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

1 Answer
Nov 29, 2017

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.