+7= 11 | 11 | Yes |
So, if = 8, LHS = RHS
Then, x = 8 is the solutions to this equation.
(v) Given 2x + 4 = 3x
Here LHS = x/4 and RHS = 12
Since RHS is natural number, x4 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 |
14 | 24/4= 6 | 12 | No |
18 | 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 | 45/4= 12 | 12 | No |
Therefore if = 48, LHS = RHS
Then, x= 48 is the solution to this equation.
(vi) Given x/4 = 12
Here LHS = (15/x ) and RHS = 3
Since RHS is a natural number, 15x must also be a natural number, so we must
Substituting values of x that are factors of 15.
(vii) Given 15/x = 3
Here LHS = (15/X) and RHS = 3
Since RHS is natural number, 15x must also be natural number, so we must
Substituting 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 | No |
Therefore if x = 5, LHS = RHS
Then, x = 5 is the solutions to this equation.
(viii) Given (x/18) = 20
Here LHS = (X/18) AND RHS = 20
Since RHS is natural number, (x/18) must be a natural number, so must
Substituting values of x that are multiples of 18.
X | LHS | RHS | Is LHS = RHS |
324 | (324/18) = 18 | 20 | No |
342 | (342/18) | 20 | No |
360 | (360/18) = 20 | 20 | Yes |
Therefore if = 360, LHS = RHS
Then, = x = 360 is the solution to this equation.