## RS Aggarwal Class 6 Solutions Chapter 9 Linear Equations in One Variable Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9B
- RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9C

**Question 1.**

**Solution:**

Let x be the given number, then

(i) 5x = 40

(ii) x + 8 = 15

(iii) 25 – x = 7

(iv) x – 5 = 3

(v) 3x – 5 = 16

(vi) x – 12 = 24

(vii) 19 – 2x = 11

(viii) \(\\ \frac { x }{ 8 } \) = 7

(ix) 4x – 3 = 17

(x) 6x = x + 5

**Question 2.**

**Solution:**

(i) 7 less than from the number x is 14.

(ii) Twice the number y is 18.

(iii) 11 increased by thrice the number x is 17.

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

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

(vi) Quotient of twice the number z and 3 is

**Question 3.**

**Solution:**

(i) The given equation is 3x – 5 = 7

Substituting x = 4, we get

L.H.S. = 3 x – 5

= 3 x 4 – 5

= 12 – 5

= 7 = R.H.S.

It is verified that x = 4 is the root of the given equation.

(ii) The given equation is 3 + 2 x = 9

Substituting x = 3, we get L.H.S. = 3 + 2x

= 3 + 2 x 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 x 2 – 8

= 10 – 8

= 2

R.H.S. = 2x – 2

= 2 x 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 x 1

= 8 – 7

= 1

= 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.

**Question 4.**

**Solution:**

(i) The given equation is y + 9 = 13

We try several values of y and find L.H.S. and the R.H.S. and stop when L.H.S. = R.H.S.

When y = 4, we have L.H.S. = R.H.S.

So y = 4 is the solution of the given equation.

(ii) The given equation is x – 1 = 10

We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 17, we hive L.H.S. = R.H.S

So x = 17 is the solution of the given equation.

(iii) The given equation is 4x = 28

We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 7, we have L.H.S. = R.H.S.

So x = 7 is the solution of the given equation.

(iv) The given equation is 3y = 36

We guess and try several values of y to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When y = 12, we have L.H.S. = R.H.S.

So y = 12 is the solution of the given equation.

(v) The given equation is 11 + x = 19

We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 8, we have L.H.S. = R.H.S.

So, x = 8 is the solution of the given equation.

(vi) The given equation is \(\\ \frac { x }{ 3 } \) = 4

We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 12, we have L.H.S. = R.H.S.

So, x = 12 is the solution of the given equation.

(vii) The given equation is 2 x – 3 = 9

We guess and try several values of x to find L.H.S. and R.H.S. and stop when

.’. When x = 6, we have L.H.S. = R.H.S.

So, x = 6 is the solution of the given equation.

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

(viii) The given equation is \(\\ \frac { 1 }{ 2 } \) x + 7 = 11

We guess and try several values of x to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When x = 8, we have L.H.S. = R.H.S.

So, x = 8 is the solution of the given equation.

(ix) The given equation is 2y + 4 = 3y (x)

We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S.

When y = 4, we have L.H.S. = R.H.S. So, y = 4 is the solution of the given equation

(x) The given equation is z – 3 = 2z – 5

We guess and try several values of z to find L.H.S. and R.H.S. and stop when L.H.S. = R.H.S

When z = 2, we have L.H.S. = R.H.S. So, z = 2 is the solution of the given equation.

Hope given RS Aggarwal Solutions Class 6 Chapter 9 Linear Equations in One Variable Ex 9A are helpful to complete your math homework.

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