The sum of the numerator and denominator of a fraction is 4 more than twice the numerator.

Let the required fraction be x/y . 

As per the question 

x + y = 4 + 2x 

⇒ y – x = 4 ……(i) 

After changing the numerator and denominator 

New numerator = x + 3 

New denominator = y + 3 

Therefore x+3/ y+3 = 2/3 

⇒3(x + 3) = 2(y + 3) 

⇒3x + 9 = 2y + 6 

⇒ 2y – 3x = 3 ……(ii) 

Multiplying (i) by 3 and subtracting (ii), we get: 

3y – 2y = 12 – 3 

⇒y = 9 

Now, putting y = 9 in (i), we get: 

9 – x = 4

⇒ x = 9 – 4 = 5 

Hence, the required fraction is 5/9 .

Let the numerator be ‘a’ and denominator be ‘b’.

Given,

Sum of the numerator and denominator of a fraction is 4 more than twice the numerator.

If the numerator and denominator are increased by 3, they are in the ratio 2 : 3.

⇒ a + b = 2a + 4

⇒ b – a = 4 -------- (1)

Also,

⇒ 3a + 9 = 2b + 6

⇒ 3a – 2b = - 3 ------(2)

Multiplying eq1 by 2 and adding to eq1

⇒ 2b – 2a + 3a – 2b = 8 - 3

⇒ a = 5

Thus b = 9

Fraction is 5/9

Let the required fraction be x/y . 

As per the question 

x + y = 4 + 2x 

⇒ y – x = 4 ……(i) 

After changing the numerator and denominator 

New numerator = x + 3 

New denominator = y + 3 

Therefore 

x+3/y+3 = 2/3 

⇒3(x + 3) = 2(y + 3) 

⇒3x + 9 = 2y + 6 

⇒ 2y – 3x = 3 ……(ii) 

Multiplying (i) by 3 and subtracting (ii), we get: 

3y – 2y = 12 – 3 

⇒y = 9 

Now, putting y = 9 in (i), we get: 

9 – x = 4 ⇒ x = 9 – 4 = 5 

Hence, the required fraction is 5/9 .