RS Aggarwal Class 6 Solutions Linear Equations In One Variable

Question 1:

Write each of the following statements as an equation:

(i) 5 times a number equals 40.

(ii) A number increased by 8 equals 15.

(iii) 25 exceeds a number by 7.

(iv) A number exceeds 5 by 3.

(v) 5 subtracted from thrice a number is 16.

(vi) If 12 is subtracted from a number, the result is 24.

(vii) Twice a number subtracted from 19 is 11.

(ix) 3 less than 4 times a number is 17.

(x) 6 times a number is 5 more than the number.

Solution:

(i) Let the required number be x.

So, five times the number will be 5x.

Therefore, 5x = 40

(ii) Let the required number be x.

So, when it is increased by 8, we get x + 8.

Therefore, x + 8 = 15

(iii) Let the required number be x.

So, when 25 exceeds the number, we get 25 – x.

Therefore, 25 – x = 7

(iv) Let the required number be x.

So, when the number exceeds 5, we get x – 5.

Therefore, x – 5 = 3.

(v) Let the required number be x.

So, thrice the number will be 3x.

Therefore, 3x – 5 = 16

(vi) Let the required number be x.

So, 12 subtracted from the number will be x – 12.

Therefore, x – 12 = 24

(vii) Let the required number be x.

So, twice the number will be 2x.

Therefore, 19 – 2x = 11

(viii) Let the required number be x.

So, the number when divided by 8 will be x8.

Therefore,x8=7

(ix) Let the required number be x.

So, four times the number will be 4x.

Therefore, 4x – 3 = 17

(x) Let the required number be x.

So, 6 times the number will be 6x.

Therefore, 6x = x + 5

Question 2:

Write a statement for each of the equations, given below:

(i) x – 7 = 14

(ii) 2y = 18

(iii) 11 + 3x = 17

(iv) 2x – 3 = 13

(v) 12y – 30 = 6

(vi) 2z3=8

Solution:

(i) 7 less than the number x equals 14.

(ii) Twice the number y equals 18.

(iii) 11 more than thrice the number x equals 17.

(iv) 3 less than twice the number x equals 13.

(v) 30 less than 12 times the number y equals 6.

(vi) When twice the number z is divided by 3, it equals 8.

Question 3:

Verify by substitution that

(i) the root of 3x – 5 = 7 is x = 4

(ii) the root of 3 + 2x = 9 is x = 3

(iii) the root of 5x – 8 = 2x – 2 is x = 2

(iv) the root of 8 – 7y = 1 is y = 1

(v) the root of z7=8 is z = 56

Solution:

(i) 3x – 5 = 7

Substituting x = 4 in the given equation:

L.H.S. : 3 x 4 – 5

or, 12 – 5 = 7 = R.H.S.

Hence, x = 4 is the root of the given equation.

(ii) 3 + 2x = 9

Substituting x = 3 in the given equation:

L.H.S. :

3 + 2 x 3

or, 3 + 6 = 9 = R.H.S.

L.H.S. = R.H.S.

Hence, x = 3 is the root of the given equation.

(iii) 5x – 8 = 2x – 2

Substituting x = 2 in the given equation:

L.H.S. :

5 x 2 – 8

or, 10 – 8 = 2

R.H.S. :

2 x 2 – 2

= 4 – 2 = 2

L.H.S. = R.H.S.

Hence, x = 2 is the root of the given equation.

(iv) 8 – 7y = 1

Substituting y = 1 in the given equation:

L.H.S. :

8 – 7 x 1

or, 8 – 7 = 1 = R.H.S.

L.H.S. = R.H.S.

Hence, y = 1 is the root of the given equation.

(v)z7=8

Substituting z = 56 in the given equation:

L.H.S. :

567=8 = R.H.S.

L.H.S. = R.H.S.

Hence, z = 56 is the root of the given equation.

Question 4:

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

(i) y + 9 = 13

(ii) x – 7 = 10

(iii) 4x = 28

(iv) 3y = 36

(v) 11 + x = 19

(vi) x3=4

(vii) 2x – 3 = 9

(viii) 12x+7=11

(ix) 2y + 4 = 3y

(x) z – 3 = 2z – 5

Solution:

(i) y + 9 = 13

We try several values of y until we get the L.H.S. equal to the R.H.S.

y L.H.S. R.H.S. Is L.H.S. = R.H.S.?
1 1 + 9 = 10 13 No
2 2 + 9 = 11 13 No
3 3 + 9 = 12 13 No
4 4 + 9 = 13 13 Yes

Therefore, y = 4

(ii) x – 7 = 10

We try several values of x until we get the L.H.S. equal to the R.H.S.

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
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, x = 17

(iii) 4x = 28

We try several values of x until we get the L.H.S. equal to the R.H.S.

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
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, x = 7

(iv) 3y = 36

We try several values of y until we get the L.H.S. equal to the R.H.S.

y L.H.S. R.H.S. Is L.H.S. = R.H.S.?
6 3 x 6 = 18 36 No
7 3 x 7 = 21 36 No
8 3 x 8 = 24 36 No
9 3 x 9 = 27 36 No
10 3 x 10 = 30 36 No
11 3 x 11 = 33 36 No
12 3 x 12 = 36 36 Yes

Therefore, y = 12

(v) 11 + x = 19

We try several values of x until we get the L.H.S. equal to the R.H.S.

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
1 11 + 1 = 12 19 No
2 11 + 2 = 13 19 No
3 11 + 3 = 14 19 No
4 11 + 4 = 15 19 No
5 11 + 5 = 16 19 No
6 11 + 6 = 17 19 No
7 11 + 7 = 18 19 No
8 11 + 8 = 19 19 Yes

Therefore, x = 8

(vi)x3=4

Since R.H.S. is an natural number so L.H.S. must also be a natural number. Thus, x has to be a multiple of 3.

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
3 33=1 4 No
6 63=2 4 No
9 93=3 4 No
12 123=4 4 Yes

Therefore, x = 12

(vii) 2x – 3 = 9

We try several values of x until we get the L.H.S. equal to the R.H.S.

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
1 2 x 1 – 3 = -1 9 No
2 2 x 2 – 3 = 1 9 No
3 2 x 3 – 3 = 3 9 No
4 2 x 4 – 3 = 5 9 No
5 2 x 5 – 3 = 7 9 No
6 2 x 6 – 3 = 9 9 Yes

Therefore, x = 6

(viii)12x+7=11

Since, R.H.S. is a natural number so L.H.S. must be a natural number. Thus, we will try values if x which are multiples of 2

x L.H.S. R.H.S. Is L.H.S. = R.H.S.?
2 22+7=8 11 No
4 42+7=9 11 No
6 62+7=10 11 No
8 82+7=11< 11 Yes

Therefore, x = 8

(ix) 2y + 4 = 3y

We try several values of y until we get the L.H.S. equal to the R.H.S.

y L.H.S. R.H.S. Is L.H.S. = R.H.S.?
1 2 x 1 + 4 = 6 3 x 1 = 3 No
2 2 x 2 + 4 = 8 3 x 2 = 6 No
3 2 x 3 + 4 = 10 3 x 3 = 9 No
4 2 x 4 + 4 = 12 3 x 4 = 12 Yes

Therefore, y = 4

(x) z – 3 = 2z – 5

We try several values of z till we get the L.H.S. equal to the R.H.S.

z L.H.S. R.H.S. Is L.H.S. = R.H.S.?
1 1 – 3 = -2 2 x 1 – 5 = -3 No
2 2 – 3 = -1 2 x 2 – 5 = -1 Yes

Therefore, z = 2


Practise This Question

How many numbers lie between 22 and 32?