Question 163903This question is from textbook Elementary and Intermediate
: 52.) Simplify each complex fraction. Reduce
1/9+1/3x
x/9-1/x this problem has a line between the two which make it one problem.
This question is from textbook Elementary and Intermediate
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
(1/9 + 1/3x)/(x/9-1/x)
Numerator: (1/9+1/3x); multiply each term by 9x/9x and we get:
((x+3)/9x)
Now denominator: (x/9-1/x); multiply each term by 9x/9x and we get
(x^2-9)/9x but (x^2-9)=(x+3)(x-3) so we have: (x+3)(x-3)/9x
Next put the numerator and denominator back together and we get:
((x+3)/9x)/((x+3)(x-3))/9x Now we can simplify this complex fraction by making the denominator equal to 1. We can do this by multiplying both the numerator and denominator by 9x/((x+3)(x-3)). When we do this, we get:
((x+3)/9x)*(9x/((x+3)(x-3)))/(((x+3)(x-3))/9x)*(9x/((x+3)(x-3))); simplifying, we have:
1/(x-3)/1 or 1/(x-3)---------------------answer
Now lets look at the general form of a complex fraction:
(a/b)/(c/d) Lets make the denominator, (c/d), equal to 1. We can if we multiply the numerator and denominator by (d/c) and we get:
(a/b)(d/c)/(c/d)(d/c) which equals
(ad)/(bc)/1=(ad)/(bc)
After we simplify your problem a little, we can say that:
a=(x+3)
b=9x
c=(x+3)(x-3) and
d=b=9x
So the answer is:(ad)/(bc) or (x+3)*9x/(9x*(x+3)(x-3)) cancel like terms
The 9x's cancel and the (x+3)'s cancel leaving 1/(x-3)
Hope this helps----ptaylor
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