Chapter 4: Simple Equations

(i) x + 3 = 0

LHS = x + 3

By substituting the value of x = 3

Then,

LHS = 3 + 3 = 6

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

LHS = x + 3

By substituting the value of x = 0,

Then,

LHS = 0 + 3 = 3

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(iii) x + 3 = 0

LHS = x + 3

By substituting the value of x = – 3,

Then,

LHS = – 3 + 3 = 0

By comparing LHS and RHS,

LHS = RHS

∴ Yes, the equation is satisfied.

(iv) x – 7 = 1

LHS = x – 7

By substituting the value of x = 7,

Then,

LHS = 7 – 7 = 0

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(v) x – 7 = 1

LHS = x – 7

By substituting the value of x = 8,

Then,

LHS = 8 – 7 = 1

By comparing LHS and RHS,

LHS = RHS

∴ Yes, the equation is satisfied.

(vi) 5x = 25

LHS = 5x

By substituting the value of x = 0,

Then,

LHS = 5 × 0 = 0

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(vii) 5x = 25

LHS = 5x

By substituting the value of x = 5,

Then,

LHS = 5 × 5 = 25

By comparing LHS and RHS,

LHS = RHS

∴ Yes, the equation is satisfied.

(viii) 5x = 25

LHS = 5x

By substituting the value of x = -5,

Then,

LHS = 5 × (-5) = – 25

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(ix) m/3 = 2

LHS = m/3

By substituting the value of m = – 6,

Then,

LHS = -6/3 = – 2

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(x) m/3 = 2

LHS = m/3

By substituting the value of m = 0,

Then,

LHS = 0/3 = 0

By comparing LHS and RHS,

LHS ≠ RHS

∴ No, the equation is not satisfied.

(xi) m/3 = 2

LHS = m/3

By substituting the value of m = 6,

Then,

LHS = 6/3 = 2

By comparing LHS and RHS,

LHS = RHS

∴ Yes, the equation is satisfied.

LHS = n + 5

By substituting the value of n = 1,

Then,

LHS = n + 5

= 1 + 5

= 6

By comparing LHS and RHS,

6 ≠ 19

LHS ≠ RHS

Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)

LHS = 7n + 5

By substituting the value of n = -2,

Then,

LHS = 7n + 5

= (7 × (-2)) + 5

= – 14 + 5

= – 9

By comparing LHS and RHS,

-9 ≠ 19

LHS ≠ RHS

Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.

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

LHS = 7n + 5

By substituting the value of n = 2,

Then,

LHS = 7n + 5

= (7 × (2)) + 5

= 14 + 5

= 19

By comparing LHS and RHS,

19 = 19

LHS = RHS

Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)

LHS = 4p – 3

By substituting the value of p = 1,

Then,

LHS = 4p – 3

= (4 × 1) – 3

= 4 – 3

= 1

By comparing LHS and RHS,

1 ≠ 13

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)

LHS = 4p – 3

By substituting the value of p = – 4,

Then,

LHS = 4p – 3

= (4 × (-4)) – 3

= -16 – 3

= -19

By comparing LHS and RHS,

-19 ≠ 13

LHS ≠ RHS

Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)

LHS = 4p – 3

By substituting the value of p = 0,

Then,

LHS = 4p – 3

= (4 × 0) – 3

= 0 – 3

= -3

By comparing LHS and RHS,

– 3 ≠ 13

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.

LHS = 5p + 2

By substituting the value of p = 0,

Then,

LHS = 5p + 2

= (5 × 0) + 2

= 0 + 2

= 2

By comparing LHS and RHS,

2 ≠ 17

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation.

Let, p = 1

LHS = 5p + 2

= (5 × 1) + 2

= 5 + 2

= 7

By comparing LHS and RHS,

7 ≠ 17

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation.

Let, p = 2

LHS = 5p + 2

= (5 × 2) + 2

= 10 + 2

= 12

By comparing LHS and RHS,

12 ≠ 17

LHS ≠ RHS

Hence, the value of p = 2 is not a solution to the given equation.

Let, p = 3

LHS = 5p + 2

= (5 × 3) + 2

= 15 + 2

= 17

By comparing LHS and RHS,

17 = 17

LHS = RHS

Hence, the value of p = 3 is a solution to the given equation.

(ii) 3m – 14 = 4

LHS = 3m – 14

By substituting the value of m = 3,

Then,

LHS = 3m – 14

= (3 × 3) – 14

= 9 – 14

= – 5

By comparing LHS and RHS,

-5 ≠ 4

LHS ≠ RHS

Hence, the value of m = 3 is not a solution to the given equation.

Let, m = 4

LHS = 3m – 14

= (3 × 4) – 14

= 12 – 14

= – 2

By comparing LHS and RHS,

-2 ≠ 4

LHS ≠ RHS

Hence, the value of m = 4 is not a solution to the given equation.

Let, m = 5

LHS = 3m – 14

= (3 × 5) – 14

= 15 – 14

= 1

By comparing LHS and RHS,

1 ≠ 4

LHS ≠ RHS

Hence, the value of m = 5 is not a solution to the given equation.

Let, m = 6

LHS = 3m – 14

= (3 × 6) – 14

= 18 – 14

= 4

By comparing LHS and RHS,

4 = 4

LHS = RHS

Hence, the value of m = 6 is a solution to the given equation.

The above statement can be written in the equation form as,

= x + 4 = 9

(ii) 2 subtracted from y is 8.

The above statement can be written in the equation form as,

= y – 2 = 8

(iii) Ten times a is 70.

The above statement can be written in the equation form as,

= 10a = 70

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

The above statement can be written in the equation form as,

= (b/5) = 6

(v) Three-fourths of t is 15.

The above statement can be written in the equation form as,

= ¾t = 15

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

The above statement can be written in the equation form as,

Seven times m is 7m.

= 7m + 7 = 77

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

The above statement can be written in the equation form as,

One-fourth of a number x is x/4.

= x/4 – 4 = 4

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

The above statement can be written in the equation form as,

6 times y is 6y.

= 6y – 6 = 60

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

The above statement can be written in the equation form as,

One-third of z is z/3.

= 3 + z/3 = 30

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.

From the question, it is given that

Number of Parmit’s marbles = m

Then,

Irfan has 7 marbles, more than five times the marbles Parmit has.

= 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having

= (5 × 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).

From the question, it is given that

Let Laxmi’s age be = y years old

Then,

Lakshmi’s father is 4 years older than three times her age.

= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father

= (3 × y) + 4 = 49

= 3y + 4 = 49

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

From the question, it is given that

Highest score in the class = 87

Let the lowest score be l.

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

= (2 × 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).

From the question, it is given that

We know that the sum of angles of a triangle is 180^o

Let the base angle be b.

Then,

Vertex angle = 2 × base angle = 2b

= b + b + 2b = 180^o

= 4b = 180^o