How do you solve 3/(2y-7) = 3/y32y7=3y?

1 Answer
Jul 16, 2016

Clear the fractions and solve a 2-step equation to get a solution of y=7y=7.

Explanation:

We first want to clear the fraction on the left side of the equation. We do that by multiplying by 2y-72y7:
3/(2y-7)=3/y32y7=3y

->3/(2y-7)(2y-7)=3/y(2y-7)32y7(2y7)=3y(2y7)

->3/cancel((2y-7))cancel((2y-7))=3/y(2y-7)

->3=(3(2y-7))/y

Now we clear the y from the right, by multiplying by y:
3=(3(2y-7))/y

->3*y=(3(2y-7))/y*y

->3*y=(3(2y-7))/cancel(y)*cancel(y)

->3y=3(2y-7)

Distributing the 3 on the right gives us:
3y=3(2y-7)
->3y=6y-21

Now we just have to solve this two-step equation, and we have y:
3y=6y-21

->-3y=-21

->y=(-21)/-3=7