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
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
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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