Difference Quotient of 2x/x-1

Question

2(x+h)/(x+h)-1 - 2x/x-1
all divided by h

2x+2h/x+h-1 - 2x/x-1
all divided by h

2x cancels out
x cancels out
1 cancels out

(2h/h)/h

2h cancels with h

result: 2/h

Answer 1

basically you do the same as the previous question you had.

[f(x+h)-f(x)]/h
f(x+h)= [2(x+h)]/(x+h-1)
f(x)= 2x/(x-1)
[[2(x+h)]/(x+h-1)]-[2x/(x-1)]/h
then simplify! (hint: find a common denominator in order to combine to one whole fraction)

Make sure you put parenthesis or brackets to distinguish your numerator from your denominator. You can only cancel the numerator and denominator if the equation is one fraction under common denominator.

Answer 2

Let me try one more time

2x+2h/x+h-1 - 2x/x-1
---------------
h

2h and h cancel

2x+2/x-1 - 2x/x-1
---------------
h

2x+2-2x/x-1
---------------
h

2x cancel out

2/x-1
---------------
h

result: 2/xh-h

Answer 3

You have to remember you have to divide the "entire" numerator. You cannot just divide an h to one part of numerator to cancel. You can only cancel when the factors in the numerator are multiplying. For example, [(2x)(h)]/h. Your x+h-1 and x-1 belongs in the denominator!

It should look like this:
[(2x+2h)/[(x+h-1)(h)]] - [(2x)/[(x-1)(h)]]

Now find a common denominator:
This is my numerators:
______(h)(x+h-1) =(x-1)(h)______

How do I make it equal? What would I multiply on each side to make it equal?

To do this you
Want:

(x+h-1)(h)(x-1)

Relate:

[(2x+2h)/[(x+h-1)(h)]]
To get my want in the denominator, I need (x-1). Therefore, I multiply both the numerator and denominator by (x-1).

[(2x)/[(x-1)(h)]]
To get my want in the denominator, I need (x+h-1). Therefore, I multiply both the numerator and denominator by (x+h-1).

You should end up with:
[[(2x+2h)(x-1)]/[(x+h-1)(h)(x-1)]] - [[(2x(x+h-1)])/[(x+h-1)(h)(x-1)]]

Then combine fractions, since both share a common denominator:
[[(2x+2h)(x-1)-(2x(x+h-1)]]/[(x+h-1)(h)(x-1)]]

Then simplify!

Answer 4

Typo:
This is my "denominators*":
______(h)(x+h-1) =(x-1)(h)______