Question 171512: simplify the complex fraction 5/6y^2/3/9y^3, 3/x-1/x+3/2/x+5/x+3 How do you do these?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! OK. Here's how you do complex fractions:
Lets look at the complex fraction (a/b)/(c/d). Now if we can make the denominator (c/d) of this complex fraction equal to 1, then we will have a simple fraction. We can do this by multiplying both the numerator and denominator by (d/c). When we do this, we get:
((a/b)*(d/c))/((c/d)*(d/c)); simplifying we have:
(ad/bc)/1 or ad/bc.
Now lets apply this to your specific problem:
(5/6y^2)/(3/9y^3); multiply numerator and denominator by (9y^3/3):
((5/6y^2)*(9y^3/3))/((3/9y^3)(9y^3/3)) and we get
(45y^3)/(18y^2)/1 cancel and simplify and we have:
5y/2
For the second part. It is sometimes good to use parens to better clarify the problem:
3/x-1/x+3/2/x+5/x+3=
(3/(x-1)/(x-3))/(2/(x+5)/(x+3))
Now here we have both the numerator and denominator as complex fractions:
We'll do the numerator first:
(3/(x-1)/(x-3))=
(3*(x-3)/(x-1))/((x-1)/(x-3))*(x-3)/(x-1))=
3(x-3)/(x-1)/1=3(x-3)/(x-1)
Now the denominator:
(2/(x+5)/(x+3))=
(2*(x+3)/(x+5))/((x+5)/(x+3)*(x+3)/(x+5))=
2(x+3)/(x+5)
Now we put the numerator and denominator back together and we still have a complex fraction:
(3(x-3)/(x-1))/(2(x+3)/(x+5))=
(3(x-3)/(x-1))*(x+5)/2(x+3))/(2(x+3)/(x+5))*(x+5)/2(x+3))=
(3(x-3)(x+5))/(2(x+3)(x-1))
Actually, using the little formula that we worked out initially:
(a/b)/(c/d)=ad/bc, you can determine the a, b , c & d of your problem and simply plug the values in. For example, look at the numerator that we worked out above:
(3/(x-1)/(x-3));
a=3
b=1
c=(x-1)
d=(x-3)
now pluggin in (ad/bc), we have
3(x-3)/1(x-1) which is what we got before.
Hope this helps---ptaylor
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