1.     Sol.

2.    (a) n + 5 = 19(n = 1)

Putting n = 1 in L.H.S.,

1 + 5 = 6

∵ L.H.S. ≠ R.H.S.,

∴ n = 1 is not the solution of given equation.

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

Putting n = -2 in L.H.S.,

7 (-2) + 5 = -14 + 5 = -9

∵ L.H.S. ≠ R.H.S.,

∴ n = -2 is not the solution of given equation.

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

Putting n = 2 in L.H.S.,

7 (2) + 5 = 14 + 5 = 19

∵ L.H.S. = R.H.S.,

∴ n = 2 is the solution of given equation.

(d) 4p – 3 = 13 (p = 1)

Putting p = 1 in L.H.S.,

4(1) – 3 = 4 – 3 = 1

∵ L.H.S. ≠ R.H.S.,

∴ p = 1 is not the solution of given equation.

(e) 4p – 3 = 13(p = -4)

Putting p= -4 in L.H.S.,

4(-4) – 3 = -16 – 3 = -19

∵ L.H.S. ≠ R.H.S.,

∴ p = -4 is not the solution of given equation.

(f) 4p – 3 = 13(p = 0)

Putting p = 0 in L.H.S.,

4(0) – 3 = 0 – 3 = -3

∵ L.H.S. ≠ R.H.S.,

∴ p = 0 is not the solution of given equation.

3.    (i) 5p + 2 =17

Putting p = -3 in L.H.S.      5(-3) + 2 =  -15 + 2 = -13

∵ -13 ≠ 17 Therefore, p =  – 3 is not the solution.

Putting p = -2 in L.H.S.      5(-2) +  2 =  -10 +  2 = -8

∵ -8 ≠ 17               Therefore, p = -2 is not the solution.

Putting p = -1 in L.H.S.       5(-1) +  2 =  -5 + 2 = -3

∵ -3 ≠ 17              Therefore, p =  -1 is not the solution.

Putting p = 0 in L.H.S.           5(0) + 2 = 0 + 2 = 2

∵ 2 ≠ 17             Therefore, p= 0 is not the solution.

Putting p =1 in L.H.S.            5(1) + 2 = 5 + 2 = 7

∵ 7 ≠ 17 Therefore, p = 1 is not the solution.

Putting p = 2 in L.H.S.           5(2) + 2 = 10 + 2 = 12

∵ 12 ≠ 17 Therefore, p = 2 is not the solution.

Putting p = 3 in L.H.S.           5(3) + 2 = 15 + 2 = 17

∵ 17 = 17 Therefore, p = 3 is the solution.

(ii) 3m – 14 = 4

Putting m = -2 in L.H.S.               3 (-2) – 14 = -6 – 14 = -20

∵ -20 ≠ 4              Therefore, m = -2 is not the solution.

Putting m =  – 1 in L.H.S.          3 (-1) – 14 = -3 – 14 = -17

∵ – 17 ≠ 4              Therefore, m = -1 is not the solution.

Putting m = 0 in L.H.S.        3(0) – 14 = 0 – 14 = -14

∵ – 14 ≠ 4  Therefore, m = 0 is not the solution.

Putting m = 1 in L.H.S.        3(1) – 14 = 3 – 14 = -11

∵ -11 ≠ 4                     Therefore, m = 1 is not the solution.

Putting m = 2 in L.H.S.       3(2) – 14 = 6 – 14 = -8

∵ -8 ≠ 4                Therefore, m = 2 is not the solution.

Putting m =3 in L.H.S.       3(3) – 14 = 9 – 14 = -5

∵ -5 ≠ 4                 Therefore, m = 3 is not the solution.

Putting m = 4 in L.H.S.      3(4) – 14 = 12 – 14 = -2

∵ -2 ≠ 4                 Therefore, m = 4 is not the solution.

Putting m = 5 in L.H.S.        3(5) – 14 = 15 – 14 = 1

∵ 1 ≠ 4              Therefore, m = 5 is not the solution.

Putting m = 6 in L.H.S.        3(6) – 14 = 18 – 14 = 4

∵ 4 = 4       Therefore, m = 6 is the solution.

4.    

5.    (i) The sum of numbers p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Two times m is 7.

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

(v) Three – fifth of the number m is 6.

(vi) Three times p plus 4 gets 25.

(vii) If you take away 2 from 4 times p, you get 18.

(viii) If you added 2 to half is p, you get 8.

6.    (i)  Let m be the number of Parmit’s marbles.

∴ 5m + 7 = 37

(ii)  Let the age of Laxmi be y years.

∴ 3y + 4 =49

(iii) Let the lowest score be 1.

∴ 21 + 7=87

(iv) Let the base angle of the isosceles triangle be b, so vertex angle = 2b.

∴ 2b + b + b = 180°    ⇒     4b = 180°          [Angle sum property of a Δ]

7.   (a) x – 1 = 0              ⇒            x – 1 + 1 = 0 + 1             [Adding 1 both sides]

x = 1

(b) x + 1 = 0             ⇒           x + 1 – 1 = 0 – 1               [Subtracting 1 both sides]

x = -1

(c) x – 1 = 5               ⇒            x – 1 + 1 = 5 + 1              [Adding 1 both sides]

x = 6

(d) x + 6 = 2             ⇒            x + 6 – 6 = 2 – 6              [Subtracting 4 both sides]

x = -4

(e) y – 4 = -7            ⇒            y – 4 + 4 = -7 + 4            [Adding 4 both sides]

y = -3

(f) y – 4 = 4              ⇒            y – 4 + 4 = 4 + 4             [Adding 4 both sides]

y = 8

(g) y + 4 = 4            ⇒            y + 4 – 4 = 4 – 4              [Subtracting 4 both sides]

y = 0

(h) y + 4 = -4            ⇒             y + 4 – 4 = -4 – 4           Subtracting 4 both sides]

y = -8

8.    

9.    (a) 3n – 2 = 46

Step I: 3n – 2 + 2 = 46 + 2    ⇒    3n = 48    [Adding 2 both sides]

Step II:                      ⇒     n = 16       [Dividing both sides by 3]

(b) 5m + 7 = 17

Step I:    5m + 7 – 7 = 17 – 7   ⇒     5m = 10  [Subtracting 7 both sides]

Step II:                          ⇒     m = 2      [Dividing both sides by 5]

(c) 

Step I:                ⇒     20p = 120    [Multiplying both sides by 3]

Step II:                    ⇒      p = 6             [Dividing both sides by 20]

(d) 

Step I:                ⇒        3p = 60      [Dividing both sides by 3]

Step II:                       ⇒          p = 20       [Dividing both sides by 10]

10. 

11.  

12.  

13.  (a) 4 = 5(p – 2)            ⇒               4 = 5 × p – 5 × 2    ⇒        4 = 5p – 10

⇒ 5p – 10 = 4                ⇒                5p = 4 + 10          ⇒       5p = 14

(b) -4 = 5(p – 2)          ⇒        -4 = 5 × p – 5 × 2     ⇒   -4 = 5p – 10

⇒ 5p – 10 = -4            ⇒                  5p = -4 + 10    ⇒         5p = 6

⇒ 

(c) 16 = -5(2 – p)        ⇒     -16 = -5 × 2-(-5) × p

⇒ -16 = -10 + 5p             ⇒             -10 + 5p = -16

⇒ 5p = -16 + 10                  ⇒            5p = -6

(d) 10 = 4 + 3(t + 2)       ⇒      10 – 4 = 3(t + 2)

⇒ 6 = 3(t + 2)                     ⇒          = t + 2

⇒ 2 = t + 2                          ⇒            2 – 2 = t

⇒ 0 = t                                 ⇒             t = 0

(e) 28 = 4 + 3 (t + 5)         ⇒             28 – 4 = 3(t + 5)

⇒ 24 = 3(t + 5)                   ⇒              = t + 5

⇒ 8 = t + 5                          ⇒             8 – 5 = t

⇒ 3 = t                                 ⇒             t = 3

(f) 0 = 16 + 4(m – 6)         ⇒           0 – 16 = 4(m – 6)

⇒ 16 = 4(m – 6)                    ⇒           

⇒ -4 = m – 6                          ⇒           -4 + 6 = m

⇒ 2 = m                                  ⇒           m = 2

14.  (a) 3 equations starting with x = 2.

(i) x = 2 Multiplying both sides by 10,              10x = 20

Adding 2 both sides                        10x + 2 = 20 + 2 = 10x + 2 = 22

(ii) x = 2 Multiplying both sides by 5           5x = 10

Subtracting 3 from both sides       5x – 3 = 10 – 3 = 5x – 3 = 7

(iii) x = 2

Dividing both sides by 5             

(b) 3 equations starting with x = -2.

(i) x = -2

Multiplying both sides by 3                          3x = -6

(ii) x = -2

Multiplying both sides by 3         3x = -6

Adding 7 to both sides     3x + 7 = -6 + 7 = 3x + 7 = 1

(iii) x = -2

Multiplying both sides by 3        3x = -6

Adding 10 to both sides     3x + 10 = -6 + 10 = 3x + 10 = 4

15.  (a) Let the number be x.

According to the question,                 8x + 4 = 60

⇒ 8x = 60- 4        ⇒         8x = 56

⇒              ⇒         x = 7

(b) Let the number be y.

According to the question,    

⇒ y = 7 × 5            ⇒       y = 35

(c) Let the number be z.

According to the question,     

(d) Let the number be x

According to the question,    2x – 11 = 15

⇒  2x = 15 + 11           ⇒      2x = 26

⇒                        ⇒     x = 13

(e) Let the number be m.

According to the question,    50 – 3m = 8

⇒ -3m = 8 – 50       ⇒          -3m = -42

                ⇒         m = 14

(f) Let the number be n.

According to the question,   

⇒  n + 19 = 8 × 5         ⇒     n + 19 = 40

⇒ n = 40 – 19                ⇒     n = 21

(g) Let the number be x.

According to the question,    

16.  (a) Let the lowest marks be y.

According to the question,   2y + 7 = 87

⇒  2y = 87-7        ⇒    2y = 80          ⇒     

⇒ y = 40

Thus, the lowest score is 40.

(b) Let the base angle of the triangle be b.

Given,           a = 40°, b = c

Since,            a + b + c = 180°              [Angle sum property of a triangle]

40°+ b + b = 180°

40° + 2b = 180°

2b =180° – 40°    ⇒         2b =140°

Thus, the base angles of the isosceles triangle are 70° each.

(c) Let the score of Rahul be x and Sachin’s score is 2 x.

According to the equation,   x + 2x = 198

Thus, Rahul’s score = 66 runs

And Sachin’s score = 2 × 66 = 132 runs.

17.  (i) Let the number of marbles Parmit has be m.

According to the question,   5m + 7 = 37

Thus, Parmit has 6 marbles

(ii) Let the age of Laxmi be y years.

Then her father’s age = (3y+ 4) years

According to question,                         3y+ 4 = 49

⇒ 3y = 49 – 4        ⇒      3y = 45

Thus, the age of Laxmi is 15 years.

(iii) Let the number of fruit trees be t.

Then the number of non-fruits tree = 3t + 2

According to the question,

Thus, the number of fruit trees are 25.

18.    Let the number be n.

According to the question,    7n + 50 + 40 = 300

Thus, the required number is 30.