## RS Aggarwal Class 7 solutions of Chapter 9 Linear Equations in One Variable Ex 9A in pdf free download

Contents

**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**

**(viii) A number divided by 8 gives 7.**

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

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

**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) = 7

(ix) 4x – 3 = 17

(x) 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) = 8**

**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.**

**Verify by substitution that:**

**(i) The root of 3 x – 5 = 7 is x = 4.**

**(ii) The root of 3 + 2x = 9 is x = 3.**

**(iii) The root of 5 x – 8 = 2x – 2 is x = 2.**

**(iv) The root of 8 – 7y = 1 is y = 1.**

**(v) The root of = 8 is z = 56**

**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 = 8

Substituting the value of z = 56, we get

L.H.S.=

= 8

= R.H.S.

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

**Question 4.**

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

**(i) y + 9= 13**

**(ii) x – 7 = 10**

**(iii) 4x = 28**

**(iv) 3y = 36**

**(v) 11 + x = 19**

**(vi) = 4**

**(v) 2x – 3 = 9**

**(viii) + 7 = 11**

**(ix) 2 y + 4 = 3y**

**(x) z – 3 = 2z – 5**

**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 = 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 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.

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