# EXERCISE – 4.1

Q1. Complete the last column of the table.

 S. No Equation Value Say, whether the equation is satisfied. ( Yes/No) (i) X + 3 = 0 X = 3 No (ii) X +3 = 0 X = 0 No (iii) X + 3 = 0 X = -3 Yes (iv) X – 7 = 1 X = 7 No (v) X – 7 = 1 X = 8 Yes (vi) 5x = 25 X = 0 No (vii) 5x = 25 X = 5 Yes width="162">(viii) 5x = 25 X = -5 No (ix) m/3 = 2 m = -6 No (x) m/3 = 2 m = 0 No (xi) m/3 = 2 m = 6 Yes

Q2. Check whether the value given in the brackets is a solution to the given equation or not:

(a)  n + 5 = 19 (n = 1)

SOLUTION:-

= 1 + 5 = 19

= 6 = 19

RHS ≠LHS

(b) 7n + 5 = 19 ( n = -2)

SOLUTION:-

= 7(-2) + 5 = 19

= (-14) + 5 = 19

= (-9) = 19

= LHS ≠ RHS

(C)  7n + 5 = 19 ( n = 2)

SOLUTION:-

= 7(2) + 5 = 19

= 14 + 5 = 19

= 19=19

= LHS=RHS

(D) 4p – 3 = 13 ( p = 1)

SOLUTION:-

= 4(1) – 3 = 13

= 4 – 3 = 13

= 1 = 13

RHS ≠ LHS

(E) 4p – 3 = 13 ( p = -4)

SOLUTION:-

= 4(-4) – 3 = 13

= (-16) – 3 = 13

= (-19) = 13

RHS ≠LHS

(F) 4p -3 = 13 (p=0)

SOLUTION:-

= 4(0) – 3 = 13

= (-3) = 13

Q3. Solve the following equations by trial and error method:

(a) 5p + 2 = 17

SOLUTION:-

p = 0

5(0) + 2 = 17

0 + 2 = 17

2 ≠ 17  LHS ≠ RHS

p = 1

5(1) + 2 = 17

5 + 2 = 17

7 ≠ 17   LHS ≠ RHS

p = 2

5(2) + 2 = 17

10 + 2 = 17

12 ≠ 17 LHS ≠ RHS

p = 3

5(3) + 2 = 17

15 + 2 = 17

17 = 17  LHS = RHS

(B) 3m – 14 = 4

SOLUTION:-

m = 1

3(1) – 14 = 4

3 -14 = 4

-11 ≠ 4   LHS ≠ RHS

m = 6

3(6) – 14 = 4

18 – 14 = 4

4 = 4   LHS = RHS

Q4. Write equations for the following statements:

(i) The sum of numbers x and 4 is 9

SOLUTION:-

X + 4 = 9

(ii) 2 subtracted from y is 8.

SOLUTION:-

Y – 2 = 8

(iii) Ten times a is 70.

SOLUTION:-

10A = 70

(iv) The numbers b divided by 5 gives 6 .

SOLUTION:-

B/5 = 6

(v) Three fourth of t is 15.

SOLUTION:-

3T /4 = 15

(vi) Seven times m plus 7 gets you 77.

SOLUTION:-

7m + 7 = 77

(vii) One – fourth of a numbers x minus 4 gives 4.

SOLUTION:-

X/4 – 4 = 4

(viii) If you take away 6 from 6 times y , you get 60.

SOLUTION:-

6Y – 6 = 60

(ix) If you add 3 to one – third of z, you get 30.

SOLUTION:-

Z/3 + 3 = 30

Q5. Write the following equations in statements forms:

(i) p + 4 = 15

SOLUTION:-

The sum of numbers p and 4 is 15.

(ii) m – 7 = 3

SOLUTION:-

7 SUBTRACTED FROM M IS 3.

(iii) 2m = 7

SOLUTION:-

Twice of number m is 7.

(iv) m/5 = 3

SOLUTION:-

The numbers m divided by 5 gives 3.

(v) (3m)/5 = 6

Three – fifth of m is 6.

(vi) 3p + 4 = 25

SOLUTION:-

Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18

SOLUTION:-

Four times p minus 2 gives you 18.

(viii) p/2 + 2 = 8

SOLUTION:-

If you add half of a number p to 2, you get 8.

Q6. Set up an equation in the following cases:

(i) Ifran says that he has 7 marbles more than five times the marbels. Parmit has. Ifran has 37 marbles. (Take m to be the numbers of Parmit marbles.)

SOLUTION:-

From the question it is given that,

Number of Parmit marbles = m

Then,

Ifran  has 7 marbles more than five times the marbles Parmit has

= ( 5 x m ) + 7 = 37

= 5m + 7 = 37

(ii) Laxmi father is 49 years old. He is 4 years older than three times Laxmi age. (Take Laxmi age to be y years.)

SOLUTION:-

From the question it is given that,

Let Laxmi age to be = y years old

Then,

Laxmi father is 4 years older than the three times of her age

= 3 x Laxmi father age + 4 = Age of Laxmi father

= (3 x y )+4 = 49

= 3y + 4 = 49

(iii) The teacher tells the class that the highest marks obtained by a students in her class is twice the lowest marks plus 7. The highest score is 87.(Take the lowest score to be l.)

SOLUTION:-

From the question it is given that,

Highest score in the class = 87

Let lowest score be l

= 2 x Lowest score + 7 = Highest score in the class

= (2 x 1 ) + 7 = 87

= 2 + 7 = 87

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let us the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees).

SOLUTION:-

From the question it is given that,

We know that, the sum of angles of a triangle is 1800

Let base angle be b

Then,

Vertex angle = 2 x  base angle = 2b

= b + b + 2b = 1800

= 4b = 1800