Question

How do you solve the rational equation `(y +2)/ y = 1 /( y - 5)`?

Answer 1

`x=2+sqrt14`

and

`x=2-sqrt14`

Explanation:

Since there are only two terms you can X-multiply:

`(y +2)/ y = 1 /( y - 5)`

X-multiply:

`(y +2)( y - 5) = y*1`

`y^2-3y-10=y`

`y^2-4y-10=0`

Now to solve we have to complete the square:

`ax^2+bx+c`

`a` must equal one (it does in your function).

`c=(b/2)^2`

the completed square is `(x+b/2)^2`

`y^2-4y=10`

your `b= -4`

`c=(-4/2)^2=4` and the square is `(x-2)^2`

So let's complete the square, remember we need to add `c` to both sides so we don't alter the equation:

`y^2-4y +c=10+c`

`y^2-4y +4=10+4`

`(x-2)(x-2)=14`

`(x-2)^2=14`

now solve for `x`:

`sqrt((x-2)^2)=+-sqrt14`

`x-2=+-sqrt14`

`x=2+-sqrt14`

Answer 2

`y=2\pm\sqrt{14}`

Explanation:

Given rational equation:
`\frac{y+2}{y}=\frac{1}{y-5}`
will have solution such that `y\ne0, y\ne5 \ or \ y \notin{0,5}`
`(y+2)(y-5)=y`
`y^2+2y-5y-10=y`
`y^2-4y-10=0`
Using quadratic formula to solve above quadratic equation,
`y=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-10)}}{2(1)}`
`y=\frac{4\pm2\sqrt{14}}{2)}`
`y=2\pm\sqrt{14}`