How would I subtract this?
[(8)/(x+h+8)] - [(8)/(x+8)]
just like in numerical fractions, form a common denominator, which is this case is
(x+h+8)(x+8)
Also your use of square brackets suggest that you have a typo and you forgot the /h at the end,
which is the usual form of this fraction
I will insert it for you
[(8)/(x+h+8)] - [(8)/(x+8)]/h
= [ (8(x+8) - 8(x+h+8) )/((x+h+8)(x+8))]/h
= [(8x+64 - 8x - 8h - 64)/((x+h+8)(x+8) ]/h
= -8h/(h(x+h+8)(x+8))
= -8/((x+h+8)(x+8))
Usually at this point you would take the Limit of that expression as h ---> 0
and you would get
-8/(x+8)^2
I'm sorry I had to pester you with a problem like that, but my algebra is extremely weak. I understand it now though. Thank you very much! :)
To subtract the given expressions, we need to have a common denominator. Here's how you can do it:
Step 1: Find the least common denominator (LCD) of the two fractions. The LCD is the smallest multiple of the denominators of the fractions. In this case, the denominators are (x+h+8) and (x+8).
Step 2: Multiply the numerator and denominator of the first fraction by the LCD.
[(8)/(x+h+8)] * [(x+8)/(x+8)]
This gives you:
[(8(x+8))/((x+h+8)(x+8))]
Step 3: Multiply the numerator and denominator of the second fraction by the LCD.
[(8)/(x+8)] * [(x+h+8)/(x+h+8)]
This gives you:
[(8(x+h+8))/((x+8)(x+h+8))]
Now that you have the two fractions with a common denominator, you can subtract them:
[(8(x+8))/((x+h+8)(x+8))] - [(8(x+h+8))/((x+8)(x+h+8))]
To combine the fractions, subtract the numerators while keeping the common denominator:
[(8(x+8) - 8(x+h+8))/((x+8)(x+h+8))]
Now, simplify the numerator:
[(8x + 64 - 8x - 8h - 64)/((x+8)(x+h+8))]
The terms 8x and -8x cancel each other out, and the terms 64 and -64 also cancel out, so we're left with:
[(-8h)/((x+8)(x+h+8))]
Thus, the simplified expression after subtracting the fractions is:
[-8h/((x+8)(x+h+8))]