Find a fraction which reduces to 2/3 when 3 is adding to both its numerator and denominator and reduces to 3/5 when 1 is added to both its numreator and denominator

Answer 1

n = numerator

d = denominator

A fraction reduces to 2 / 3 when 3 is adding to both its numerator mean:

( n + 3 ) / ( d + 3 ) = 2 / 3

Reduces to 3 / 5 when 1 is added to both its numreator and denominator mean:

( n + 1 ) / ( d + 1 ) = 3 / 5

Now:

( n + 3 ) / ( d + 3 ) = 2 / 3 Multiply both sides by 3

3 ( n + 3 ) / ( d + 3 ) = 2 Multiply both sides by d + 3

3 ( n + 3 ) = 2 ( d + 3 )

3 * n + 3 * 3 = 2 d + 2 * 3

3 n + 9 = 2 d + 6

( n + 1 ) / ( d + 1 ) = 3 / 5 Multiply both sides by 5

5 ( n + 1 ) / ( d + 1 ) = 3 Multiply both sides by d + 1

5 ( n + 1 ) = 3 ( d + 1 )

5 * n + 5 * 1 = 3 d + 3 * 1

5 n + 5 = 3 d + 3

Now you must solve system of 2 equations with 2 unknow:

3 n + 9 = 2 d + 6

5 n + 5 = 3 d + 3

3 n + 9 = 2 d + 6
-
5 n + 5 = 3 d + 3
_______________
- 2 n + 4 = - d + 3 Subtract 3 to both sides

- 2 n + 4 - 3 = - d + 3 - 3

- 2 n + 1 = - d Multiply both sides by - 1

2 n - 1 = d

d = 2 n - 1

Replace this value in equation:

3 n + 9 = 2 d + 6

3 n + 9 = 2 ( 2 n - 1 ) + 6

3 n + 9 = 2 * 2 n - 2 * 1 + 6

3 n + 9 = 4 n - 2 + 6

3 n + 9 = 4 n + 4 Subtract 3 n to both sides

3 n + 9 - 3 n = 4 n + 4 - 3 n

9 = n + 4 Subtract 4 to both sides

9 - 4 = n + 4 - 4

5 = n

n = 5

d = 2 n - 1

d = 2 * 5 - 1

d = 10 - 1

d = 9

numerator = 5

denominator = 9

Your fraction:

5 / 9

Proof:

( n + 3 ) / ( d + 3 ) = ( 5 + 3 ) / ( 9 + 3 ) = 8 / 12 = 4 * 2 / 4 * 3 = 2 / 3

( n + 1 ) / ( d + 1 ) = ( 5 + 1 ) / ( 9 + 1 ) = 6 / 10 = 2 * 3 / 2 * 5 = 3 / 5