Question

Simply the expression. Write your answer in lowest terms and leave it in factored form

2y/y(3y-1) + 1/ 3y-1

Answer 1

to make the denominators the same, mutliply \(\frac{1}{3y-1}\)
top and bottom by \(\frac{y}{y}\) to get \(\frac{y}{y(3y-1)}\)

now that the denominators are the same, add up in the numerator

Answer 2

So then it's just 2y+1= 3y/y(3y-1)

Answer 3

the "final answer" is \(\frac{3y}{y(3y-1)}=\frac{3}{3y-1}\)