NCERT solutions for class 7 maths 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.

 

(i) 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) 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.

 

(i) 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) 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) 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) From the question, it is given that,

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

Let the base angle be b.

Then,

Vertex angle = 2 × base angle = 2b

= b + b + 2b = 180o

= 4b = 180o

(a) 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) 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) 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) 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) 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) 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.

 

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

= x + 4 = 9

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

= y – 2 = 8

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

= 10a = 70

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

= (b/5) = 6

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

= ¾t = 15

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

Seven times m is 7m.

= 7m + 7 = 77

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

6 times y is 6y.

= 6y – 6 = 60

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

One-third of z is z/3.

= 3 + z/3 = 30

 

(i) The sum of numbers p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Twice of number m is 7.

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

(v) Three-fifth of m is 6.

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

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

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

(a) Now, we have to divide both sides of the equation by 3.

Then, we get

= 3l/3 = 42/3

= l = 14

(b) Now, we have to multiply both sides of the equation by 2.

Then, we get

= b/2 × 2= 6 × 2

= b = 12

(c) Now, we have to multiply both sides of the equation by 7.

Then, we get

= p/7 × 7= 4 × 7

= p = 28

(d) Now, we have to divide both sides of the equation by 4.

Then, we get

= 4x/4 = 25/4

= x = 25/4

(e) Now, we have to divide both sides of the equation by 8.

Then, we get

= 8y/8 = 36/8

= x = 9/2

(f) Now, we have to multiply both sides of the equation by 3.

Then, we get

= (z/3) × 3 = (5/4) × 3

= x = 15/4

(g) Now, we have to multiply both sides of the equation by 5.

Then, we get

= (a/5) × 5 = (7/15) × 5

= a = 7/3

(h) Now, we have to divide both sides of the equation by 20.

Then, we get

= 20t/20 = -10/20

= x = – ½

 

(a) First, we have to add 2 to both sides of the equation.

Then, we get

= 3n – 2 + 2 = 46 + 2

= 3n = 48

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3n/3 = 48/3

= n = 16

(b) First, we have to subtract 7 from both sides of the equation.

Then, we get

= 5m + 7 – 7 = 17 – 7

= 5m = 10

Now,

We have to divide both sides of the equation by 5.

Then, we get

= 5m/5 = 10/5

= m = 2

(c) First, we have to multiply both sides of the equation by 3.

Then, we get

= (20p/3) × 3 = 40 × 3

= 20p = 120

Now,

We have to divide both sides of the equation by 20.

Then, we get

= 20p/20 = 120/20

= p = 6

(d) First, we have to multiply both sides of the equation by 10.

Then, we get

= (3p/10) × 10 = 6 × 10

= 3p = 60

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3p/3 = 60/3

= p = 20

 

(a) Now,

We have to divide both sides of the equation by 10.

Then, we get

= 10p/10 = 100/10

= p = 10

(b) First, we have to subtract 10 from both sides of the equation.

Then, we get

= 10p + 10 – 10 = 100 – 10

= 10p = 90

Now,

We have to divide both sides of the equation by 10.

Then, we get

= 10p/10 = 90/10

= p = 9

(c) Now,

We have to multiply both sides of the equation by 4.

Then, we get

= p/4 × 4 = 5 × 4

= p = 20

(d) Now,

We have to multiply both sides of the equation by – 3.

Then, we get

= – p/3 × (- 3) = 5 × (- 3)

= p = – 15

(e) First, we have to multiply both sides of the equation by 4.

Then, we get

= (3p/4) × (4) = 6 × 4

= 3p = 24

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3p/3 = 24/3

= p = 8

(f) Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = -9/3

= s = -3

(g) First, we have to subtract 12 from both sides of the equation.

Then, we get

= 3s + 12 – 12 = 0 – 12

= 3s = -12

Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = -12/3

= s = – 4

(h) Now,

We have to divide both sides of the equation by 3.

Then, we get

= 3s/3 = 0/3

= s = 0

(i) Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

(j) First, we have to add 6 to both sides of the equation.

Then, we get

= 2q – 6 + 6 = 0 + 6

= 2q = 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

(k) First, we have to subtract 6 from both sides of the equation.

Then, we get

= 2q + 6 – 6 = 0 – 6

= 2q = – 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = – 6/2

= q = – 3

(l) First, we have to subtract 6 from both sides of the equation.

Then, we get

= 2q + 6 – 6 = 12 – 6

= 2q = 6

Now,

We have to divide both sides of the equation by 2.

Then, we get

= 2q/2 = 6/2

= q = 3

(a) We have to add 1 to both sides of the given equation.

Then, we get

= x – 1 + 1 = 0 + 1

= x = 1

(b) We have to subtract 1 from both sides of the given equation.

Then, we get

= x + 1 – 1 = 0 – 1

= x = – 1

(c) We have to add 1 to both sides of the given equation.

Then, we get

= x – 1 + 1 = 5 + 1

= x = 6

(d) We have to subtract 6 from both sides of the given equation.

Then, we get

= x + 6 – 6 = 2 – 6

= x = – 4

(e) We have to add 4 to both sides of the given equation.

Then, we get

= y – 4 + 4 = – 7 + 4

= y = – 3

(f) We have to add 4 to both sides of the given equation.

Then, we get

= y – 4 + 4 = 4 + 4

= y = 8

(g) We have to subtract 4 from both sides of the given equation.

Then, we get

= y + 4 – 4 = 4 – 4

= y = 0

(h) We have to subtract 4 from both sides of the given equation.

Then, we get

= y + 4 – 4 = – 4 – 4

= y = – 8

 

(a) Let us divide both the sides by 2,

= (2(x + 4))/2 = 12/2

= x + 4 = 6

By transposing 4 from LHS to RHS it becomes -4

= x = 6 – 4

= x = 2

(b) Let us divide both the sides by 3,

= (3(n – 5))/3 = 21/3

= n – 5 = 7

By transposing -5 from LHS to RHS it becomes 5

= n = 7 + 5

= n = 12

(c) Let us divide both the sides by 3,

= (3(n – 5))/3 = – 21/3

= n – 5 = -7

By transposing -5 from LHS to RHS it becomes 5

= n = – 7 + 5

= n = – 2

(d) Let us divide both the sides by -4,

= (-4(2 + x))/ (-4) = 8/ (-4)

= 2 + x = -2

By transposing 2 from LHS to RHS it becomes – 2

= x = -2 – 2

= x = – 4

(e) Let us divide both the sides by 4,

= (4(2 – x))/ 4 = 8/ 4

= 2 – x = 2

By transposing 2 from LHS to RHS it becomes – 2

= – x = 2 – 2

= – x = 0

= x = 0

 

(a) Let us divide both the sides by 5,

= 4/5 = (5(p – 2))/5

= 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= (4/5) + 2 = p

= (4 + 10)/ 5 = p

= p = 14/5

(b) Let us divide both the sides by 5,

= – 4/5 = (5(p – 2))/5

= – 4/5 = p -2

By transposing – 2 from RHS to LHS it becomes 2

= – (4/5) + 2 = p

= (- 4 + 10)/ 5 = p

= p = 6/5

(c) By transposing 4 from RHS to LHS it becomes – 4

= 16 – 4 = 3(t + 2)

= 12 = 3(t + 2)

Let us divide both the sides by 3,

= 12/3 = (3(t + 2))/ 3

= 4 = t + 2

By transposing 2 from RHS to LHS it becomes – 2

= 4 – 2 = t

= t = 2

(d) By transposing 4 from LHS to RHS it becomes – 4

= 5(p – 1) = 34 – 4

= 5(p – 1) = 30

Let us divide both the sides by 5,

= (5(p – 1))/ 5 = 30/5

= p – 1 = 6

By transposing – 1 from RHS to LHS it becomes 1

= p = 6 + 1

= p = 7

(e) By transposing 16 from RHS to LHS it becomes – 16

= 0 – 16 = 4(m – 6)

= – 16 = 4(m – 6)

Let us divide both the sides by 4,

= – 16/4 = (4(m – 6))/ 4

= – 4 = m – 6

By transposing – 6 from RHS to LHS it becomes 6

= – 4 + 6 = m

= m = 2

 

 

(a) First equation is,

Multiply both sides by 6

= 6x = 12 … [equation 1]

Second equation is,

Subtracting 4 from both sides,

= 6x – 4 = 12 -4

= 6x – 4 = 8 … [equation 2]

Third equation is,

Divide both sides by 6

= (6x/6) – (4/6) = (8/6)

= x – (4/6) = (8/6) … [equation 3]

(b) First equation is,

Multiply both sides by 5

= 5x = -10 … [equation 1]

Second equation is,

Subtracting 3 from both sides,

= 5x – 3 = – 10 – 3

= 5x – 3 = – 13 … [equation 2]

Third equation is,

Dividing both sides by 2

= (5x/2) – (3/2) = (-13/2) … [equation 3]

 

 

 

(a) By transposing (5/2) from LHS to RHS it becomes -5/2

Then,

= 2y = (37/2) – (5/2)

= 2y = (37-5)/2

= 2y = 32/2

Now,

Divide both sides by 2,

= 2y/2 = (32/2)/2

= y = (32/2) × (1/2)

= y = 32/4

= y = 8

(b) By transposing 28 from LHS to RHS it becomes -28

Then,

= 5t = 10 – 28

= 5t = – 18

Now,

Divide both sides by 5,

= 5t/5= -18/5

= t = -18/5

(c) By transposing 3 from LHS to RHS it becomes -3

Then,

= a/5 = 2 – 3

= a/5 = – 1

Now,

Multiply both sides by 5,

= (a/5) × 5= -1 × 5

= a = -5

(d) By transposing 7 from LHS to RHS it becomes -7

Then,

= q/4 = 5 – 7

= q/4 = – 2

Now,

Multiply both sides by 4,

= (q/4) × 4= -2 × 4

= a = -8

(e) First we have to multiply both the sides by 2,

= (5x/2) × 2 = – 5 × 2

= 5x = – 10

Now,

We have to divide both the sides by 5,

Then we get,

= 5x/5 = -10/5

= x = -2

(f) First we have to multiply both the sides by 2,

= (5x/2) × 2 = (25/4) × 2

= 5x = (25/2)

Now,

We have to divide both the sides by 5,

Then we get,

= 5x/5 = (25/2)/5

= x = (25/2) × (1/5)

= x = (5/2)

(g) By transposing (19/2) from LHS to RHS it becomes -19/2

Then,

= 7m = 13 – (19/2)

= 7m = (26 – 19)/2

= 7m = 7/2

Now,

Divide both sides by 7,

= 7m/7 = (7/2)/7

= m = (7/2) × (1/7)

= m = ½

(h) By transposing 10 from LHS to RHS it becomes – 10

Then,

= 6z = -2 – 10

= 6z = – 12

Now,

Divide both sides by 6,

= 6z/6 = -12/6

= m = – 2

(i) First we have to multiply both the sides by 2,

= (3l/2) × 2 = (2/3) × 2

= 3l = (4/3)

Now,

We have to divide both the sides by 3,

Then we get,

= 3l/3 = (4/3)/3

= l = (4/3) × (1/3)

= x = (4/9)

(j) By transposing -5 from LHS to RHS it becomes 5

Then,

= 2b/3 = 3 + 5

= 2b/3 = 8

Now,

Multiply both sides by 3,

= (2b/3) × 3= 8 × 3

= 2b = 24

And,

Divide both sides by 2,

= 2b/2 = 24/2

= b = 12

 

(a) Let us assume the lowest score is x

From the question it is given that,

The highest score is = 87

Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7

5/2 of the number = (5/2) x

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

Then,

= 2x + 7 = Highest score

= 2x + 7 = 87

By transposing 7 from LHS to RHS it becomes -7

= 2x = 87 – 7

= 2x = 80

Now,

Divide both the sides by 2

= 2x/2 = 80/2

= x = 40

Hence, the lowest score is 40

(b) From the question it is given that,

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

Let base angle be b

Then,

= b + b + 40o = 180o

= 2b + 40 = 180o

By transposing 40 from LHS to RHS it becomes -40

= 2b = 180 – 40

= 2b = 140

Now,

Divide both the sides by 2

= 2b/2 = 140/2

= b = 70o

Hence, 70o is the base angle of an isosceles triangle.

(c) Let us assume Rahul’s score is x

Then,

Sachin scored twice as many runs as Rahul is 2x

Together, their runs fell two short of a double century,

= Rahul’s score + Sachin’s score = 200 – 2

= x + 2x = 198

= 3x = 198

Divide both the sides by 3,

= 3x/3 = 198/3

= x = 66

So, Rahul’s score is 66

And Sachin’s score is 2x = 2 × 66 = 132

 

(i) Let us assume number of Parmit’s marbles = m

From the question it is given that,

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 has

= (5 × m) + 7 = 37

= 5m + 7 = 37

By transposing 7 from LHS to RHS it becomes -7

= 5m = 37 – 7

= 5m = 30

Divide both the sides by 5

= 5m/5 = 30/5

= m = 6

So, Permit has 6 marbles

(ii) Let Laxmi’s age be = y years old

From the question it is given that,

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

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

= (3 × y) + 4 = 49

= 3y + 4 = 49

By transposing 4 from LHS to RHS it becomes -4

= 3y = 49 – 4

= 3y = 45

Divide both the sides by 3

= 3y/3 = 45/3

= y = 15

So, Lakshmi’s age is 15 years.

(iii) Let the number of fruit tress be f.

From the question it is given that,

3 × number of fruit trees + 2 = number of non-fruit trees

= 3f + 2 = 77

By transposing 2 from LHS to RHS it becomes -2

=3f = 77 – 2

= 3f = 75

Divide both the sides by 3

= 3f/3 = 75/3

= f = 25

So, number of fruit tree was 25.

 

Let us assume the number is x.

Take me seven times over and add a fifty = 7x + 50

To reach a triple century you still need forty = (7x + 50) + 40 = 300

= 7x + 50 + 40 = 300

= 7x + 90 = 300

By transposing 90 from LHS to RHS it becomes -90

= 7x = 300 – 90

= 7x = 210

Divide both sides by 7

= 7x/7 = 210/7

= x = 30

Hence, the number is 30.

(a) Let us assume the required number be x

Eight times a number = 8x

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

= 8x + 4 = 60

By transposing 4 from LHS to RHS it becomes – 4

= 8x = 60 – 4

= 8x = 56

Divide both side by 8,

Then we get,

= (8x/8) = 56/8

= x = 7

(b) Let us assume the required number be x

One-fifth of a number = (1/5) x = x/5

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

= (x/5) – 4 = 3

By transposing – 4 from LHS to RHS it becomes 4

= x/5 = 3 + 4

= x/5 = 7

Multiply both side by 5,

Then we get,

= (x/5) × 5 = 7 × 5

= x = 35

(c) Let us assume the required number is x

Three-fourths of a number = (3/4) x

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

= (3/4) x + 3 = 21

By transposing 3 from LHS to RHS it becomes – 3

= (3/4) x = 21 – 3

= (3/4) x = 18

Multiply both sides by 4,

Then we get,

= (3x/4) × 4 = 18 × 4

= 3x = 72

Then,

Divide both sides by 3,

= (3x/3) = 72/3

= x = 24

(d) Let us assume the required number is x

Twice a number = 2x

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

= 2x –11 = 15

By transposing -11 from LHS to RHS it becomes 11

= 2x = 15 + 11

= 2x = 26

Then,

Divide both sides by 2,

= (2x/2) = 26/2

= x = 13

(e) Let us assume the required number is x

Thrice the number = 3x

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

= 50 – 3x = 8

By transposing 50 from LHS to RHS it becomes – 50

= – 3x = 8 – 50

= -3x = – 42

Then,

Divide both sides by -3,

= (-3x/-3) = – 42/-3

= x = 14

(f) Let us assume the required number is x

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

= (x + 19)/5 = 8

Multiply both sides by 5,

= ((x + 19)/5) × 5 = 8 × 5

= x + 19 = 40

Then,

By transposing 19 from LHS to RHS it becomes – 19

= x = 40 – 19

= x = 21

(g) Let us assume the required number is x

5/2 of the number = (5/2) x

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

= (5/2) x – 7 = 23

By transposing -7 from LHS to RHS it becomes 7

= (5/2) x = 23 + 7

= (5/2) x = 30

Multiply both sides by 2,

= ((5/2) x) × 2 = 30 × 2

= 5x = 60

Then,

Divide both the sides by 5

= 5x/5 = 60/5

= x = 12