SOLUTION: I am very lost on this equation...can someone please help explain this to me. (x-7+5/x-1)/(x-3+1/x-1) do the denominators of x-1 cancel out???

Question 247175: I am very lost on this equation...can someone please help explain this to me.
(x-7+5/x-1)/(x-3+1/x-1) do the denominators of x-1 cancel out???

Found 2 solutions by College Student, richwmiller:
Answer by College Student(505) (Show Source): You can put this solution on YOUR website!
Thank you for re-posting it. I'm not sure about how the problem should look like. Should look like this???

If so, this is how you solve it. When two fractions are divided by each other, is the same as multiplying by the reciprocal of the second one. Thus we would be able to re-write the expression like this:

In this case, we may cancel (x-1), so we get:

Now simplify to:

Can we simplify it further? Yes.
Any number divided by itself will always equal 1, thus our final answer is .

Done! :)
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
There are many problems with the way you present this problem.
It is not an equation as you wrote it. Nothing is set equal to it. It is an expression.
let's begin by simplifying
x-7+5=x-2 and x-3+1=x-2
Should I assume that (x-1) is the denominator and not just x?
(x-2)/(x-1))/((x-2)/(x-1)) equals one because not only do (x-1) cancel out but also ((x-2) and part of the denominator (x-2)
Notice the difference
((x-2)/(x-1))/(x-2)/(x-1)
here the numerator (x-2) and part of the denominator (x-2) cancel out leaving 1/(x-1)^2
It is hard to tell what your expression was meant to be.

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