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

1 Answer
Nov 29, 2017

Answer:

#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.