Question

How do you solve `(4b+2)/(b^2-3n)+(b+2)/b=(b-1)/b`?

Answer 1

`n=(b*(7b+2))/9`

Explanation:

Given :
`((4b+2)/(b^2-3n))+((b+2)/b)=(b-1)/b`
`((4b+2)/(b^2-3n))=((b-1)/b)-((b+2)/b)`
`(4b+2)/(b^2-3n)=(b-1-b-2)/b=-(3/b)`
Cross multiplying,
`(b)(4b+2)=(-3)(b^2-3n)`
`4b^2+2b=-3b^2+9n`

`9n=7b^2+2b`
`n=(b*(7b+2))/9`