Question

Question #09dbc

Answer 1

start off by writing out
`3/(2x-5)=7/(3-5x)`

crossmultiply
`(3)(3-5x)`
`(7)(2x-5)`

you now have
`9-15x=14x-35`

combine like terms
-add `35` to both sides*

you now have
`44-15x=14x`

add `15x` to both sides to get x on its own side
`44=29x`

`x=44/29`
(`1 15/29` as a mixed fraction)

WORK:
`3/(2x-5)=7/(3x-5)`

`9x-15=14x-35`

`44-15x=14x`

`44=29x`

`44/29=x`

TO CHECK, plug in `44/29` for `x`

*there are different ways of doing this; you could subtract 9 from both sides, you could add 15x to both sides, or you could subtract 14x from both sides