Question

How do you combine `(6y+5)/(5y-25)-(y+2)/(y-5)`?

Answer 1

See the entire solution process below:

Explanation:

To subtract these two fractions they must be over a common denominator. To create a common denominator multiply the fraction on the right by the appropriate form of `1` to not change the value of the function but to create a common denominator:

`(6y + 5)/(5y - 25) - (5/5 xx (y + 2)/(y-5)) ->`

`(6y + 5)/(5y - 25) - (5 xx (y + 2))/(5 xx (y-5)) ->`

`(6y + 5)/(5y - 25) - ((5 xx y) + (5 xx 2))/((5 xx y) - (5 xx 5)) ->`

`(6y + 5)/(5y - 25) - (5y + 10)/(5y - 25)`

We can next subtract the numerators over the common denominator:

`((6y + 5) - (5y + 10))/(5y - 25) ->`

`(6y + 5 - 5y - 10)/(5y - 25) ->`

`(6y - 5y + 5 - 10)/(5y - 25) ->`

`((6 - 5)y + (5 - 10))/(5y - 25) ->`

`(1y - 5)/(5y - 25) ->`

`(y - 5)/(5y - 25)`

We can now factor the denominator and cancel common terms:

`(y - 5)/((5 xx y) - (5 xx 5)) ->`

`(y - 5)/(5(y - 5)) ->`

`color(red)(cancel(color(black)(y - 5)))/(5color(red)(cancel(color(black)((y - 5))))) ->`

`1/5`

However, from the original expression: `(5y - 25) != 0` and `(y-5) != 0` therefore, `y != 5`

Answer 2

`1/y`

Explanation:

`(6y+5)/color(red)((5y-25)) -(y+2)/(y-5)" "larr` factorise

`=(6y+5)/color(red)(5(y-5)) -(y+2)/(y-5)" "larr` find the LCD

`=(6y +5 - 5(y+2))/(5(y-5))" "larr` make equivalent fractions

`=(6y+5-5y-10)/(5(y-5))" "larr`remove brackets

`=(y-5)/(y(y-5))" "larr` collect like terms

`=cancel((y-5))/(ycancel((y-5)))" "larr` cancel like factors

`1/y`