Question

How do you solve `2/(y+2)+3/y=-y/(y+2)`?

Answer 1

`x = 1.5 or -6`

Explanation:

`2/(y+2) + 3/y = -y/(y+2)`

find LCM of `y` and `y+2`

factors of `y`: `y, 1`

factors of `y+2`: `y+2, 1`

multiply together `y , 1 and (y+2)`:

`y * 1 * (y+2) = y(y+2)`

multiply denominators to get `y(y+2)`:

`(2y)/(y(y+2)) + (3(y+2))/(y(y+2)) = -y^2/(y(y+2))`

multiply everything by `y(y+2)`

hence, `2y + 3(y+2) = -(y^2)`

multiply out brackets:

`2y + 3y + 6 = -(y^2)`

`5y + 6 = -(y^2)`

multiply by `-1`:

`-(5y+6)=y^2`

`6-5y=y^2`

add `5y`:

`y^2+5y = 6`

complete the square:

halve the `y` term `(2.5)`.

square it `(6.25)` and add to both sides.

`5x + 6.25 + x^2 = 6 + 6.25`

`x^2+5x+6.25 = 12.25`

factorise:
`(x + 2.5)(x + 2.5) = 12.25`

calculate the square root of the right side `(3.5 or -3.5)`

`therefore x+2.5 = 3.5 or -3.5`

subtract `2.5`:

`x = 1.5 or -6`