RS Aggarwal Class 6 Solutions Linear Equations In One Variable

RS Aggarwal Class 6 Solutions Chapter 9

Equations which are having variables with the highest power as 1 are called linear equations in one variable. Learning the concepts of linear equations is very much important to understand the concepts of linear equations in two variables which will be taught in higher classes. To practice questions on linear equations in one variable student are advised to follow RS Aggarwal class 6 solutions chapter 9 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

At which of the times shown in the options, the hour hand and the minute hand of your clock will be perpendicular to each other?