Question

How do you solve `5/(y - 3) = (y + 7) /(2y - 6) + 1` and find any extraneous solutions?

Answer 1

`y=3` and it is extraneous. Basically, no real solutions.

Explanation:

First, multiply both sides by `(y-3)(2y-6)` to get rid of the fractions.

`5/(y-3)=(y+7)/(2y-6)+1 => (y-3)(2y-6)xx5/(y-3)=(y-3)(2y-6)xx(y+7)/(2y-6)+1(y-3)(2y-6)`

Which is equal to `(2y-6)5=(y-3)(y+7)+(y-3)(2y-6)`

Distribute them out to get `10y-30=y^2+4y-21+2y^2-12y+18`

Combine like terms: `10y-30=3y^2-8y-3`

Use the inverse operation to "clear out" the left side of the equation.
`10y-30=3y^2-8y-3` => `0=3y^2-18y+27`

When we factor this, we get `0=3(y-3)^2` The only solution to this equation is when `0=y-3` which means that `y=3`.

To see whether this is extraneous, we plug the value in the equation.
`5/(y-3)=(y+7)/(2y-6)+1` becomes `5/0=(y+7)/0+1` , indicating that the equation is undefined. Therefore, the solution is extraneous.