maths問題

Given 4x-5y = 0 => 4x = 5y => x = 5y/4
(3x-y):(2x+3y)
= (15y/4 - y):(10y/4 + 3y)
= (11y/4):(22y/4)
= 1:2

Given a:b = 2:3 & b:c = 4:5 => a = 2b/3 & c = 5b/4
(3a-b):(b+2c)
= (2b-b):(b+5b/4)
= b:(9b/4)
= 4:9

Given 2x = 3y = 4z => x = 3y/2 & z = 3y/4
(x-y+z):(x+y-z)
= (3y/2 - y + 3y/4) : (3y/2 + y - 3y/4)
= (5y/4) : (7y/4)
= 5:7

Given a:b = 3:5 & b:c = 3:2 => a = 3b/5 & c = 2b/3
(2a+3b-c)/(3a-2b+c)
= (6b/5 + 3b - 2b/3)/(9b/5 - 2b + 2b/3)
= (53b/15) / (7b/15)
= 53/7

OR
2a + 3b - c/3a - 2b + c
= 6b/5 + 3b - (2b/3)/(9b/5) - 2b + 2b/3
= 28b/15 + b - 10/27
= 43b/15 - 10/27