Question

In each of the following ,make y the subject and hence find the value of y when a=2, b=3 ,and c=4.

i.) (2a-7)/6y=(3b-5)/2c
ii.) 3-34y + 2a= (3b-5)/(2c+1)

Answer 1

Part i

`(2a-7)/(6y) = (3b-5)/(2c)`

`y = ( 2c(2a-7))/(6(3b-5)) = (c(2a-7))/(9b-15)`

Substituting,

`y = (4(2(2)-7))/(9(3)-15) = ( -12)/(12) = -1`

Answer: -1

Part ii

I'm not sure that one's typed in correctly but I'll solve it as written.

`3-34y + 2a = (3b-5)/(2c + 1)`

`34y = 3+2a-(3b-5)/(2c + 1)`

`y = \frac 1 {34}\left(3+2a-(3b-5)/(2c + 1) \right)`

We're not asked to simplify it so I wont. Substituting,

`y = \frac 1 {34}\left(3+2(2)-(3(3)-5)/(2(4) + 1) \right) = \frac 1 {34}(7-4/9) = (59)/(306)`

Answer: 59/306