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```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

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.
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