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## Chapter 4 Simple Equations Exercise 4.1

Question 1: Complete the last column of the table.

 S. No. Equation Value Say, whether the equation is satisfied. (Yes/No) 1 x + 3 = 0 x = 3 = 3 + 3 ≠ 0 Therefore, the equation is not satisfied. 2 x + 3 = 0 x = 0 = 0 + 3 ≠ 0 Therefore, the equation is not satisfied. 3 x + 3 = 0 x = -3 = -3 + 3 = 0 Therefore, the equation is satisfied. 4 x – 7 = 1 x = 7 = 7 - 7 ≠ 1 Therefore, the equation is not satisfied. 5 x – 7 = 1 x = 8 = 8 - 7 = 1 Therefore, the equation is satisfied. 6 5x = 25 x = 0 = 5 x 0 ≠ 25 Therefore, the equation is not satisfied. 7 5x = 25 x = 5 = 5 x 5 = 25 Therefore, the equation is satisfied. 8 5x = 25 x = -5 = 5 x -5 ≠ 25 Therefore, the equation is not satisfied. 9 (m/3) = 2 m = – 6 = -6/2 ≠ 2 Therefore, the equation is not satisfied. 10 (m/3) = 2 m = 0 = 0/3 ≠ 2 Therefore, the equation is not satisfied. 11 (m/3) = 2 m = 6 = 6/2 = 3 Therefore, the equation is satisfied.

Question 2: Check whether the value given in the brackets is a solution to the given equation or not:
(a) n + 5 = 19 (n = 1)

 LHS RHS 1 + 5 19 6 19

6 ≠ 19
Therefore, the value n = 1 is not a solution to n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)

 LHS RHS 7 x (-2) + 5 = -14 + 5 19 -9 19

-9 ≠ 19
Therefore, the value n = -2 is not a solution to 7n + 5 = 19.

(c) 7n + 5 = 19 (n = 2)

 LHS RHS 7 x 2 + 5 = 14 + 5 19 9 19

9 = 19
Therefore, the value n = 2 is a solution to 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)

 LHS RHS 4 x 1 - 3 13 1 13

1 ≠ 13
Therefore, the value p = 1 is not a solution to 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)

 LHS RHS 4 x (-4) - 3 13 = -16 - 3 13

-19 ≠ 13
Therefore, the value p = -4 is not a solution to 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)

 LHS RHS 4 x 0 - 3 13 = 0 - 3 13

-3 ≠ 13
Therefore, the value p = 0 is not a solution to 4p – 3 = 13.

Question 3: Solve the following equations by trial and error method:
(i) 5p + 2 = 17

 p 5p +2 = 17 0 = 5 x 0 + 2 = 0 + 2 = 2 No 1 = 5 x 1 + 2 = 5 + 2 = 7 No 2 = 5 x 2 + 2 = 10 + 2 = 12 No 3 = 5 x 3 + 2 = 15 + 2 = 17 Yes

(ii) 3m – 14 = 4

 m 3m - 14 = 4 0 = 3 x 0 - 14 = 0 - 14 = -14 No 1 = 3 x 1 - 14 = 3 - 14 = -11 No 2 = 3 x 2 - 14 = 6 - 14 = -8 No 3 = 3 x 3 - 14 = 9 - 14 = -5 No 4 = 3 x 4 - 14 = 12 - 14 = -2 No 5 = 3 x 5 - 14 = 15 - 14 = 1 No 6 = 3 x 6 - 14 = 18 - 14 = 4 Yes

Question 4: Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
x + 4 = 9

(ii) 2 subtracted from y is 8.
y - 2 = 8

(iii) Ten times a is 70.
10a = 70

(iv) The number b divided by 5 gives 6.
b/5 = 6

(v) Three-fourth of t is 15.
3/4 t = 15

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

(vii) One-fourth of a number x minus 4 gives 4.
1/4 x - 4 = 4

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

(ix) If you add 3 to one-third of z, you get 30.
1/3 z + 3 = 30

Question 5: Write the following equations in statement forms:
(i) p + 4 = 15
The sum of numbers p and 4 is 15.

(ii) m – 7 = 3
7 subtracted from m is 3.

(iii) 2m = 7
Twice of number m is 7.

(iv) m/5 = 3
The number m divided by 5 gives 3.

(v) (3m)/5 = 6
Three-fifth of m is 6.

(vi) 3p + 4 = 25
Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18
Four times p minus 2 gives you 18.

(viii) p/2 + 2 = 8
If you add half of a number p to 2, you get 8.

Question 6: Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)

Number of marbles Parmit has = m
Irfan has 7 marbles more than five times the marbles Parmit has
= 5 x number of Parmit’s marbles + 7 = Total number of marbles Irfan having
= (5 x m) + 7 = 37
= 5m + 7 = 37

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

Let Laxmi’s age be = y years old
Laxmi’s father is 4 years older than three times of her age
= 3 x Laxmi’s age + 4 = Age of Laxmi’s father
= (3 x y) + 4 = 49
= 3y + 4 = 49

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

Highest score in the class = 87
Let the lowest score be l.
= 2 x lowest score + 7 = highest marks in the class
= (2 x l) + 7 = 87
= 2l + 7 = 87

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