RS Aggarwal 2021-2022 for Class 6 Maths Solutions Chapter 9- Linear Equations in One Variable
RS Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable Exercise 9A is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily. RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.
Rs Aggarwal Class 6 Math Solution Chapter 9- Linear Equations in One Variable
Exercise 9A
1. Write each of the following statements as an equation:
(i) 5 times a number equals 40.
Solution:
Let the number be x
So, 5 times the number will be 5x.
∴ 5x = 40
(ii) A number increased by 8 equals 15.
Solution:
Let the number be x
So, when it is increased by 8, we get x + 8.
∴ x + 8 = 15
(iii) 25 exceeds a number by 7.
Solution:
Let the number be x
So, when 25 exceeds the number, we get 25 − x.
∴ 25 − x = 7
(iv) A number exceeds 5 by 3.
Solution:
Let the number be x
So, when the number exceeds 5, we get x − 5.
∴ x − 5 = 3
(v) 5 subtracted from thrice a number is 16.
Solution:
Let the number be x
So, thrice the number will be 3x.
∴ 3x − 5 = 16
(vi) If 12 is subtracted from a number, the result is 24.
Solution:
Let the number be x
So, 12 subtracted from the number will be x − 12.
∴ x − 12 = 24
(vii) Twice a number subtracted from 19 is 11.
Solution:
Let the number be x
So, twice the number will be 2x.
∴ 19 − 2x = 11
(viii) A number divided by 8 gives 7.
Solution:
Let the number be x
So, the number when divided by 8 will be `frac\{x}{8}`.
∴ `frac\{x}{8}` = 7
(ix) 3 less than 4 times a number is 17.
Solution:
Let the number be x
So, four times the number will be 4x.
∴ 4x − 3 = 17
(x) 6 times a number is 5 more than the number.
Solution:
Let the number be x
So, 6 times the number will be 6x.
∴ 6x = x + 5
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) `frac\{2z}{3}` = 8
Solution:
(i) 7 taken away from the number x is 14.
(ii) Twice the number y is 18.
(iii) 11 added to thrice the number x is 17.
(iv) 3 less than twice the number x is 13.
(v) 30 less than 12 times the number y is 6.
(vi) When twice the number z is divided by 3, it equals 8.
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 `frac\{z}{7}` = 8 is z = 56
Solution:
(i) The given equation is 3x – 5 = 7
Substituting x = 4, we get
L.H.S. = 3x – 5
= 3 × 4 – 5
= 12 – 5
and R.H.S. = 7
It is verified that x = 4 is the root of the given equation.
(ii) The given equation is 3 + 2x = 9
Substituting x = 3, we get
L.H.S. = 3 + 2x
= 3 + 2 × 3
= 3 + 6 = 9
= R.H.S.
It is verified that x = 3 is the root of the given equation.
(iii) The given equation is 5x – 8 = 2x – 2
Substituting x = 2, we get
L.H.S. = 5x – 8
= 5 × 2 – 8
= 10 – 8
= 2
R.H.S. = 2x – 2
= 2 × 2 – 2
= 4 – 2
= 2
L.H.S. = R.H.S.
Hence, it is verified that x = 2, is the root of the given equation.
(iv) The given equation is 8 – 7y = 1
Substituting y = 1, we get
L.H.S. = 8 – 7y
= 8 – 7 × 1
= 8 – 7
= 1
L.H.S = R.H.S.
Hence, it verified that y = 1 is the root of the given equation.
(v) The given equation is `frac\{z}{7}` = 8
Substituting the value of z = 56, we get
L.H.S.= `frac\{56}{7}`
= 8
= R.H.S.
Hence, it is verified that z = 56 is the root of the given equation.
4. Solve each of the following equation by trail and error method:
(i) y + 9 = 13
(ii) x – 7 = 10
(iii) 4x = 28
(iv) 3y = 36
(v) 11 + x = 19
(vi) `frac\{x}{3}` = 4
(vii) 2x – 3 = 9
(viii) `frac\{1}{2}`x + 7 = 11
(ix) 2y + 4 = 3y
(x) z – 3 = 2z – 5
Solution:
(i) `y` + 9 = 13
We try different value of `y` until we get the L.H.S. equal to the R.H.S
`y` = 4 satisfy the equation. So `y` = 4 is right answer.
(ii) `x` – 7 = 10
We try different value of `x` until we get the L.H.S. equal to the R.H.S
`x` = 17 satisfy the equation. So `x` = 17 is right answer.
(iii) 4x = 28
We try different value of `x` until we get the L.H.S. equal to the R.H.S
`x` = 7 satisfy the equation. So `x` = 7 is right answer.
(iv) 3y = 36
We try different value of `y` until we get the L.H.S. equal to the R.H.S
`y` = 12 satisfy the equation. So `y` = 12 is right answer.
(v) 11 + x = 19
We try different value of `x` until we get the L.H.S. equal to the R.H.S
`x` = 8 satisfy the equation. So `x` = 8 is right answer.
(vi) `frac\{x}{3}` = 4
We try different value of `x` until we get the L.H.S. equal to the R.H.S
`x` = 12 satisfy the equation. So `x` = 12 is right answer.
(vii) 2x – 3 = 9
We try different value of `x` until we get the L.H.S. equal to the R.H.S
`x` = 6 satisfy the equation. So `x` = 6 is right answer.
(viii) `frac\{1}{2}`x + 7 = 11
We try different value of `x` until we get the L.H.S. equal to the R.H.S.
`x` = 8 satisfy the equation. So `x` = 8 is right answer.
(ix) 2y + 4 = 3y
We try different value of `y` until we get the L.H.S. equal to the R.H.S.
`y` = 4 satisfy the equation. So `y` = 4 is right answer.
(x) z – 3 = 2z – 5
We try different value of `z` until we get the L.H.S. equal to the R.H.S.
`z` = 2 satisfy the equation. So `z` = 2 is right answer.
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